From dennizin@pcstarnet.com Tue Apr 27 08:27:25 EDT 1999 Article: 170676 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!ix.netcom.com!news From: dennizin@pcstarnet.com (Denny Anderson) Newsgroups: rec.running Subject: Re: Cadence? Date: Tue, 27 Apr 1999 07:57:23 GMT Organization: Netcom Lines: 199 Message-ID: <37256b82.3943696@206.214.99.10> References: <29275-37196987-17@newsd-231.iap.bryant.webtv.net> <371BD9FF.60FC@jps.net> <371c0c35.10440547@206.214.99.10> <7g2iec$fve$1@news4.svr.pol.co.uk> NNTP-Posting-Host: mon-ca8-28.ix.netcom.com Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-NETCOM-Date: Tue Apr 27 2:57:03 AM CDT 1999 X-Newsreader: Forte Agent 1.5/32.452 Xref: news1.radix.net rec.running:170676 "Telaren" wrote: >Denny Anderson wrote in message <371c0c35.10440547@206.214.99.10>... > >[snip] > >>But, I dare anyone to run at a constant cadence for all speeds. CAN'T >>be done! At least not efficiently. Efficiency is the objective of >>finding the right cadence to match the s/l. > >I always thought that the cadence stays the same, but the stride length >changes depending on the elevation of the ground - i.e. shorter uphill and >longer downhill. > >Telaren Many people think that. But, studies have shown it just ain't so! About 1 1/2 yrs back I posted an edited version of a famous study regarding "Stride Length". I repost it here for all those who missed it! [Quote from my original edited summary] As promised, excerpts from the Runner's world article by Dr. Peter Cavanagh & Keith R. Williams; July 1979 Issue, pg. 62. I quote and edit at the same time: ....... means text omitted before and ~~~, means text omitted after. { } means I interject (probably something useless). "Bill Rodgers~~~ has an average {whatever that means} leg length of 36.8 inches. When he's running~~~ at around 5 min. per mile, Bill takes strides 65.8 inches long - about 1.8 times his leg length. .......there is the provocative question: Is Bill running at his most efficient stride length or would a change in stride length enable him to run a 2:05 marathon?" "... We must warn you at the outset that there is no easy self-test to determine your best stride length. However, we believe that evidence presented in this article will allow a more informed approach to this aspect of distance-running style." "It may not surprise you to learn that Bill Rodgers is running at his most efficient stride length. ~~~ By subjecting Bill to an experimental situation {treadmill}~~~ we determined that his performance could *not* be improved by changing his stride length....." "But what about your stride length? Should it be the same as Bill Rodgers' or should it be some multiple of your leg length? To begin with, many readers~~~ probably can't run one mile at 5 min per mile, let alone 26. This brings us to our first point: {{{Note ref. below to S/L and S/F being co-dependent}}} 1. Stride length depends on running speed and gradient variance." " Running faster does not simply involve taking more strides in a given time but, ~~~ it is a combination of longer strides and greater stride frequency. Increases in stride length {Editorial; I will henceforth use* SL = stride length*, and *SF = stride frequency*} are actually dominant characteristics at distance running speeds~~~. For example, SL increases at a rate 5 times greater than SF between 7 min per mile and 5 min per mile. ~~~~ Speeds slower than 8 min per mile have never been examined~~~~." " 2. Everyone has an optimal SL at a given speed." "Sometime when you are on a training run try to run at an 8 min/mile pace {I wish} with strides only 2 ft long. You will soon notice that this is a very inefficient way to run. A quick check of your heart rate will confirm this. Now go to the other extreme. Keeping your speed the same, take the longest strides that you possibly can. In a short time, your fatigue will tell you that this too is an inefficient way to run and the pounding that your legs experience will force you to stop. ~~~" {This is so cool ~ I did exactly that 2 days ago; it's just so!}. ......".At both the short and long extremes of the SL, energy cost is high and consequently efficiency is low. Somewhere between these two extremes there is an optimal point - a SL which is most efficient for *you* at this speed of running. ~~~" ......" It is not easy to determine when you have found your optimal SL. However, with proper equipment: 3. Optimal SL can be measured - but not easily." ....."Ten male subjects we used in ~~ (an) experiment were a fit group of long distance and marathon runners who were serious about their running, but certainly not elite runners. ~~~ Subjects were instructed ~~~ to vary their SL's by 7 increments over a range of 20% of leg length shorter to 20% of leg length longer than their chosen SL." ~~~ ......" All of the 10 subjects in our experiment showed a U-shaped curve relating efficiency and SL indicating the existence of an optimal condition {i.e., the lowest point in the curve, between very short & very long SL}. One of the most important inferences we can make from the experiment is: 4. You will probably locate your optimal SL subconsciously." ....." The exact process by which runners find their optimal SL is open to some debate. It is likely, over a long period of time, that a runner tries various SL's and gradually closes in on the optimal SL by trial and error." ~~~ ....."There is the possibility that some runners will never run at their most efficient SL's. ~~~ Even in the presence of the overall efficiency of the group of (test) subjects, there is a further important point to be made: 5. There are substantial individual differences in response to altered SL." .... "Most of our discussion to this point assumed that the more physiologically efficient the runner can be, the better." ~~~ ...." At this point, the runner may be intimidated enough about the question of SL to have decided to quit running until he can get to the nearest lab and have his optimal SL determined. If such is the case, we suggest that at this stage you go out for a run to convince yourself that running is possible even in the presence of uncertainty about your optimal SL. You might also take with you the following analgesic thought; 6. Small deviations from optimal SL's have a minor effect on efficiency." ~~~ ....."in most subjects that we have studied~~~ if the SL was changed by as much as 2 inches, in the typical subject, there was still no detectable change in efficiency~~~. That means that runners could vary their SL's from optimal during a race as a means of relaxation without incurring an efficiency penalty." " There are also runners who are particularly insensitive to SL changes. One of our runners, for example, showed only a 1% decrement in efficiency while shortening his SL by over 6 inches. He is probably an exception to the general patterns we have outlined."~~~~ " Runners know that many of their physical characteristics are subject to change as a function of time, level of training, injury, etc. We must therefore assume that: 7. Your optimal SL may change as you grow older." {Tell me about it!} "Definite changes in muscle fiber composition have been detected during aging {flash! hold the presses!} and it is highly probable that this will influence the optimal SL.~~~ No experiments have yet been conducted to determine optimal SL's at various stages in a runner's career."~~~ ....."it's time to synthesize these various pieces of evidence in to some statements of practical importance. It should be clear now that a sound piece of advice is: 8. SL should not be arbitrarily imposed on a runner." ....." This means that optimal SL is not merely a function of body size. In fact, in our study the tallest subject had the shortest optimal SL.~~~ Individual variation is so great that any generalized statement about optimal SL for a given speed of running for an athlete of a given size would be unsound."~~~ ....." An obvious further corollary of the large individual differences that we have encountered leads us to answer one of our initial questions: 9. Do not pattern your stride length after someone else's SL." "This advice holds true whether you are copying Bill Rodgers or your running partner. The latter is important because there is a tendency for runners who train together to match each other stride-for-stride during certain periods of the run. This is profitable for the runner who is running at his optimal SL. However, it may put the other runner at a distinct disadvantage."~~~ ....." Our results indicate clearly that there is *no* relationship between leg length and optimal SL."~~~ ....."You may have noticed that the text is liberally supplied with the words 'probably', 'maybe', 'likely', and 'possibly'. This leads to the following point: 10. There is still much to be learned about stride length." ....."You may be one of the unfortunate group who is running with a non-optimal SL. If so, you should take comfort in the one major benefit of non-optimized exercise -- the increased energy cost means that you are getting a better workout. {Had to go for a one-liner, didn't he!} That's All Folks. A lot of editing, and I couldn't replicate the graphs, but the essence is there. [End Quote from edited posting] zzzzzzzzzzzzzzzzzzzzzzzz From bobg@Radix.Net Thu Apr 15 16:13:54 EDT 1999 Article: 169135 of rec.running Path: news1.radix.net!saltmine.radix.net!not-for-mail From: bobg@Radix.Net (Robert Grumbine) Newsgroups: rec.running Subject: Re: treadmill running effort vs. road running effort Date: 15 Apr 1999 11:59:38 -0400 Organization: RadixNet Lines: 93 Message-ID: <7f52da$j1u$1@saltmine.radix.net> References: <3715192F.3B8C93E2@teron.com> <7f37ju$4e8$1@camel21.mindspring.com> <7f3hc9$k6s@ds2.acs.ucalgary.ca> <7f4reo$r2s$1@camel19.mindspring.com> NNTP-Posting-Host: saltmine.radix.net Xref: news1.radix.net rec.running:169135 Status: RO In article <7f4reo$r2s$1@camel19.mindspring.com>, HappyTrails wrote: >The concern I have about treadmill running has nothing to do with air >resistance, I've got to believe it is negligible. Why do you _have_ to believe it is negligible? I've encountered some good headwinds that reminded me it was far from negligible. Since I notice even a 5 m/s headwind, and 5 m/s is an 80 second 400 -- something withing a lot of runner's reach -- it is probably even an issue for no wind. >I wish there was a >physicist who could address the dynamics of running to propel oneself >forward vs. running to maintain a static position relative to a moving >surface. Unless you're doing things on the treadmill that can't be done on pavement, there isn't a big difference. (If you're somehow letting the treadmill _carry_ your foot back, then there could be a difference I suppose, but if you do that you're probably going to do a face plant in the near future). If you're carrying yourself on the siderails, again there could be a significant difference; but if you're doing that you hardly expect the two to be comparable. As another poster noted take a frame of reference moving steadily with your (the runner's) center of mass. Steady motion (of the COM) is a running form that is recommended and even seems to be achieved by the better runners at least. Given that frame of reference, what do you see? 1) Runner's arms saw back and forth. The back and forth involves changes of speed and direction so he's applying force. -- arms don't care about treadmills so no change 2) Runner's legs a) cycling back and forth, including b) a ballistic launch upwards. b--) You have to go upwards or you're walking; the launch puts you in the air long enough to haul the trailing leg forward. Again, this is true regardless of treadmill vs. pavement. a--) The legs moving forward and backward. Again, if this is at the same rate/stride length, the work by the legs is the same. They have to be accelerated through the cycle, regardless of whether the foot is touching a stationary or moving surface. Your comment about 'propel forward' is a good one for coaching or thinking to yourself on running form, but it isn't the physics. The major physics point is that (barring frictional terms) you're going to move in a straight line at a steady speed _unless_ you apply some force. Energetically, the major force applied in running is the one that makes you leave the ground. (Hence the comments about not bounding, not bouncing, wanting to glide, ...) If you try to apply the force in your running motion (after accelerating to your steady pace) strictly in the vertical, the mechanics of the legs are such that this means you apply a lot of collateral _braking_ force to your horizontal motion. Energetics: The power requirement for overcoming wind is U * rho*A*Cd*(W-U)^2 where U is your speed relative to the ground, W is the wind speed (taken positive in the direction you're running), A is the cross sectional area you present to the wind, rho is the density of the air, and Cd is a drag coefficient. rho and Cd are about 1, A is ... call it 0.5. Let's take 3 m/s for the runner's speed (an 9:33 1600) and a 5 m/s head wind. Power requirement is then 3*.5*8^2 or 96 W. At a 12 m/s head wind (~25 mph) this goes to 337 W. At zero wind it is 14 W. First ballpark comparison: ballpark athlete can sustain about 1000 W power output. Even a modest breeze at a modest running speed is 10% of potential power output. Second comparison: The rule of thumb caloric expenditure is 100 Kcal per mile. That's about 400 KJ/mile, and our 3 m/s runner is taking 533 seconds to complete a mile, for about 780 W. (Ref: 1000 W would be 7 minute mile pace given these assumptions/roundings) Q: Where, exactly, does all the extra energy go? I believe the answer is 'muscle dissipation', i.e. the fact that your muscles aren't springy enough to absorb all the energy from each touchdown that you had released in launching yourself. But I find this very unsatisfying as an explanation since it leaves no destription of how running mechanics can affect the loss term, nor how it depends (if it does?) on the speed of motion. Help? Ah well. He did ask for a physics answer. Unfortunately the physics needs to know about the biology as well. -- Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links. Sagredo (Galileo Galilei) "You present these recondite matters with too much evidence and ease; this great facility makes them less appreciated than they would be had they been presented in a more abstruse manner." Two New Sciences From dek@bnl.gov Fri Apr 16 11:27:55 1999 Return-Path: Received: from mail1.radix.net (mail1.radix.net [209.48.224.31]) by saltmine.radix.net (8.8.7/8.8.7) with ESMTP id LAA03140 for ; Fri, 16 Apr 1999 11:27:54 -0400 (EDT) Received: from magpie.prod.itd.earthlink.net (magpie.prod.itd.earthlink.net [209.178.63.8]) by mail1.radix.net (8.9.3/8.9.3) with ESMTP id LAA05442 for ; Fri, 16 Apr 1999 11:27:52 -0400 (EDT) Received: from bnl.gov (david@sdn-ar-002nyhempP228.dialsprint.net [168.191.55.244]) by magpie.prod.itd.earthlink.net (8.8.7/8.8.5) with ESMTP id IAA18050 for ; Fri, 16 Apr 1999 08:27:36 -0700 (PDT) Sender: david@sprintmail.com Message-ID: <371756B2.98DE65DC@bnl.gov> Date: Fri, 16 Apr 1999 11:26:42 -0400 From: "David E. Kahana" Reply-To: dek@bnl.gov X-Mailer: Mozilla 4.04 [en] (X11; I; Linux 2.2.3 i686) MIME-Version: 1.0 Newsgroups: rec.running To: Robert Grumbine Subject: Re: treadmill running effort vs. road running effort References: <3715192F.3B8C93E2@teron.com> <7f37ju$4e8$1@camel21.mindspring.com> <7f3hc9$k6s@ds2.acs.ucalgary.ca> <7f4reo$r2s$1@camel19.mindspring.com> <7f52da$j1u$1@saltmine.radix.net> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: RO Content-Length: 2409 Very nice initial analysis Robert. I happen to be a physicist too. I would only add that another factor could be that some springiness exists in the treadmill that doesn't exist on the road. This would lead to a less inelastic collision of the downward moving body with the treadmill than occurs with the road. So, some energy might be stored in the elastic deformation of the treadmill which was retransmitted to the body on takeoff. We all know it is easier to jump up and down on a trampoline than on asphalt! Robert Grumbine wrote: > Q: Where, exactly, does all the extra energy go? I believe the > answer is 'muscle dissipation', i.e. the fact that your muscles aren't > springy enough to absorb all the energy from each touchdown that you > had released in launching yourself. But I find this very unsatisfying > as an explanation since it leaves no destription of how running mechanics > can affect the loss term, nor how it depends (if it does?) on the > speed of motion. > > Help? > > Ah well. He did ask for a physics answer. Unfortunately the > physics needs to know about the biology as well. Wild speculations: Good question: the energy lost from inelasticity of ground-body interaction I think has to be dissipated somewhere in the body. Energy is conserved, period. (v/c is small for most runners so we'll neglect relativistic effects). Certainly muscles/tendons/bones are somewhat lossy ... not returning all the energy that is stored in their elastic deformation. Of course this is very complex, muscles are also burning energy through much of their range of motion. Internal friction in the body is clearly too a factor. Impact also creates vibration all through the legs, bones and all, all the way up the torso. Vibrations end up as heat, heat flows into the extracellular space, into the blood and out to the skin, to the lungs and out with the breath as well. It is radiated and evaporated away. Is it possible, with perfectly elastic lossless legs, in a vacuum, to run an entire marathon, by taking only the first step? This is the only input of kinetic energy. Then you just bounce along ... I think not, friction will be needed so you can provide the initial forward impetus. So some work must be done too against the friction of the road. It seems to me there are various places here where running mechanics could affect the loss term. cheers, - dave k. From nelsonl@direct.ca Thu Apr 15 16:14:17 EDT 1999 Article: 169143 of rec.running Path: news1.radix.net!delphi!news1!sunqbc.risq.qc.ca!cyclone.bc.net!newsfeed.direct.ca!newsgate.direct.ca!not-for-mail From: "Nelson Lee" Newsgroups: rec.running References: <3715192F.3B8C93E2@teron.com> <7f37ju$4e8$1@camel21.mindspring.com> <7f3hc9$k6s@ds2.acs.ucalgary.ca> <7f4reo$r2s$1@camel19.mindspring.com> <7f52da$j1u$1@saltmine.radix.net> Subject: Re: treadmill running effort vs. road running effort Lines: 25 X-Newsreader: Microsoft Outlook Express 4.72.3110.5 X-Mimeole: Produced By Microsoft MimeOLE V4.72.3110.3 Message-ID: Date: Thu, 15 Apr 1999 11:00:58 -0700 NNTP-Posting-Host: 216.66.140.58 X-Trace: newsgate.direct.ca 924199245 216.66.140.58 (Thu, 15 Apr 1999 11:00:45 PDT) NNTP-Posting-Date: Thu, 15 Apr 1999 11:00:45 PDT Organization: Canada Internet Direct, Inc. Xref: news1.radix.net rec.running:169143 Status: RO Robert Grumbine wrote in message <7f52da$j1u$1@saltmine.radix.net>... >In article <7f4reo$r2s$1@camel19.mindspring.com>, >HappyTrails wrote: >>The concern I have about treadmill running has nothing to do with air >>resistance, I've got to believe it is negligible. > > Why do you _have_ to believe it is negligible? I've encountered some >good headwinds that reminded me it was far from negligible. Since I >notice even a 5 m/s headwind, and 5 m/s is an 80 second 400 -- something >withing a lot of runner's reach -- it is probably even an issue for >no wind. > [snip] There's a good section on the effect of wind in Jack Daniel's book Running Formula. A headwind of as lttle as 2 m/s (4.5 mph) requires an increase in oxygen consumption (VO2) of ~5% as compared to calm air. A headwind of 5m/s increases VO2 by ~15%. Also, the equivalent tailwind is not as beneficial to runners (ie VO2 decrease is not as great). So air resistance is a bigger factor than one may think. Nelson From marathonman@mindspring.com Fri Apr 16 09:04:59 EDT 1999 Article: 169183 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!news.maxwell.syr.edu!cpk-news-hub1.bbnplanet.com!news.gtei.net!firehose.mindspring.com!not-for-mail From: marathonman@mindspring.com (Sam Callan) Newsgroups: rec.running Subject: Re: treadmill running effort vs. road running effort Date: Wed, 14 Apr 1999 02:46:51 GMT Organization: MindSpring Enterprises Lines: 22 Message-ID: <37140058.436426884@news.mindspring.com> References: <3715192F.3B8C93E2@teron.com> <7f37ju$4e8$1@camel21.mindspring.com> <7f3hc9$k6s@ds2.acs.ucalgary.ca> <7f4reo$r2s$1@camel19.mindspring.com> Reply-To: marathonman@mindspring.com NNTP-Posting-Host: cf.cd.d7.23 X-Server-Date: 16 Apr 1999 02:49:19 GMT X-Newsreader: Forte Free Agent 1.1/32.230 Xref: news1.radix.net rec.running:169183 Status: RO "HappyTrails" wrote: >The concern I have about treadmill running has nothing to do with air >resistance, I've got to believe it is negligible. I wish there was a >physicist who could address the dynamics of running to propel oneself >forward vs. running to maintain a static position relative to a moving >surface. Robert handles the physics far better than I can (or want to). If you are running at 10 mph on a treadmill at 0% grade, you are using less energy than running 10 mph outside at 0% grade because on the treadmill you are not moving air out of the way since you are moving vertically and not horizontally on the treadmill. Raising the grade of the treadmill 1-2% can better simulate the energetics of running outside. For all practical purposes, there is not much difference in terms of exercise benefits. Someone else noted that treadmillers often hold onto the rails while walking. There actually is a formula for this somewhere but I have no idea where without a lit search but I think Balke did the research. zzzzzzzzzzzzzzzzzzzzzz From bobg@Radix.Net Thu May 13 15:52:09 EDT 1999 Article: 172850 of rec.running Path: news1.radix.net!saltmine.radix.net!not-for-mail From: bobg@Radix.Net (Robert Grumbine) Newsgroups: rec.running Subject: How far do you fall? Date: 13 May 1999 15:51:54 -0400 Organization: RadixNet Lines: 11 Message-ID: <7hfagq$5p6$1@saltmine.radix.net> NNTP-Posting-Host: saltmine.radix.net Xref: news1.radix.net rec.running:172850 ... that is, does anyone know or have a good estimate of how far up and down your center of mass moves in running? Feel free to qualify it depending on runner pace or efficiency. I was looking again at the energetics of running, and the answer to this question may simplify things significantly. -- Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links. Sagredo (Galileo Galilei) "You present these recondite matters with too much evidence and ease; this great facility makes them less appreciated than they would be had they been presented in a more abstruse manner." Two New Sciences From shpyrko@fas.harvard.edu Sat May 15 11:24:59 EDT 1999 Article: 172856 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!baron.netcom.net.uk!netcom.net.uk!newspump.monmouth.com!newspeer.monmouth.com!logbridge.uoregon.edu!news.fas.harvard.edu!not-for-mail From: Oleg Shpyrko Newsgroups: rec.running Subject: Re: How far do you fall? Date: 13 May 1999 20:18:53 GMT Organization: Harvard University, Cambridge, Massachusetts Lines: 39 Message-ID: <7hfc3d$4ve$1@news.fas.harvard.edu> References: <7hfagq$5p6$1@saltmine.radix.net> NNTP-Posting-Host: ice4.fas.harvard.edu User-Agent: tin/pre-1.4-980117 (UNIX) (OSF1/V4.0 (alpha)) Xref: news1.radix.net rec.running:172856 Robert - I don't know the number, but I could suggest a way of estimating it. There must be some research data on how much oxygen is needed to run at small incline (say .2) compared to no incline at the same speed. Assuming most of the energy is lost in bringing COM up and down, the loss of energy is mgh per step if incline=0, and an additional mg*sin(alpha)*length_of_step per step for incline=alpha. Just an idea... The original assumption is probably wrong, however. COM going up and down is significant, but probably dissipation of energy in the muscles and air resistance is comparable as well, and needs to be taken into account. Question for you - suppose we are talking about COM movement only, do you think energy a runner supplies to bring COM up and down by height h is mgh or 2*mgh? As any bodybuilder can tell you, bringing weight down costs just as much energy if done SLOWLY. On the other hand we do lose some of that second mgh, that's the reason shoes loose their cushioning and feet and knees have to absorb the rest of the shock, often leading to injuries. Too many unknown variables... :) Oleg Robert Grumbine wrote: : ... that is, does anyone know or have a good estimate of how far : up and down your center of mass moves in running? Feel free to : qualify it depending on runner pace or efficiency. I was looking : again at the energetics of running, and the answer to this question : may simplify things significantly. : -- : Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links. : Sagredo (Galileo Galilei) "You present these recondite matters with too much : evidence and ease; this great facility makes them less appreciated than they : would be had they been presented in a more abstruse manner." Two New Sciences -- Oleg From kahana@sprintmail.com Sat May 15 11:25:04 EDT 1999 Article: 172887 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!news.maxwell.syr.edu!newsfeed.berkeley.edu!newsfeed1.earthlink.net!nntp.earthlink.net!posted-from-earthlink!not-for-mail From: "David E. Kahana" Newsgroups: rec.running Subject: Re: How far do you fall? Date: Thu, 13 May 1999 20:10:36 -0400 Content-Transfer-Encoding: 7bit References: <7hfagq$5p6$1@saltmine.radix.net> <7hfc3d$4ve$1@news.fas.harvard.edu> X-Posted-Path-Was: not-for-mail Content-Type: text/plain; charset=us-ascii X-ELN-Date: 14 May 1999 00:10:44 GMT X-ELN-Insert-Date: Thu May 13 17:15:21 1999 Organization: EarthLink Network, Inc. Lines: 84 Mime-Version: 1.0 Reply-To: kahana@sprintmail.com NNTP-Posting-Host: sdn-ar-002nydparp225.dialsprint.net Message-ID: <373B69FC.9AB32C8D@bnl.gov> X-Mailer: Mozilla 4.04 [en] (X11; I; Linux 2.2.3 i686) Xref: news1.radix.net rec.running:172887 Oleg Shpyrko wrote: > Robert - I don't know the number, but I could suggest a way of > estimating it. There must be some research data on how much > oxygen is needed to run at small incline (say .2) compared to > no incline at the same speed. > Assuming most of the energy is lost in bringing COM up and down, > the loss of energy is mgh per step if incline=0, and an additional > mg*sin(alpha)*length_of_step per step for incline=alpha. > Just an idea... > The original assumption is probably wrong, however. COM going up > and down is significant, but probably dissipation of energy in the > muscles and air resistance is comparable as well, and needs to be taken > into account. > Question for you - suppose we are talking about COM movement only, > do you think energy a runner supplies to bring COM up and down > by height h is mgh or 2*mgh? As any bodybuilder can tell you, > bringing weight down costs just as much energy if done SLOWLY. > On the other hand we do lose some of that second mgh, that's the > reason shoes loose their cushioning and feet and knees have to > absorb the rest of the shock, often leading to injuries. > Too many unknown variables... > :) > Oleg > > Robert Grumbine wrote: > : ... that is, does anyone know or have a good estimate of how far > : up and down your center of mass moves in running? Feel free to > : qualify it depending on runner pace or efficiency. I was looking > : again at the energetics of running, and the answer to this question > : may simplify things significantly. > > : -- > : Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links. > : Sagredo (Galileo Galilei) "You present these recondite matters with too much > : evidence and ease; this great facility makes them less appreciated than they > : would be had they been presented in a more abstruse manner." Two New Sciences > > -- > Oleg Answer to Oleg's last question: on takeoff the body is given kinetic enery upward, roughly equal to mgh, where h is the final height reached by the COM above the point it was at on takeoff. At the top of the trajectory, all of this kinetic energy has become gravitational potential energy mgh. It is converted back to kinetic energy as the CM falls back down, and this happens with no effort on the part of the runner: he falls freely. The energetics of weight lifting is quite different because there is no free fall involved, unless you have just dropped the weight. So you do work on the way down as well as up. My discussion neglects air resistance (surely negligible for the up/down motion, which is not fast), and takes no account of the efficiency of the body in imparting the upward velocity. It also does not discuss what happens next, on landing, to the kinetic energy of downward motion, Some of this energy is stored in the muscles/bones/tendons, and some is dissipated. So the body might do some additional work too in braking the downward fall, after the foot makes contact with the ground. After the downward motion of the CM is finished, some of the downward kinetic energy (mgh) has been converted to elastic energy, that is, to stretching of the muscles tendons and bones, and some has been dissipated. The actual work done per step in producing CM upward motion must therefore be somewhat bigger than mgh, due to inefficiency, but most likely is smaller than 2*mgh, at least so I would guess. mgh is an absolute lower bound, but what I have said is not really a proof that it is less than 2*mgh. You also have to worry about forward and backward oscillation of the CM, if such occurs, and assuming we are working in an inertial frame following the average velocity of the runner up the hill. To Robert: there is a book I saw in Borders medical section recently, I think it was called either `Kinetics of motion' or `Kinesiology of motion.' It had figures on the up down motion of the CM in walking gait, but not running. The number was on the order of 2-3 cm, and was kept this low by an extremely complex motion of the pelvis, knees, and legs. For running, this is a lower bound, therefore. I'll see if I can find any better data for you. I am certain this must have been measured. cheers, - dave k. From shpyrko@fas.harvard.edu Sat May 15 11:25:17 EDT 1999 Article: 172921 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!newsfeed.direct.ca!logbridge.uoregon.edu!news.fas.harvard.edu!not-for-mail From: Oleg Shpyrko Newsgroups: rec.running Subject: Re: How far do you fall? Date: 14 May 1999 04:21:58 GMT Organization: Harvard University, Cambridge, Massachusetts Lines: 85 Message-ID: <7hg8d6$9qg$1@news.fas.harvard.edu> References: <7hfagq$5p6$1@saltmine.radix.net> <7hfc3d$4ve$1@news.fas.harvard.edu> <373B69FC.9AB32C8D@bnl.gov> NNTP-Posting-Host: login3.fas.harvard.edu User-Agent: tin/pre-1.4-980117 (UNIX) (OSF1/V4.0 (alpha)) Xref: news1.radix.net rec.running:172921 David E. Kahana wrote: Hey David, I am a physisist so you didn't have to go thru extended explanation how kinetic energy transforms into gravitational and back. :) I enjoyed it nevertheless :) The thing is - it's not clear to me that the number is closer to mgh rather than to 2mgh. Probably for less efficient runners it is mgh, but I bet for more efficient it is closer to 2mgh. h itself is considerably smaller for efficient runners, therefore they do not "fall" from height h, but rather raise and lower themselves. If anyone saw Boston or Olympics coverage, Fatuma Roba, IMPHO one of the most efficient and smooth runners out there, almost doesn't have that impact, it's all very smooth - just a transitional stage after transitional stage. For efficient runners both wind and energy dissipation in muscles could be quite a considerable factor. For example, if you try swinging your hands while standing still, imitating running (I know it sounds stupid but it's actually quite useful excersice to improve running form) you'd be surprised how after only a couple of minutes of such intensive swinging you'd be out of breath. You waste plenty of energy, and the same goes for legs as well, even without moving your COM. For efficient runners h can be quite small and this dissipation issue could very well leading factor. And finally - the wind. Difficult to say for sure, but strong winds can throw marathon times off by 2-3 minutes easily even for 2:10 marathoners. In other words one could probably find an equivalent to the slope situation by finding similar difference in times and therefore calculate the energy. Wind-in-the-back situation is somewhat masked by lack of cooling effect (like this year at Boston), so nobody really knows how serious wind effects really are and how fast one could run if there'd be no resistance (in vacuum for example :) hehe) But it's an interesting topic for discussion and I am curious if anyone knows something for sure :) Oleg : Answer to Oleg's last question: on takeoff the body is given kinetic enery : upward, roughly equal to mgh, where h is the final height reached by the COM : above the point it was at on takeoff. At the top of the trajectory, all of : this kinetic energy has become gravitational potential energy mgh. It is : converted back to kinetic energy as the CM falls back down, and this happens : with no effort on the part of the runner: he falls freely. The energetics of : weight lifting is quite different because there is no free fall involved, : unless you have just dropped the weight. So you do work on the way down as : well as up. My discussion neglects air resistance (surely negligible for the up/down : motion, which is not fast), and takes no account of the efficiency of the : body in imparting the upward velocity. : It also does not discuss what happens next, on landing, to the : kinetic energy of downward motion, Some of this energy is stored : in the muscles/bones/tendons, and some is dissipated. So the : body might do some additional work too in braking the : downward fall, after the foot makes contact with the ground. : After the downward motion of the CM is finished, some of the : downward kinetic energy (mgh) has been converted to elastic energy, : that is, to stretching of the muscles tendons and bones, and some has been dissipated. : The actual work done per step in producing CM upward motion must : therefore be somewhat bigger than mgh, due to inefficiency, but most : likely is smaller than 2*mgh, at least so I would guess. mgh is an : absolute lower bound, but what I have said is not really a proof that it is : less than 2*mgh. You also have to worry about forward and backward : oscillation of the CM, if such occurs, and assuming we are working : in an inertial frame following the average velocity of the runner up the : hill. : To Robert: there is a book I saw in Borders medical section recently, : I think it was called either `Kinetics of motion' or `Kinesiology of : motion.' It had figures on the up down motion of the CM in walking : gait, but not running. The number was on the order of 2-3 cm, and : was kept this low by an extremely complex motion of the pelvis, knees, : and legs. For running, this is a lower bound, therefore. I'll see : if I can find any better data for you. I am certain this must have : been measured. : cheers, : - dave k. -- Oleg From kahana@sprintmail.com Sat May 15 11:27:46 EDT 1999 Article: 173130 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!news.maxwell.syr.edu!newsfeed1.earthlink.net!nntp.earthlink.net!posted-from-earthlink!not-for-mail From: "David E. Kahana" Newsgroups: rec.running Subject: Re: How far do you fall? Date: Sat, 15 May 1999 08:01:09 -0400 Content-Transfer-Encoding: 7bit References: <7hfagq$5p6$1@saltmine.radix.net> <7hfc3d$4ve$1@news.fas.harvard.edu> <373B69FC.9AB32C8D@bnl.gov> <7hg8d6$9qg$1@news.fas.harvard.edu> To: Oleg Shpyrko X-Posted-Path-Was: not-for-mail Content-Type: text/plain; charset=us-ascii X-ELN-Date: 15 May 1999 12:01:43 GMT X-ELN-Insert-Date: Sat May 15 05:05:06 1999 Organization: EarthLink Network, Inc. Lines: 226 Mime-Version: 1.0 Reply-To: kahana@sprintmail.com NNTP-Posting-Host: sdn-ar-001nyhempp162.dialsprint.net Message-ID: <373D6205.F589F401@bnl.gov> X-Mailer: Mozilla 4.04 [en] (X11; I; Linux 2.2.3 i686) Xref: news1.radix.net rec.running:173130 Oleg Shpyrko wrote: > David E. Kahana wrote: > > Hey David, I am a physisist so you didn't have to go thru extended > explanation how kinetic energy transforms into gravitational and back. > :) > I enjoyed it nevertheless :) > The thing is - it's not clear to me that the number is closer to mgh > rather than to 2mgh. Probably for less efficient runners it is mgh, > but I bet for more efficient it is closer to 2mgh. h itself is > considerably smaller for efficient runners, therefore they do not > "fall" from height h, but rather raise and lower themselves. > If anyone saw Boston or Olympics coverage, Fatuma Roba, IMPHO one of the > most efficient and smooth runners out there, almost doesn't have that > impact, it's all very smooth - just a transitional stage after > transitional stage. > > For efficient runners both wind and energy dissipation in muscles could > be quite a considerable factor. For example, if you try swinging your > hands while standing still, imitating running (I know it sounds stupid > but it's actually quite useful excersice to improve running form) > you'd be surprised how after only a couple of minutes of such intensive > swinging you'd be out of breath. You waste plenty of energy, and the same > goes for legs as well, even without moving your COM. For efficient runners > h can be quite small and this dissipation issue could very well leading > factor. > And finally - the wind. Difficult to say for sure, but strong winds can > throw marathon times off by 2-3 minutes easily even for 2:10 marathoners. > In other words one could probably find an equivalent to the slope > situation by finding similar difference in times and therefore calculate > the energy. Wind-in-the-back situation is somewhat masked by lack of > cooling effect (like this year at Boston), so nobody really knows > how serious wind effects really are and how fast one could run > if there'd be no resistance (in vacuum for example :) hehe) > But it's an interesting topic for discussion and I am curious if > anyone knows something for sure :) > Oleg Oleg: Sorry, I can be pretty thick sometimes :-) I should have picked up on your sophistication in asking about whether it was 2*mgh. I took it to be a naive question. Glad you enjoyed my elementary physics lectures in any case :) It is a very interesting subject for discussion, and I don't know anything at all for sure about it, but I won't let that stop me from talking :) This is going to be a very long post, so to all those who don't really want to follow extensive theoretical and possibly wrong discussions, I apologize herewith and advise you to quit reading now or after the next couple of paragraphs. Reading much further may result in permanent psychosis. There is risk of loss in trading futures, and past performance does not necessarily guarantee future results ... oops, that belongs in another newsgroup. On further reflection and reading what you say, I agree with you, in your sense, that it could easily be closer to 2*mgh or even more. But this might become, if we aren't careful, to some extent a semantic question because it will be extremely hard, without experiments, to separate the work done against internal friction in the body into components which produce forward motion and up/down motion as well as sideways motion and torsional motion of all the body parts. As you say, watching elite long distance runners such as Fatuma Roba one gets the distinct impression that all of the motion is a very smooth transition between different postures. I would agree with you that dissipation is large, and your example of waving arms for 10-20 min is one I promise to try, as soon as I can get to a place where noone can observe me and possibly have me hauled away for creating a public disturbance :) But let me proceed even before that in a simple-minded fashion to estimate some of the numbers, using almost pure reasoning and freshman physics. We can get a weak upper bound on the total work just by considering the total energy used in running a mile: a canonical measured figure being 110kCal. Now I will make some assumptions. First of all, forget the hill. For me, at my ten minute pace, and striding about 120/min, I cover the 1600m distance in 1200 steps of 1.33 m each. 110kCal is 462000J, so we have 385J/step. A weak upper bound to the work producing upward motion is thus 385 J/step. Now if every last Joule goes into up down motion at each step, and I weigh 70 kg, then what is h? Answer: h= 385J/70x9.8N=0.56 m ! No way am I jumping that high. So the assumptions were very bad indeed, particularly the one about all work going into up/down motion. We have a clear energy budget surplus here, because, I will assert, without doing any measurements at all, that the height I reach with each step is less than 10cm. Assume it is 10cm. Then I do 68.6 J of work every step, in raising my CM to 10 cm height (in your sense there may well be more, but just consider what it costs to raise the CM). This leaves 316.4 J. So the energy going into increasing mgh at each step is only some 18% of the total. What is the free fall time if h=10cm? An easy calculation gives t_f = 0.29s = 2 * 0.143s, and my time per step is 0.5s under the conditions given. Therefore the foot will be in contact with the ground for t_c =0.21s or 40% of the step time, and will move through a horizontal distance of d_c = 0.56 m. The main new assumption here was that all of the vertical CM motion occurs after the foot leaves the ground. This is possibly a very bad assumption, I agree, since it assumes that the leg and the pelvis move perfectly so as to maintain the CM at a constant height during contact of the foot with the ground. Maybe if the CM amplitude h is 10cm, then the motion is really something like 7cm before, 3cm after, or even 2cm/8cm as you suggest for the very efficient elite runners (for whom h might be much smaller too). But none of that would affect the energy going into raising the CM, only reducing the total amplitude would. So I'ld guess that some 20% of the total energy per step is in play for this type of improvement in efficiency. By the way, this free fall time strikes me as too long given my own subjective experience of my own running. I'ld say that at the 10:00 pace, I am on the ground more than I am in the air. This would suggest that h is smaller than 10cm even for an inefficient runner like myself, or that as you say some of the motion occurs during contact with the ground. Now here's yet another diversion from my main argument, and a possible explanation of other experimental facts: This 18% factor could help to explain another empirical observation, that the total work required to go a given distance is roughly independent of the speed. Why? Because to go faster, I must either increase step length at constant step frequency, or increase step frequency at constant step length, or some combination of the two. Increasing step frequency will give me the same number of steps to go a mile in less time, so the mgh work will be the same if h is independent of step frequency. If however, h increases or decreases a little with step frequency, the mgh work is only 18% of the total, and the total mgh work will not get much larger. Increasing step length will require less steps to go a mile, and therefore, possibly, less mgh work. But with greater step length it is possible that h increases, I would guess. You would argue it decreases or is constant, in an efficient runner. To me this is counterintuitive, but it may very well be true. In either case, the mgh work seems to be a relatively small fraction of the total, and so we would have no great mystery about the relatively weak dependence of total energy expenditure for a fixed distance on speed of running. Back to the main discussion: Where does the extra 316.4 J go? I imagine, into work required to maintain forward and other motions against internal friction of the body, horizontal work against the road, and friction of the road surface, and air resistance. Can we estimate some of these factors? I will try to. The least certain and most difficult to estimate is internal friction of the body so let's deal with the others first. Let's assume no wind to start. What is the work against air resistance? I have frontal surface area of about 2m x 0.5m = 1m^2. I am running 1600m/10:00, so the effective wind velocity is 2.67 m/s. Rather than doing hydrodynamical calculations, let me just assume that the total volume of air which I traverse in one step is all accelerated from 0 m/s to 2.67 m/s as I push through it. This ought to give an order of magnitude estimate of the air resistance work, though I would agree one can critcise this calculation very easily. But you'll have a hard time making it bigger. I would say that it gives close to an upper bound on the air resistanc. I also did this calculation assuming I was a perfect sphere with same frontal area and using Stoke's law. The answer for the viscous work comes out infinitesimally small in that calculation because the viscosity of air is tiny, only 181 micropoise. Please don't ask me to solve the Navier-Stokes equation for a runner in still air! What is the total volume of air I traverse in one step? At my 1.33m step length it is 1.33 m x 1 m^2, or 1.33 cubic meters, which is 1330 liters. At standard temperature and pressure, one mole of a gas occupies 22.4 liters. So we have roughly 60 moles of air, which has an average molecular weight of let's just say 30 (forget the 20% O2, pretend it is all N2). So 1800 grams of air are accelerated to 2.67 m/s, at which time the air has a kinetic energy of 1/2 * 1.8 * (2.67)^2 or 6.416 J. Subtract this from the remaing 316.4 and we have 310 J left. At 5:00 per mile, this would be 24 J. With a 10 mph headwind this becomes 85 J, and is greater than the mgh work. Wind can be very important indeed, as you said. But at 10:00 pace in still air, air resistance is dominated by mgh work. Now, what about work done against the road. Again I have to make a few more questionable assumptions :) Let me get the average vertical force by calculating the downward velocity of the body at the moment the foot makes contact. This is v = 1.4 m/s, assuming h of 10 cm. This velocity is reversed by the time the body takes off again, after t_c = 0.21s. therefore F_v = 934 N, for a mass of 70 kg. I will assume that this force is the only one, and it is exerted downward along the leg, which I treat as a rigid pole attached to a rigid foot by a hinge joint with zero friction. So the knee is neglected :) Let's say that the body takes off when the leg reaches an angle of 75 degrees to the foot. (Leg is 1m long and step distance is 0.56m, of which 0.28 is in front of and 0.28 is in back of the CM in my fanatsy picture of the runner). Then the maximum horizontal force is 261 N. The average might be more like 130. So the work done against the road would be 73 J. Out of the 385 we started with, we now have about 237 J left. All of this must be dissipated against the internal friction of the body. This is a rather amazing feature of running if my analysis is anywhere near reality. It would seem that most of the energy generated is dissipated in the body, simply in the process of positioning the limbs, just so that the next step can be taken. The percentages were, for my 10:00 pace, h=10cm example with no wind. mgh = 18% air resistance = 1.5% road = 19% internal friction of body = 61.5% cheers, - dave k. From dennizin@pcstarnet.com Sat May 15 11:49:05 EDT 1999 Article: 172911 of rec.running Path: news1.radix.net!outfeed1.news.cais.net!nntp.abs.net!newsfeed.enteract.com!ix.netcom.com!news From: dennizin@pcstarnet.com (Denny Anderson) Newsgroups: rec.running Subject: Re: How far do you fall? Date: Fri, 14 May 1999 03:03:54 GMT Organization: Netcom Lines: 44 Message-ID: <373b9292.6690138@206.214.99.10> References: <7hfagq$5p6$1@saltmine.radix.net> NNTP-Posting-Host: app-wi5-24.ix.netcom.com Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-NETCOM-Date: Thu May 13 10:03:52 PM CDT 1999 X-Newsreader: Forte Agent 1.5/32.452 Xref: news1.radix.net rec.running:172911 bobg@Radix.Net (Robert Grumbine) wrote: > ... that is, does anyone know or have a good estimate of how far >up and down your center of mass moves in running? Feel free to >qualify it depending on runner pace or efficiency. I was looking >again at the energetics of running, and the answer to this question >may simplify things significantly. Oleg's analysis acknowledged, here's a different way to look at the energy relationships. Consider you are pushing a moderately (not excessively) heavy vehicle on a level surface. It has round wheels and good bearings. Shouldn't take a whole lot of effort to keep it moving at a reasonable rate. Now, replace the wheel assemblies with ones that have a slightly eccentric axis. As the wheels turn the vehicle translates AND moves slightly vertically at each wheel revolution. To maintain an identical speed will require a bit more effort, since the CofG is moving up and down. In all things mechanical, the energy expended (up) is never completely returned (down) due to entropy (losses). Next, install wheels with an even greater eccentricity at the axis of rotation. Repeat above - but add more force to maintain speed. So, as Oleg said, the vertical component of force to raise the CofG (COM) is wasted energy. The higher the CofG goes the lower the input energy that remains for horizontal speed. An answer to your question about estimating how high CofM moves (my answer) is that *it depends*. Depends on how efficiently one moves horizontally. Some bound like a Springbok, and others glide (Ozzie?) like an Ostrich . Actually an Ostriche's running form is pretty efficient, if one looks closely. Lots of mass to carry on those spindly legs (but i digress). That's why I have long harped on learning to minimize vertical motion when running. It's a matter of conservation of energy. Denny Anderson To ERR is human... to ZIN, divine! < dennizin@pcstarnet.com > From bobg@Radix.Net Sat May 15 11:50:16 EDT 1999 Article: 173148 of rec.running Path: news1.radix.net!saltmine.radix.net!not-for-mail From: bobg@Radix.Net (Robert Grumbine) Newsgroups: rec.running Subject: Re: How far do you fall? Date: 15 May 1999 11:48:58 -0400 Organization: RadixNet Lines: 73 Message-ID: <7hk51a$c96$1@saltmine.radix.net> References: <7hfagq$5p6$1@saltmine.radix.net> <373B69FC.9AB32C8D@bnl.gov> <7hg8d6$9qg$1@news.fas.harvard.edu> <373D6205.F589F401@bnl.gov> NNTP-Posting-Host: saltmine.radix.net Xref: news1.radix.net rec.running:173148 In article <373D6205.F589F401@bnl.gov>, David E. Kahana wrote: [deletia of some interesting points] >Now if every last Joule goes into up down motion at each step, >and I weigh 70 kg, then what is h? Answer: h= 385J/70x9.8N=0.56 m ! >No way am I jumping that high. Yes, this was the kind of calculation I'd made. I was hoping to get _observed_ values of h, as I was reasonably confident from horizon watching that I (and other casual distance runners) wasn't going substantially more than 10 cm up. (I'm no so sure about sprinters.) >This 18% factor could help to explain another empirical observation, that >the total work required to go a given distance is roughly independent >of the speed. Why? Because to go faster, I must either increase step >length at constant step frequency, or increase step frequency at >constant step length, or some combination of the two. This is another of the interesting points. We do have the observation, posted occasionally in the group (I believe Denny Anderson is the one who found the good article on this point) that there is an optimal stride rate/length combination for a given runner at a given speed. (Optimal in an energetic sense). The interesting feature being that the optimum is a rather broad one, but U shaped one. This says some things about how the energy is being lost. What, exactly, I don't know yet, but something. >Can we estimate some of these factors? I will try to. The >least certain and most difficult to estimate is internal >friction of the body so let's deal with the others first. Let's >assume no wind to start. What is the work against air resistance? >I have frontal surface area of about 2m x 0.5m = 1m^2. I am running Life is simpler than this. Dissipation can be taken out of 1/2*rho*Cd*A*U^3 (U^2 gives the drag force, the extra U converts to power loss) where rho is the density of air (approximately 1 kg/m^3, A is area perpendicular to the wind, and Cd is the great fudge factor, 'Coefficient of drag'. It varies between about .3 and 1.5 for ordinary shapes, under ordinary speeds. For a flat plate (my hydrodynamic approximation to the body) it is about 1. For your U and A, with rho = 1, I get 9.5 W, substantially the same as your estimate. Stokes' law failed because the motion is nowhere near laminar. UL/nu = Re, which must be under about 2000 for Stokes to apply. U = 2.67, L is in the vicinity of 1, and nu is around 1E-5. So your Re is over 200,000. (Actually stokes is more limited than that, but over 2000 you're turbulent, not just outside the realm of good approximation.) >internal friction of body = 61.5% Indeed. Thanks for the work on road consideration. In some loose estimates, I find that considering the muscles as springs, the ballpark spring constant is something like 50 kN/m (that assumed total dissipation, which isn't true). But how's this for an ancillary conclusion: The major difference between very fast runners and slower ones is not fast twitch vs. slow twitch (etc) _per_ _se_, but low-friction vs. high friction muscles. As yet unsupported, but an interesting point for research (Sam, does this ring any bells?) -- Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links. Sagredo (Galileo Galilei) "You present these recondite matters with too much evidence and ease; this great facility makes them less appreciated than they would be had they been presented in a more abstruse manner." Two New Sciences From dek@bnl.gov Mon May 17 07:15:53 1999 Return-Path: Received: from mail1.radix.net (mail1.radix.net [209.48.224.31]) by saltmine.radix.net (8.8.7/8.8.7) with ESMTP id HAA25097 for ; Mon, 17 May 1999 07:15:52 -0400 (EDT) Received: from magpie.prod.itd.earthlink.net (magpie.prod.itd.earthlink.net [209.178.63.8]) by mail1.radix.net (8.9.3/8.9.3) with ESMTP id HAA12173 for ; Mon, 17 May 1999 07:15:51 -0400 (EDT) Received: from bnl.gov (david@sdn-ar-002nyhempP288.dialsprint.net [168.191.118.218]) by magpie.prod.itd.earthlink.net (8.8.7/8.8.5) with ESMTP id EAA17552 for ; Mon, 17 May 1999 04:15:18 -0700 (PDT) Sender: david@sprintmail.com Message-ID: <373FFA33.4939FBB5@bnl.gov> Date: Mon, 17 May 1999 07:14:59 -0400 From: "David E. Kahana" Reply-To: kahana@sprintmail.com X-Mailer: Mozilla 4.04 [en] (X11; I; Linux 2.2.3 i686) MIME-Version: 1.0 Newsgroups: rec.running To: Robert Grumbine Subject: Re: How far do you fall? References: <7hfagq$5p6$1@saltmine.radix.net> <373B69FC.9AB32C8D@bnl.gov> <7hg8d6$9qg$1@news.fas.harvard.edu> <373D6205.F589F401@bnl.gov> <7hk51a$c96$1@saltmine.radix.net> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Status: R Content-Length: 13783 Robert Grumbine wrote: > In article <373D6205.F589F401@bnl.gov>, > David E. Kahana wrote: > > >Now if every last Joule goes into up down motion at each step, > >and I weigh 70 kg, then what is h? Answer: h= 385J/70x9.8N=0.56 m ! > >No way am I jumping that high. > > Yes, this was the kind of calculation I'd made. I was hoping to > get _observed_ values of h, as I was reasonably confident from > horizon watching that I (and other casual distance runners) wasn't > going substantially more than 10 cm up. (I'm no so sure about > sprinters.) > I'm still looking in vain for any observed values. None of the books I havediscusses this sort of basic energy budget in detail, including even some by world reknowned physiologists. They certainly must have thought of things in these terms, but do not appear to have set it down in writing if they did. I find this bizarre, but I am just beginning to look at the literature. At worst there are some famous early stroboscopic photos, if I recall, from MIT labs, of a man running, and horse galloping, with a grid in the background. They proved conclusively that the horse had all four legs off the ground in a gallop, which was a point of controversy at one time. If you could get yourself a copy of those photos, you'ld have a data point. Or get a video of yourself against some appropriate background. I may try to do that myself. > This is another of the interesting points. We do have the observation, > posted occasionally in the group (I believe Denny Anderson is the one > who found the good article on this point) that there is an optimal > stride rate/length combination for a given runner at a given speed. > (Optimal in an energetic sense). The interesting feature being that > the optimum is a rather broad one, but U shaped one. This says some > things about how the energy is being lost. What, exactly, I don't > know yet, but something. > This is certainly food for thought. I will look up Denny's posts for the article. > Life is simpler than this. Dissipation can be taken out of > 1/2*rho*Cd*A*U^3 > Thanks for this! My life, at least will be simpler :) > Stokes' law failed because the motion is nowhere near laminar. > UL/nu = Re, which must be under about 2000 for Stokes to apply. > U = 2.67, L is in the vicinity of 1, and nu is around 1E-5. So > your Re is over 200,000. (Actually stokes is more limited than > that, but over 2000 you're turbulent, not just outside the realm > of good approximation.) > Thanks too for this reminder, I was wondering about it for sometime. Terminal velocity for a man was coming out near the speed of light. But it should have been pretty clear what was wrong. I have done some relativistic hydrodynamics, so there is no excuse for me here. > >internal friction of body = 61.5% > > Indeed. Thanks for the work on road consideration. In some loose > estimates, I find that considering the muscles as springs, the ballpark > spring constant is something like 50 kN/m (that assumed total dissipation, > which isn't true). > > But how's this for an ancillary conclusion: The major difference > between very fast runners and slower ones is not fast twitch vs. slow > twitch (etc) _per_ _se_, but low-friction vs. high friction muscles. > As yet unsupported, but an interesting point for research (Sam, does > this ring any bells?) > This is a very interesting suggestion, and certainly one which might seem to be justified by the numbers we are working with at the moment. Along those lines I have a couple of corrections, but they don't so far really affect the basic conclusion that a very large fraction of the energy really is dissipated in body motion which does no direct work of propulsion. Here is another long discussion, with a corrected table at the end for the energy budget: This is a little embarrassing. On thinking a bit over the weekend, I noticed a tacit assumption I made about the runner in all my calculations. Namely, he is dead, or at least his heart was beating without doing any work at all, and his respirations cost him no energy. So in the large term which I called `internal friction in the body' there should be at least two further subdivisions called `cardiovascular work' and `work of breathing'. This nicely illustrates the difference between a physiologist and a physicist. You see, I was quite happy calling the work of the heart and lungs `internal dissipation.' Everyone should rejoice that I am not in charge at the nearest emergency room :) I can also correct some of my input parameters using data from my most recent run. Here's a report on actual running parameters for my most recent run. It was a beautiful evening yesterday for running here on LI, temperature about 50 Fahrenheit, no wind, clear skies, Venus high up in the sky and about as bright as I've seen it in recent times. I wanted to do some measurements, so I purposefully ran only about 1/2 my normal distance. In short, I ran 2.6 miles in 24:30. This was a bit better than the 10:00 pace I'ld guessed at before: about 9:25 per mile, including two brief stops under two lampposts to estimate my step frequency. My actual step frequency turned out to be 50 steps in something near 20 sec, or about 2.5/sec. This is about 150/min rather than the 120/min I guessed before. My respirations were deep but rather slow: in every three steps, out every three steps, or about 25/min. I was not short of breath at all, and I'ld say my effort was easy for the first 2.2 miles and moderate for the last 0.4 mi. My heart rate, which I measured immediately after finishing by counting my carotid pulse for 30 sec, was 154 bpm. My blood pressure I couldn't get until I went inside, probably some 2 min later, and it was at that time 145/85. In another five minutes these dropped to 94bpm and 120/80. I am going to estimate, arbitrarily, that my BP was 160/90 during the run. This was a very enjoyable run, I felt like going much faster, but stopped myself because I have planned some speed work for the middle of the week. OK, there you have it. I will go back and recalculate all of the energetics for myself, but I'm not going to present the numbers here. I doubt that they are accurate enough in any case that the differences should be taken very seriously. But to me h=10cm seems less likely now, since the step time has been reduced to 0.4s, and the free fall time was already 0.29s. Clearly at higher step rates and fixed velocity, we begin to get a limit on the maximum possible h, if it is spent in free fall, maybe pointing to the correctness of Oleg's views about h. I spent much of the weekend trying to learn enough physiology to estimate the work of breathing, and the cardiovascular work. I must admit that so far I have failed miserably, especially for the work of breathing. I thought, at first, it can't be that hard to get a ballpark figure, it's mostly just the diaphragm doing the work. So then I looked up the anatomy in a very nice book: Anatomy of Movement, by Blandine Calais-Germain. The muscles involved in respiration are according to her, and in no particular order: (1) diaphragm (2) transversus thoracis (3) internal and external intercostals (4) levatores costarum The origins and insertions of these muscles are all over the thorax. I won't attempt to list them. The diaphragm is especially complicated, having no bony insertion, but inserting instead on the `central tendon', which has a very complex three-leaved shape. It forms a dome whose top moves from the fourth intercostal space on expiration to the sixth on inspiration. The abdominal muscles are supposed to cooperate in respiration too, relaxing with inspiration, contracting with expiration and deforming the viscera at constant volume. All of this should be happening while you are running, and pushing all of the stuff up and down. I am not going to attempt to model all of this for the moment. Give me a week or two to think of a simple minded approach :) Maybe I can use some of the time we have for supernovae calculations on the latest supercomputers :) These minor complexities will also make a very precise experimental determination of the path of the CM of the body during running quite a difficult task, I would guess. Probably a couple of percent is the best that can be done. I will therefore just take an estimate for the work of breathing from David L. Costill's, Inside Running, which is a beautifully written book, but unfortunately leaves many of the basic parameters which a physicist would like to know with respect to the kinematics and energetics of running totally undefined. Since he seems to be a well-respected authority I can only hope that that is not because these parameters are unknowable, but just because they are unknown. On page 46 he states that: `During distance races, where the respiratory volume may average 100 to 120 liters per minute, as much as 9% of the energy used by the body will be needed for respiration [52] .' Reference [52] is Nielsen, M., `Die Respirationsarbeit bei Korperruhe and bei Muskelarbeit'. Skand. Arch. Physiol., 74:299-366 (1936). Costill thinks in any case that this work is probably not a limiting factor in distance running. I will have to brush up on my Deutsch. For cardiovascular work, I can do only a little bit better. I can at least estimate the PV, or external work of the left ventricle, and can guess at the volume parameters using my own heart, since I have the report of a recent echocardiogram. I will neglect the work of the right ventricle, and the atria. I also have no idea of the efficiency of the myocardium, so I will just ignore it. So this will just be a lower bound on the work of pumping the blood. Relevant echocardiographic parameters of my heart at rest were: LVID (ES): 33 mm LVID (ED): 50 mm EF: 66% In fact my BP and HR were 130/80 and 94bpm during the echo, as I was tired and a bit nervous. Normal for me is about 110/70 and 60 bpm. I also wish he would have included another dimension of the LV in the report as it would make me happier about estimating the volume, but that's all the data I have. Using a cylindrical shape with length equal to its diameter to approximate my LV, I find the end systolic and and diastolic LV volumes to be: V(ES) = 28 mL V(ED) = 98 mL The stroke volume is thus 70 mL, which gives me an ejection fraction of 71%, in reasonable agreement with what my cardiologist has calculated in the echo report. My cardiac output then works out to 6.6 L/min which seems rather high considering I was just lying on my side, and usual at rest would be about 5L/min. However it's probably just that sympathetic nervous activity was resulting in a higher heart rate and contractility than usual. To get the PV work I need the area inside the PV curve for the LV, which is nicely discussed in `Cardiovascular Physiology' by David E. Mohrman and Lois Jane Heller. Basically there is an isovolumetric contraction starting at end-diastole when the LV has finished filling and beginning at the end-diastolic pressure in the LV which is on the order of 20 mmHg. This contraction continues until the aortic valve opens, which happens when the pressure in the LV is equal to the pressure in the aorta, or the diastolic BP (90 mmHg for me when I am running). Then the pressure continues to rise in the LV and aorta, while the LV contracts, until the systolic pressure (160 mmHg) is reached. Eventually the outflow into the aorta seems to slow down, because there then follows a brief period of decline in the LV pressure, below the systolic but while the LV is still contracting. Presumably this occurs after the aortic valve has closed and it lasts until the LV finally reaches the end systolic volume. At this point an isovolumetric relaxation of the LV occurs, producing a rapid pressure drop in the LV until the mitral valve finally opens, when the LV pressure drops below the pressure in the left atrium, and diastolic filling begins. The pressure at this time seems to be on the order of 10 mm Hg. There is then an expansion of the LV as it is filled with blood, both passively and actively as the left atrium beats, until the end-diastolic volume and pressure are reached again. Roughly, the area in the PV plane looks like a rectangle with a parabola on top. You can try to use two triangles to better approximate the area of the top section and I did. I worked out the area as best I could without knowing some of the precise bounding curves, which are somewhat complex on the top and the bottom. I found, using the same stroke volume calculated in the resting state, and my estimated pressures during exercise: PV = 7955.0 mmHg x mL = 1.06 J. Now of course my stroke volume is probably larger in exercise, so let's adjust this to 1.5 J. Nevertheless it is not an overwhelmingly large number. Even if I multiply it by 2 for the two sides of the heart (surely a large overestimate since the pulmonary circulation is at low pressure) this is still only 3 J /beat. Have I got something wrong in the units or conceptually am I way off? Anyone? If not, then let me assume a large inefficiency of the myocardium, on the order of 5 to 1, in producing external work. Then the work of the heart would be 15 J / beat. It just happens that my heart was beating almost exactly once per step in last night's run. So on my former scale of energies the work of the heart would be 15 J / 385 J = 4% of total per step. A revised table of energy use in running might thus be: mgh = 18% air resistance = 1.5% road = 19% cardiovascular work = 4% breathing work = 9% internal friction of muscles = 48.5% However I am less than happy with the cardiovascular and breathing estimates here :) cheers, - dave k. From pro_man@for_d.com Mon May 17 13:45:24 EDT 1999 Article: 173421 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!news1.tor.metronet.ca!cyclone.news.idirect.com!island.idirect.com!nntp.abs.net!dca1-hub1.news.digex.net!intermedia!lynx.unm.edu!jobone!dailyplanet.srl.ford.com!eccws1.dearborn.ford.com!news From: Y-Rotation Newsgroups: rec.running Subject: Re: How far do you fall? Date: Mon, 17 May 1999 12:40:53 -0400 Organization: Totally_Disorganized_Place Lines: 24 Message-ID: <37404695.1F04E96@for_d.com> References: <7hfagq$5p6$1@saltmine.radix.net> NNTP-Posting-Host: 19.3.20.146 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-Mailer: Mozilla 4.04C-4.04f4.0 [en] (X11; I; IRIX64 6.4 IP30) Xref: news1.radix.net rec.running:173421 It depends on the amount of leg and arm lift and on how heavy both are in relation to the rest of the body. The CG location changes because the spacial distribution of mass is affected in temporal fashion by the position of runner's legs/arms. As the leg goes up and down so does the CG. Efficient runner's torso does not move up and down, though the location of his CG does. pete Robert Grumbine wrote: > > ... that is, does anyone know or have a good estimate of how far > up and down your center of mass moves in running? Feel free to > qualify it depending on runner pace or efficiency. I was looking > again at the energetics of running, and the answer to this question > may simplify things significantly. > > -- > Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links. > Sagredo (Galileo Galilei) "You present these recondite matters with too much > evidence and ease; this great facility makes them less appreciated than they > would be had they been presented in a more abstruse manner." Two New Sciences From bobg@Radix.Net Mon May 17 08:49:06 EDT 1999 Article: 173369 of rec.running Path: news1.radix.net!saltmine.radix.net!not-for-mail From: bobg@Radix.Net (Robert Grumbine) Newsgroups: rec.running Subject: Re: How far do you fall? Date: 17 May 1999 08:47:19 -0400 Organization: RadixNet Lines: 50 Message-ID: <7hp34n$1np$1@saltmine.radix.net> References: <7hfagq$5p6$1@saltmine.radix.net> <373D6205.F589F401@bnl.gov> <7hk51a$c96$1@saltmine.radix.net> <373FFA33.4939FBB5@bnl.gov> NNTP-Posting-Host: saltmine.radix.net Xref: news1.radix.net rec.running:173369 In article <373FFA33.4939FBB5@bnl.gov>, David E. Kahana wrote: > >I'm still looking in vain for any observed values. None of the books >I have discusses this sort of basic energy budget in detail, including >even some by world reknowned physiologists. They certainly must have >thought of things in these terms, but do not appear to have set it down >in writing if they did. I find this bizarre, but I am just beginning >to look at the literature. I'd bet that in the literature the consideration has been made. This is just the kind of thing I like to fool around with independantly, regardless of whether it has been analyzed before. Time was, I'd have assumed it hadn't been done (that is, back in my physics class-inspired days of 'physics superiority'). But contact with the biomechanics and paleontology groups in grad school taught me better regarding some other points, so I'll concede the rest now too. It is just entertaining to work some things out for yourself. You're working a lot with particular values. Handy for some things, but since the 100 (or 110, or whatever) cal/mile figure is to hold across a broad range of paces/stride lengths/stride rates, I've just taken some simple figures: 1 m stride, 180 strides/min, 3 m/s. They're not too far gone at least (that's about a 9 minute mile, a figure which many should be able to reach for a while at least). A couple of neat things fell out of considering the springs. I'm going to write it for my web page as it gets a bit specialized for the group (I assume this isn't a venue in which integrating ODE's is considered routine). But interesting (and entirely unsurprising to conventional wisdom) conclusion : It is best to run with a smooth stride. The physics goes on to define 'smooth stride' as one in simple harmonic motion (given certain assumptions). Higher frequency motion than the fundamental adds greatly to the dissipation. The higher, the worse in proportion to the square of the frequency (and the square of the amplitude). Things like 'an extra push at the end' are higher frequency components to the motion. I too missed things like the dissipation in circulation (which I really shouldn't have, this was a research area of the guy who taught my first fluid dynamics class). Keep us both out of the ER. (Heart, what's that? It does work?) :-) But let's not forget to get out and run! -- Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links. Sagredo (Galileo Galilei) "You present these recondite matters with too much evidence and ease; this great facility makes them less appreciated than they would be had they been presented in a more abstruse manner." Two New Sciences From dennizin@pcstarnet.com Wed May 19 11:49:48 EDT 1999 Article: 173518 of rec.running Path: news1.radix.net!dca1-feed2.news.digex.net!dca1-hub1.news.digex.net!intermedia!cyclone.news.idirect.com!island.idirect.com!newsfeed.enteract.com!ix.netcom.com!news From: dennizin@pcstarnet.com (Denny Anderson) Newsgroups: rec.running Subject: Re: Re: How far do you fall? Date: Tue, 18 May 1999 04:29:16 GMT Organization: Netcom Lines: 96 Message-ID: <3742da08.1892327@206.214.99.10> References: <7hfagq$5p6$1@saltmine.radix.net> <373B69FC.9AB32C8D@bnl.gov> <7hg8d6$9qg$1@news.fas.harvard.edu> <373D6205.F589F401@bnl.gov> <7hk51a$c96$1@saltmine.radix.net> NNTP-Posting-Host: app-wi5-12.ix.netcom.com Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-NETCOM-Date: Mon May 17 11:29:18 PM CDT 1999 X-Newsreader: Forte Agent 1.5/32.452 Xref: news1.radix.net rec.running:173518 bobg@Radix.Net (Robert Grumbine) wrote: .................deletia (to stop the headache already).............. > This is another of the interesting points. We do have the observation, >posted occasionally in the group (I believe Denny Anderson is the one >who found the good article on this point) that there is an optimal >stride rate/length combination for a given runner at a given speed. >(Optimal in an energetic sense). The interesting feature being that >the optimum is a rather broad one, but U shaped one. This says some ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Not broad! More later :~) >things about how the energy is being lost. What, exactly, I don't >know yet, but something. Well, you smoked me out. The math / analysis sent me for cover (sand bags, etc.) until the smoke cleared the field. Besides, all my books are in the garage and I don't feel like digging through musty boxes. Yes, I've posted many times regarding the study by Cavanaugh & Williams, as reported in a 1978 (79?) issue of Runners World. Bill Rogers was one of several subjects in the test. Short Take: Stride Length and Stride Rate are linked. The assumption must be made that the runner is either a) experienced and efficient/economical, or b) inexperienced - and efficient/economical. Hmm... looks like there's a common link. The result indicated that as running speed ( V ) increases, USUALLY the Stride Length (SL) increases at 5 to 1 (% increase) ratio compared to Stride Rate, or Steps/Minute (s/m). Long Take: The caveat is that no runner was tested at a speed slower than 8 min/mile (4.94 min/km). Therefore, aging's effect on that relationship was not investigated. My personal tests, regarding SL vs. s/m (at age 58), is that there is about a 2.5:1 to 3:1 ratio of % change when I vary speed. Tonight I slugged out three Tempo miles at my 10km pace (around 7:45; stop smirking!), and s/m was 194, 196, 196. Each measured over 60 seconds. Don't bother with SL math. Teeny steps! Miles Lakin calculated that at 185 s/m. Haille Geb...well here it is: *Here's my low down on this with a very simple calculation: * *1) Haille Gebreselassie running a 10Km with 185 turnover (actually, *I've not verified this for myself, but I think Charlie has) at 2.65 *min/Km = 6' 8" * *2) Ordinary runner 10Km with 185 turnover at 4.00 min/Km, stride *length *= 4' 5" Each individual has an optimum combined SL and s/m for a given running speed. Per Cavanaugh & Williams. That "U" shaped curve mentioned above, is very steep! Basic high school physics example, of stable equilibrium, is a bowl shaped depression into which a ball is rolled. Eventually it will come to rest at the bottom (real world). The steeper the slope, the more strongly attracted to the equilibrium position. Cavanaugh, et al's curve was also very steep. It was based on a plot of SL vs.s/m - the rusultant curve being efficiency. In general, they determined that varying the SL more than about 2" from the optimum (on a tread mill they had runners maintain const. s/m) caused efficiency to degrade. Yet also observed that one subject could deviate as much as 6" inches w/o measurable loss of efficiency. He was the tallest in the group. They didn't know what to make of it. > ...............more deletia, followed by inanity; mine.............. > Stokes' law failed because the motion is nowhere near laminar. >UL/nu = Re, which must be under about 2000 for Stokes to apply. >U = 2.67, L is in the vicinity of 1, and nu is around 1E-5. So >your Re is over 200,000. (Actually stokes is more limited than >that, but over 2000 you're turbulent, not just outside the realm >of good approximation.) Wasn't the original analysis based on a sphere? Seems a cylinder would be more applicable. In which case Re ~ 10,000 -150,00. Still well beyond Stokes laminar limits. Heh, heh. Just kidding. Found an old text in my bookcase. Took 1/2 an hour to recall the what N-S equations were about. My old head is getting more viscous every year. ........... deletia............... Denny Anderson To ERR is human... to ZIN, divine! < dennizin@pcstarnet.com > From kahana@sprintmail.com Thu May 20 08:12:55 EDT 1999 Article: 173792 of rec.running Path: news1.radix.net!tor-nx1.netcom.ca!news-out.uswest.net!newsfeed.cwix.com!198.138.0.5!newshub.northeast.verio.net!netnews.com!news-feed.fnsi.net!newsfeed1.earthlink.net!nntp.earthlink.net!posted-from-earthlink!not-for-mail From: "David E. Kahana" Newsgroups: rec.running Subject: Re: How far do you fall? Date: Wed, 19 May 1999 19:47:49 -0400 Content-Transfer-Encoding: 7bit References: <7hfagq$5p6$1@saltmine.radix.net> <37404695.1F04E96@for_d.com> X-Posted-Path-Was: not-for-mail Content-Type: text/plain; charset=us-ascii X-ELN-Date: 19 May 1999 23:47:53 GMT X-ELN-Insert-Date: Wed May 19 16:55:03 1999 Organization: EarthLink Network, Inc. Lines: 43 Mime-Version: 1.0 Reply-To: kahana@sprintmail.com NNTP-Posting-Host: sdn-ar-001nyhempp264.dialsprint.net Message-ID: <37434DA5.8A594115@bnl.gov> X-Mailer: Mozilla 4.04 [en] (X11; I; Linux 2.2.3 i686) Xref: news1.radix.net rec.running:173792 Y-Rotation wrote: > It depends on the amount of leg and arm lift and on how heavy both are > in relation to the rest of the body. > The CG location changes because the spacial distribution of mass is > affected in temporal fashion by the position of runner's legs/arms. > As the leg goes up and down so does the CG. > > Efficient runner's torso does not move up and down, though the location > of his CG does. > > pete > Yes, this is definitely part of the story, maybe most of it, but I don't think it can be all of it. The trouble is, both legs are off the ground for a finite time. During that time you are in free fall. So the whole body is accelerating downwards then at least, and you can bet that the torso is as well. What could stop it from doing so? The only thing I could imagine is some kind of aerofoil effect producing enough lift to keep the torso up, untill the leading foot touches down again. I can't believe that is the case, though I've often wished I had winged feet like Hermes :) Of course, your description would be practically correct, if the time in the air is reduced to a very small fraction of the step time. So that is all we need to know to check this. How long are both feet off of the ground? That alone will tell us the extent of the CM vertical motion during free fall. In any case, we can still define this as mgh work, even though the real motion could be much closer to what you describe. That would mean even more work is done really, in just repositioning the limbs. But your point is well taken, and I can't help but be reminded of the high jump, in which CM passes *below* the bar, for the most efficient jumpers. cheers, - dave k.