25 replies to this topic

[PhysicsGuy]

I’ve been interested in finding out how much power in watts a runner can develop.  

There are actually two cases that I was interested in; one was for sprinting and the other was for running at a constant speed for an extended period of time, such as, for a 5K.

The second one is a bit complicated, but the case of the sprinter is pretty easy if you look at the kinetic energy after they leave the blocks.

The kinetic energy is half the mass times the velocity squared.  It’s measured in joules.  If you divide that by the time, it gives you the power in joules per second or watts.

Usain Bolt is a good example.  He weighs 198 lbs or 90 kg.  He runs the 100 m in about 10 sec, so his average speed is about 10 m/s.

If he starts from a dead stop and accelerates to his average speed in about 3 seconds, then his kinetic energy at 3 seconds into the sprint is 0.5 x 90 x 10^2 or 4500 joules.  If he generates that much energy in 3 seconds, he is generating about 1500 watts.  That’s the same as 2 horsepower.

You can also calculate how much power a stair-runner generates when they run up a tall building.  Then the energy generated would be the potential energy of the runner at the top of the building.  (The potential energy at the bottom is zero.)  The potential energy is the runner’s weight multiplied by gravity and by the height of the climb.

The Taipei 101 is a race to the top of a tall building.  The race is 91 floors or 390 meters.  If a 150 lb or 68 kg runner reaches the top of that building, they have generated 68 x 9.8 x 390 or 260,000 joules.  If they do that in 16 mins or 960 seconds, then they are generating about 270 watts of power.  That’s about a third of a horsepower.

Some cyclists train with hubs on their bikes that measure their power output.  A cyclist may actually train at the 270 watt level, so about 250 to 300 watts is pretty reasonable for a stair racer.

I think that’s interesting,

Ted

http://members.aol.c...odel/SM/SM.html

[tscm]

Measuring power output usually involves force plates in a sport science setting, which give readings for velocity and force in both thehorizontal and vertical directions, these are termed Ground Reaction Forces (GRF) and they act vs the musculature when producing force against the ground. Otherwise power output is expressed as joint power, this study http://www.sportmeda...ower Sprint.pdf , for instance, shows a peak power of 3070 Watts at the hip for two elite sprinters (sub 10.26). For the non-elite sprinters in the same study (sub 10.70), peak hip power was 2134 Watts so there's definitely a difference there, and undoubtedly a huge difference compared with the likes of Usain Bolt. If you're interested in this stuff maybe try http://www.elitetrack.com/articles/ especially the "sport science" section, http://www.elitetrac...dingofpower.pdf might also interest you.

[OscarZarate]

tedjan, on Aug 30 2008, 02:46 PM, said:

I’ve been interested in finding out how much power in watts a runner can develop.  

There are actually two cases that I was interested in; one was for sprinting and the other was for running at a constant speed for an extended period of time, such as, for a 5K.

The second one is a bit complicated, but the case of the sprinter is pretty easy if you look at the kinetic energy after they leave the blocks.

The kinetic energy is half the mass times the velocity squared.  It’s measured in joules.  If you divide that by the time, it gives you the power in joules per second or watts.

Usain Bolt is a good example.  He weighs 198 lbs or 90 kg.  He runs the 100 m in about 10 sec, so his average speed is about 10 m/s.

If he starts from a dead stop and accelerates to his average speed in about 3 seconds, then his kinetic energy at 3 seconds into the sprint is 0.5 x 90 x 10^2 or 4500 joules.  If he generates that much energy in 3 seconds, he is generating about 1500 watts.  That’s the same as 2 horsepower.

You can also calculate how much power a stair-runner generates when they run up a tall building.  Then the energy generated would be the potential energy of the runner at the top of the building.  (The potential energy at the bottom is zero.)  The potential energy is the runner’s weight multiplied by gravity and by the height of the climb.

The Taipei 101 is a race to the top of a tall building.  The race is 91 floors or 390 meters.  If a 150 lb or 68 kg runner reaches the top of that building, they have generated 68 x 9.8 x 390 or 260,000 joules.  If they do that in 16 mins or 960 seconds, then they are generating about 270 watts of power.  That’s about a third of a horsepower.

Some cyclists train with hubs on their bikes that measure their power output.  A cyclist may actually train at the 270 watt level, so about 250 to 300 watts is pretty reasonable for a stair racer.

I think that’s interesting,

Ted

http://members.aol.c...odel/SM/SM.html

Indeed it is!

WOW. Is there more to read about this? I was thinking that it would be easier to calculate the power an Marathoner could develop ... but after what you said ...  

Cheers,

[PhysicsGuy]

Hi Tscm,

Those are very interesting links.  They are calculating the instantaneous or peak joint power.  As you pointed out, that analysis is normally based on force plate and video data.  Values in excess of 3500 watts are common.

I was interested in the continuous power over longer periods of time.

From the spring mass model it looks like Usain Bolt is expending an average power of about 800 to 900 watts in mechanical leg-spring compression.  I am sure that would include the much higher instantaneous wattage segments that you pointed out.

I was interested in calculating how much continuous mechanical power is used to move accelerate the sprinter's limbs.  If Mr. Bolt roughly generates 1500 watts and the model can "explain" about 800 watts, then I wonder if he expends about 400 watts driving his legs.

I believe that the calculation can be done my modeling the swing leg as a two-segment pendulum driven by hip torque.  The folded swing legs have a resonant step rate in the area of 100 steps per minute.  A sprinter normally runs with a step rate in the area of 200 steps per minute.  It takes a lot of power to swing the legs at the higher step-rate.  That's what I am interested in calculating.

Hi Oscar,

There is a truly interesting website on the power expenditure of different sports at:

http://members.aol.c...ews/energy.html

Regards,

Ted

[Gasher]

Ted,

Enjoyed reading the article you put a link to in your first post. Most of it went right over my head but I thought of great interest was the section on Efficiency. There has been much discussion on this board about the 'correct' cadence/stride length when running so great to see the results of tests done in this area......

For those that can't be bothered to read it, stick with your PSF (preferred stride frequency) which will naturally increase with speed.

[PhysicsGuy]

Gasher, on Aug 30 2008, 06:51 PM, said:

Ted,

For those that can't be bothered to read it, stick with your PSF (preferred stride frequency) which will naturally increase with speed.

Hi Gasher,

You are totally correct about the PSF.  That is the step rate or frequency to use to minimize you O2 or power consumption at a particular speed.

I did make one interesting finding recently.  It relates to the "natural" swing rate of a runner's leg when it is almost straight compared to when the knee is bent and the leg is folded up under the butt.  The folded leg swings faster than the straight leg.  I think some people refer to this as the "heel kick", but I am not sure what that means.

A runner increases their speed by swinging their legs faster and taking longer steps.  Increased speed seems to come from increasing both step rate (leg swing rate) and step length.

It turns out that it is requires less power to swing a leg faster if its “natural” swing rate is already fast.  To some extent that would explain why sprinters seem to have a higher heel kick than slower runners.

Some people recommend that a runner train by using a higher step rate than their PSF.  I'm not sure that helps a runner increase their "continuous output speed", but I can see how using a higher heel kick or lift would require less power expenditure than letting the shank/foot move on it's own.  This is just an interesting thought.  I don't know if the saved power output would go into increased speed.  It could go into something silly like head bobbing

By the way you can actually see this "natural" swing rate change if you measure your leg swing rate while it is straight and then when your lower leg is folded up next to the thigh.  I recommend using a belt to keep the lower leg folded-up next to the thigh.  The folded leg swings faster than the straight leg.

It's an interesting experiment.

Ted

[SoLucky]

Some time ago I applied a text book definition of Horse Power (550 foot pounds per sec) to a couple of my regular hill training venues. With my body weight at ~145lbs, and making no allowance for distance travelled, the results were:

(1) Run on a 2.55k sealed road with elevation gain ~200 metres.
Perceived effort say 85%, time ~16 min, apparent HP required ~0.18.
(2) Strong walk on a 1.0k track with intermittent steps, elevation gain ~200metres.
Perceived effort say 95%, time ~12 min, apparent HP required ~0.24.

I realise that energy actually expended would be significantly greater but it appears that the calculation of required HP forms a reasonable indicator of activity intensity.
The comparison with cycling seems valid. On the sealed road mentioned above, I am able to stay with or pass weaker cyclists, but can only hang on to serious club riders for a short distance. Better runners can stay with the competitive cyclists for much longer. The steeper the hill, the easier it is for the man on foot to pace the man on wheels.

[PhysicsGuy]

Hi SoLucky,

Those were pretty good efforts.  It's actually very difficult to find situations where a runner can measure his or her power output.

Basically it requires a situation where the runner has a known potential (height) energy and kinetic (speed) energy at the beginning and end of measured time period.  

The first case is very good for measuring the power put into increasing your potential energy over a 16 min period.  If you weigh 145 lbs (4.5 slugs) and climb 200 m (650 ft) in 16 min (960 sec), the change in potential energy would be: m x g x h = 4.5 x 32 x 650 = 93,600 ft-lb of potential energy.  That would be 98 ft-lb/sec or about a fifth of a horsepower as you pointed out.

If you add in the change in kinetic energy you would need the terminal velocity; 2.55 km/16 min = 0.16 km/min = 9 ft/sec.  The kinetic energy at the finish would be 0.5 x m x velocity^2 = 0.5 x 4.5 x 9^2 = 183 ft-lb of kinetic energy.  To get the power expended in generating this kinetic energy, the kinetic energy would have to be divided by the time.  Unfortunately, that would be 182 ft-lb/960 sec = 0.2 ft-lb/sec.  That's tiny compared to the power put into potential energy of 98 ft-lb/sec.  I'm sure you can see why this type of calculation may be lacking.  It's just missing all the "sweat" power put into the 16 min run up the 2.25 km slope.

I just thought of a neat method that you could easily use to measure your maximum horsepower.  You could run up a not-too-long flight of stairs and measure the time it takes.  I think that would work.

If you assume that all the energy goes into increasing your potential energy and measure the time, you could get a pretty good estimate of you maximum power output.

In the metric system gravity is 9.8 m/s^2; you would have to use your mass in kilograms.  In the foot-lb system gravity is 32 ft/s^2, you have to convert your mass to slugs by dividing your weight in pounds by 32, so 145 lbs is 4.5 slugs.

In either case the change in potential energy would be m x gravity x (vertical height of stairway).  Of course the power would be the potential energy change divided by the time.

It might be a good idea to do this on a stairway of about 15 to 30 steps.

If you or anyone else does this measurement, would you or they please post the results here.  It would be interesting to see if the power output works out to be about 300 watts or half a horsepower.

Ted

[SoLucky]

tedjan, on Sep 1 2008, 01:01 AM, said:

If you or anyone else does this measurement, would you or they please post the results here.  It would be interesting to see if the power output works out to be about 300 watts or half a horsepower.

Ted

Ted,
I had a couple of trial runs this am up a stairwell of my 5 level apartment block, there are 10 flights of stairs, each with seven steps averaging 7.125 inches. Total vertical lift of 41.5 ft. The time on my watch was a steady 6 secs per floor, 30 secs in total. On my calculation, with a body weight of 145 lbs, that yields a result of ~0.36HP.
I had a second attempt using one arm on the handrail to help me to swing around from one flight to another and that time stopped my watch at 25 secs, an apparent requirement of 0.43 HP.
I think that on stairs a strong runner (not me) would readily exceed the one half horse power level in a 30 second episode. It wouldn’t surprise me to see calculated results in the 65 to 75% range.
It will be interesting to see the figures from a range of different runners.
I’ll try and locate a local stairway which doesn’t force frequent changes of direction.

[PhysicsGuy]

Wow!  That is so very interesting.  There has been a good deal of research on this in the area of man-powered flight.  One method of quantifying this information is to calculate the power output in watts divided by the mass of the subject.  A group at MIT offered 3.5 watts/kg as a typical value for continuous output.

Of course this is more a maximum power output, not a continuous power output measurement.  In any event, the power output value of 0.43 HP over 145 lbs would be 320 watts over 66.8 kg or 4.8 watts/kg.

When I use the spring mass model on myself I get values in the area of 3.5 watts/kg.  When I run it on Usain Bolt, I get ~8.5 watts/kg.

I often wonder if there is a method that the non-elite runner could use to measure his or her fitness and compare it with that of others.  The only problem is that this is not running.  It may be a good method for measuring fitness, but it not the same as running.  We should figure what it is measuring.

Setting standards for the total stair height or climbing method (single or double) might refine this tool.  Maybe it would be useful to start another thread with a more appropriate title so we could pool data using a standardized method.

If you were setting standards for such a measurement, do you think that single, double or triple stair stepping should be used?  

Do you think there could be a standard range for the height or time length of the measurement?  Not everyone has access to a six-floor stairway.  

If the time measurement is too short, the measurement uncertainty in the start and stop times will corrupt the total time measurement.  I would guess that the uncertainty in the start and stop times is about +/-0.5 sec.  If the total time was 6 seconds, that would introduce a total measurement uncertainty of +/-1 sec or 1/6 or 17%.  That seems to be too much.  

Maybe the total time should be about 20 seconds.  Then the measurement uncertainty should be 1/20 or 5%.  That would imply that the standard should be about three flights.  Unfortunately, that may mean that the runner will have to use a multi-level stairway.  As you pointed out, it takes additional time to turn at each level.

Perhaps the recommended stairway should be continuous and about 20-seconds in length.  That might be a typical stairway at a school stadium.  Unfortunately, the stadiums, that I have seen, have longer "step runs" than the standard stair step.  Maybe the stair should have a normal run and rise.  It might be possible to find continuous 20-second standard stairways at a shopping center or mall.  If it takes about 20 seconds to climb a stairway in a mall, then the only issue to resolve is the issue of single, double or triple step climbing.  What do you think?

Ted

[SoLucky]

Ted,
You’re right, stair climbing tests can only measure stair climbing ability. I doubt that they would show a very reliable relationship to performance in any running event, from the 100m to 42.2k.
However, it is an effective adjunct to training, and for an individual, it does give a quantifiable measure of useful leg strength and power output.
While it would seem that testing procedure should specify standard step height and stepping span (1,2 or 3 steps at a time), runners themselves don’t come in standard weights, heights or leg lengths. They should be able to use the climbing pattern that best suits their characteristics and current ability.
A factor that really would muddy the water would be using your arms to assist in propelling yourself upwards.
In relation to timing, I agree anything less than at least 20 secs may suffer disproportionately from self timing errors.

[SoLucky]

Ted
Further on standardising step span or height for comparative testing, a 5000m time trial does not specify stride length or frequency. It is purely a matter of getting from point A to point B on a level surface. Similarly with a stair climbing test, it is simply a matter of elevating yourself from X to Y using only your legs as propulsion.
In the context of stair climbing, an abbreviated calculation to determine Horse Power from metric inputs could read as follows: (elevation gain in metres)/(time in seconds)*(weight in kg)/76. Run some figures and advise me if I am wrong.
It does lead to a happy coincidence for somebody weighing about 76 kg, their calculated required HP is simply (elevation gained in metres)/(time in seconds).

[PhysicsGuy]

SoLucky,

This is a "go".  I'm working on some text for a posting here and at few other forums.  I'll pool the data and update a file of the results at a website.

Some runners will have trouble with the arithmetic, so I'll include the word equations for calculating the power in three units: watts/kg, total watts and total horsepower.  That part of the instructions would look like this:

If you measure the step heights in inches, your weight in pounds, use:

Watts/Kg = 2.33 x Step_Count x Step_height(in.) / time(sec)

Watts = 0.113 x Step_Count x Step_height(in.) x Weight(lbs.) / time(sec)

HP = 0.0001515 x Step_Count x Step_height(in.) x Weight(lbs.) / time(sec)

If you measure step heights in centimeters, your weight in kilograms, use:

Watts/Kg = 0.01 x Step_Count x Step_height(cm) / time(sec)

Watts = 0.01 x Step_Count x Step_height(cm) x Weight(kg) / time(sec)

HP = 0.0001315 Step_Count x Step_height(cm) x Weight(kg) / time(sec)

We will only be analyzing the watts/kg data.  This will allow everyone to compare his or her maximum power output with that of runners of different weights.

In addition you can compare your power output in watts with those shown half-way down the page at:

http://members.aol.c...ews/energy.html

We will graph the watts/kg for all the runners every few days at:

http://members.aol.c.../PowerStudy.gif

To do this you should post three pieces of data below:

1. Your maximum power output in watts/kg from the stair climb measurement.  Remember no hands except to help you turn on each level.
2. Your expected, present day, finish time for a 5K if you where to run a 5 K right now.
3. You're approximate weekly mileage.


SoLucky, regarding your equation, I think they might get confused with the order-of-operations in:

(elevation gain in metres)/(time in seconds)*(weight in kg)/76

It might be put as:

(elevation gain in metres)*(weight in kg)/((time in seconds)/76)

But, they will get confused in the denominator.  What about moving the 76 up to the numerator as 0.01315 and putting it first as in:

HP = 0.01315 * (elevation gain in metres)*(weight in kg)/(time in seconds)

Do you think that people would make errors in computing the total height?  

I thought it might be better to use the number of steps * the step height in inches or centimeters?  Some make errors very easily, get frustrated and give up when the results don't look right.

SoLucky, there is an excellent free utility for converting units at:

http://joshmadison.c...ert-for-windows

I have been using it for years.  Plus, you can add conversions as you see fit.

There is a great story on the incidence of diseases in runners vs. non-runners at:

http://news.bbc.co.u...lth/7554293.stm

The right side-bar has some good info, also.

Ted

[SoLucky]

Quote

Do you think that people would make errors in computing the total height?
I thought it might be better to use the number of steps * the step height in inches or centimeters? Some make errors very easily, get frustrated and give up when the results don't look right.

The easier and more convenient that you can make it, the greater the sample size that you will collect.
Are you also going to capture age and gender?

[HillsAths1]

The reason I go running is to escape all the annoying numbers I deal with every day at work, now its starting to haunt me on My Favorite Website Cool Running, will there be no end to all this?

[Martin D]

tedjan, on Sep 1 2008, 01:01 AM, said:

The first case is very good for measuring the power put into increasing your potential energy over a 16 min period.  If you weigh 145 lbs (4.5 slugs) and climb 200 m (650 ft) in 16 min (960 sec), the change in potential energy would be: m x g x h = 4.5 x 32 x 650 = 93,600 ft-lb of potential energy.  That would be 98 ft-lb/sec or about a fifth of a horsepower as you pointed out.

Good heavens! I've never seen anyone actually use slugs in a calculation before. You're not a NASA contractor are you, http://www.cnn.com/T...mars.metric.02/  .

I'm old enough to have grown up with yards and acres, and even got exposed to chains and hundredweight, but never slugs. How many BTU would that be (or would it be BTU/hour)?

Martin

[PhysicsGuy]

SoLucky,

I’m working on the text and the layout.  

Gender makes sense.  If I get enough data, I’ll be able to split the data by gender.  I think that some people will be reluctant to give their age because of their concern with anonymity.

Ted

[PhysicsGuy]

Martin,

Yes, I worked for Honeywell, Guidance and Navigation.  Our MIMU was on that probe.

http://en.wikipedia....ent_Unit_(MIMU)

It was an unfortunate error.  The values for angular momentum in the MKS and FPS system are pretty close, so it wasn’t obvious that the numbers were out of range.  The article pointed out that congress never wanted to pay for the full metric conversion.  It’s expensive and complicated when you have to convert years of developed hardware, standards, software, documentation, test equipment and procedures from one system to the other.  Every change has to be tested to death and triple checked to make sure the product still produces measurements of the same accuracy.

Slugs, of course.  A slug is the same as 2 stone 4.

Ted

[Martin D]

OK, I'm a nerd, I find this stuff fascinating.

Ted, I think your estimate for Bolt might be reasonably close. Science of Sport has 10 m splits for Bolt,
http://www.sportssci...e-analysis.html. Peak speed came after about 6 seconds, but most of the acceleration was in the first 3.

Top level track sprint cyclists can produce about 1800 W, e.g., http://www.springerl...3q42x5533754r0/. These were 6 second tests. I remember seeing a graph of power output over a flying 200m in a magazine years ago. Power peaked just before the start and
pretty much decreased over the whole distance until they were almost coasting at the finish.

Here's some other miscellaneous estimates

The Empire State Building run up is a 320 m climb and Paul Crake's record is 9:33. Assuming he's 60 kg gives 330 W. It'd be interesting to know exactly how much time is spent climbing compared to getting round the corners of the landings.

The power estimates at http://members.aol.c...ews/energy.html seem to be based on energy usage, rather than useful mechanical work delivered. They're going to be a lot higher than estimates from climbing because they also include the heat generated in the muscles. If I remember correctly, net efficiency is only something like 20%.

The results at http://www.sportmeda...ower Sprint.pdf seem to be peak power rather than an average over the stride cycle, so not surprisingly are also much higher than your Bolt estimate.

Stage 16 of this years Giro D'Italia http://www.cyclingne...stages/giro0816 was a 13 km time trial, climbing 1086 m (with a gravel road at the top for extra fun!). The winner did it in 40:26. Assuming 70 kg for rider and bike gives 310 W. There should only be a small extra cost for wind resistance at this speed.

The Mt Washington Road Race in the US climbs 4650 feet (1420 m) in 7.6 miles. The running record is 56:41 by Jonathan Wyatt (multiple world mountain running champion). He's apparently 63 kg which gives 260 W. The record for the bike race on the same course is only slightly faster, 49:24.

Cyclists have been estimated to produce over 400 W in the world hour record, http://www.pponline..../encyc/0801.htm

[Colin]

tedjan, on Aug 30 2008, 02:46 PM, said:

Usain Bolt is a good example.  He weighs 198 lbs or 90 kg.  He runs the 100 m in about 10 sec, so his average speed is about 10 m/s.

If he starts from a dead stop and accelerates to his average speed in about 3 seconds, then his kinetic energy at 3 seconds into the sprint is 0.5 x 90 x 10^2 or 4500 joules.  If he generates that much energy in 3 seconds, he is generating about 1500 watts.  That’s the same as 2 horsepower.

The actual power used by Bolt is much of a muchness... but your calculations are based on flawed assumptions.  

Not going to lose much sleep over it, so not looking for a 'correct' answer, but.....

Martin D, on Sep 3 2008, 08:38 PM, said:

Ted, I think your estimate for Bolt might be reasonably close. Science of Sport has 10 m splits for Bolt,
http://www.sportssci...e-analysis.html. Peak speed came after about 6 seconds, but most of the acceleration was in the first 3.

...if we used the figures quoted by Science of Sport, then we could get many different answers.

eg...after 1.85sec he covered 10m and had a terminal velocity of between 19km/h and 35km/h (averages for first and second 10m intervals). Take 27km/h then his power is calculated by your formula at 1370 watts.
At 20m it would be 1660W, 30m 1470W, 40m 1310W, 50m 1170W and 60m (top speed) 1060W.

In other words "how long is a piece of string?"  We don't know the correct answer by this method because other forces are acting against his acceleration.


As for the link and comment on 'preferred stride frequency' , thanks for that (and Gasher for pointing out) ...exactly what I have been cautioning against where people were trying to emulate a 'magic cadence' of 180 (or any other number).
The most effecient cadence and stride length is different at different speeds, and both will (need to ) increase at higher speeds.

cheers

[PhysicsGuy]

Hi Martin D,

Thanks for the link to the Sports Scientists article on Mr. Bolt.  They made a common error in the velocity determination.  I posted a note and pointed them to a good reference.

I subscribe to the BIOMCH-L and the Yahoo SportsScience lists.  I didn’t know about Sports Scientists.  That’s a good site.  If you know of any other analytical sports sites, post their info here.

You are correct about the numbers on the energy site.  By the way you can convert ml of O2/(kg-min) to watts/kg by dividing by 3.  So, if a runner consumes 60 ml of O2/(kg-min), they are metabolically consuming about 20 watts/kg.

Dr. Ed Coyle measured the Lance Armstrong’s efficiency at about 22% (See Fig 1).

http://jap.physiolog...print/98/6/2191

Thanks for your input,

Ted

[Colin]

tedjan, on Sep 4 2008, 11:26 AM, said:

Thanks for the link to the Sports Scientists article on Mr. Bolt.  They made a common error in the velocity determination.  I posted a note and pointed them to a good reference.

They didn't... they said this:

Quote

we have been sent the split times for each 10m interval from Bolt's race , which we've used to calculate average speed for each 10m interval.

As another poster already replied to you.

I therefore just estimated the instantaneous speeds as midway between average speeds...not accurate but suffice for explanation of the fallacy of using the kinetic energy at a particular point, to determine the power to that point.

Likewise, VO2 etc are inaccurate estimators when we don't know all or include all the counter acting forces , energy sources used etc etc .

As I said "much of a muchness'.

cheers

[PhysicsGuy]

There was posting on the BIOMCH-L today that cited a draft paper on Usain Bolt.  

http://arxiv.org/PS_...0809.0209v2.pdf

After the abstract you might look at the (6.) Conclusions on page 4.  I think you may find other sections of interest.

I like the use of a single spline in the section (4.) on Motion Profiles.

Ted

[SoLucky]

Ted,
thanks for posting the BIOMCH-L link. I am intrigued by the amount of information that that those guys have distilled primarily from image analysis. And some cunning maths to establish instantaneous speeds.

[PhysicsGuy]

SoLucky,

Thank you.

As an aside, if you ever need a best-fitting polynomial through a set of data points, you can let Excel do all the work.  Just graph the data, select the set of points and then use Chart/Add Trendline.  Under the “Type” tab check the polynomial box and select the order (2 to 6).  Under the Options tab, check the “Display equation on chart” box.  Its a great way to get a “best fit” with very little work.

Ted

[SoLucky]

Good one.