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Running in Circles, a Physics problem?

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MJuric

Mechanical
Oct 28, 2003
10
US
A while back I got some info here about calcualting centrifugal force. Taken from a discussion that took place in march. Since then I've been messing around with a problem that is making me nuts.

I'm trying, so far unsuccessfully to figure out a little problem. I'm having a debate with another fellow about the benefits of a running product pictured and explained here.

The disagreement is based on the idea that the runner in the track will have to somehow accelerate faster (move the legs faster or more explosively) than a similar runner on flat land wearing extra mass.

My argument, albeit probably flawed is this. F=M*A, it shouldn't matter whether or not the runner has to overcome twice the acceleration or twice the mass. His response was the track is similar to someone bench-pressing the mass at twice the acceleration. In other words the bench presser would have to move the arms faster in order to
accelerate the force to get the 2x acceleration vs the same person benching two times the mass. My example was to take the same person bench-pressing on one planet with 2x the mass and then the same person on another planet with 1x the mass and 2x the gravitational force. In this scenario both situations would have the same force with the same
speed of arm movement etc.

Intuitively, although my intuition has been off with this type of stuff many times, I suspect that the situation is more similar to my scenario than his, however he seems to have a much stronger grasp than I do on the physics of the situation. My reasoning is the acceleration
in the circular track is provided via the velocity of the runner running around the inside of the track. The centrifugal force created,is counteracted by the force created by the legs to next to the next
step. I also suspect that the path/trajectory taken by the runner is similar to that of a runner on a flat plane, difference being the main mass of the runner will always be perpendicular to the centripetal force rather than the gravitational force.

I've been messing with this thing so long now it's turning into a mud puddle so any help, links, info, comments, disparaging remarks are welcome.

~Matt
 
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I suspect anyone running on such a device would by crippled for several weeks. As you left the vertical plane, the angular forces on your legs/ankles would be devastating. Imagine if you will running within a cylinder, similar to the mouse wheel. If you maintained speed to be stationary at the level plane you would simulate a flate track. As you positioned yourself higher up the incline, you would simulate running up an ever increasing grade. A very good workout. You could slow or speed up your pace to work, recover, or steady state. Now imagine running on the side of a hill. You would approach it in a semi arc to maintain some upright position, but the arc would also be designed to bring you off the slope quickly. Try it, you'll feel the strain endured.
 
One big problem with the dish is balancing the forces. Your body mass will be accelerated straight down by gravity and towards the center by centripital acceleration. The end result is that your feet and joints have to resist some really odd vectors. The acceleration being countered is neither normal to gravity nor normal to the centripital force, it is an intermediate vector.

You would have to be running pretty fast for the centripital acceleration to be high enough to double the total acceleration (therefore the force for a given mass) since the centripital acceleration is normal to the circle and the net acceleration vector is somewhere in between. To give you a total acceleration of twice gravity you would need centripital acceleration of 17 m/s^2. If you assume the radius of the dish is 2 m and the runner masses 100kg, then the runner must be traveling 6 m/s (21 km/hr) to double the force.

The bench press example doesn't hold water. There is no velocity term in f=ma, so if you bench press 600 kg on the moon you would exert the same force (and do the same work) by bench pressing 100 kg on earth regardless of the pace of work.

David Simpson, PE
MuleShoe Engineering
 
As far as being crippling yes, both of us involved in the arguement have discussed the potential for, mostly knee issues to arise. Not good for long distance applications but not crippling. The purpose of this application is supposedly for sprint distance runners anyway.

If work is equal to F*D or M * r The only difference, Unless of course I'm missing something again, between running in the track and on flat land is that the runner in the track is doing slightly less work because the moment is following a smaller arc distance than the distance the force is being applied for the runner on flat land at similar speeds. I believe, though can't prove, that both runners have a similar rise and fall in there gait and both woudl be traveling the same distance at the same velocity. So I guess I'm not sure how work and energy woudl help me out here, unless I'm misunderstanding.

~Matt
 
Correct me if I'm wrong here. It's been awhile since I've had to mess with this stuff.

F=M*A
Bench press 2kg(Mass) on earth
F=2kg*9.8M/s^2
F=19.6Nm

Bench 1kg on Planet Zolton were G=19.6M/s^2
F=19.6Nm

As far as calculating forces induced by the track I understand that the centripal force will be in addition to some factor of gravity. This will be dependant on the speed of the runner. However in the interest of ease of calculation, and my own personal sanity, I've been working with the calculations as if gravity didn't exist and all forces were centripal and inline with the runner. Granted these will not be real world results but my goal here is to try and figure out if the effects of running in this thing are significantly different from running and flat land with weights on.

~Matt
 
MJuric
I didn't really understand your doubts and position.
Anyway it is clear to me that, when the athlet runs at a radius in the track and at a constant speed such that:
1) the vector composition of its weight and of the centrifugal force is along his body vertical axis,
and
2) the radius where he runs is such that the local slope of the track is normal to his body axis,
then he will feel (and spend work) almost the same as if he was running on a planet with an appropriate increased gravity (or if he was wearing weights).
I said almost because of course also the distribution of weights over the body has importance (e.g. the outer arm is subject to a higher centrifugal force than the inner one, also the feet with respect to the head, etc.), but this is a second order effect.
A way of reasoning that could help in figuring this out is the following:
1) take a person standing on the track at a given radial position
2) now make the track rotate about its axis: at a given speed the person will feel comfortably standing as the composition of centrifugal force with weigth will pass through its feet: this is exactly the same we feel when standing in a train making a curve on an inclined track, our legs have to carry an increased weight
3) now suppose the person starts to move circumferentially and that the track rotation slows down such that the revolutions per minute of the person remain constant: if we discard the effect of moving the legs and jumping in the run (suppose that person has air cushions or roller skate under its shoes and a small rocket on the back), nothing changes from the point of view of the runner.

prex

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>MJuric
>I didn't really understand your doubts and position.
>Anyway it is clear to me that,

Understand and agree with everything to here. As does the person I'm arguing with does, I believe.

>3) now suppose the person starts to move circumferentially >and that the track rotation slows down such that the >revolutions per minute of the person remain constant: if we >discard the effect of moving the legs and jumping in the run >(suppose that person has air cushions or roller skate under >its shoes and a small rocket on the back), nothing changes >from the point of view of the runner.

As you stated above the forces or weights on the runner are the same when the runner is not in motion. My argument lies in the problem of whats going on during the runners motion. So far I believe I have convinced my self that that the only motion that really matters is that of the runner moving up and down. (Counter acting gravitational forces on the flat run and centripal forces in the track) The question is under the following criteria:

1) The runner in the track is 1Mass
2) The track is spinning at a speed to able to create a for of 2A or 2G
3) The runner on the flat ground is 2Mass
4) The runner on the flat ground is under 1A or 1G
5) Both runners are jumping the same height in the same time

Does...

1) The runner in the track have to exert more force in order to than the person on flat land? Since F=MA Then no.
2) Does the runner in the track have to do more work than the person on the flat to jump. Since W=FD then no.
3) Does the runner in the track have to create more power than the person on flat. Since P=w/t then no.
4) Does the runner in the track have to move differently than the runner on the flat in order to jump. Since both runners are creating similar power, than muscle recruitment patterns can be the same.

Now if any of the above are incorrect please let me know. And on to the next problem.

Assuming all of the criteria above plus:

1) Each runner is moving at the same velocity, thus covering the same distance.
2) The runner in the track is now running in a stationary track at a velocity great enough to create 2A or 2G
3)the runners are still going thru the same "rise" and "fall" in their running gait that was taking place in the jumping example in the first problem.

Does...

1) The runner in the track have to exert more force in order to than the person on flat land? Since F=MA Then no.
2) Does the runner in the track have to do more work than the person on the flat to jump. Here I'm not sure. If the runner is running in a curved surface then does the runner have to travel further verticaly in order to get to the next step. Almost like climbing a hill. Or is this "extra work" what is causing the force of acceleration? Clear as mud? IOW If one foot of the runner is at the quadrant of the circle and the runner takes one step, the runner will, in relationship to the starting point, have gone forward and up some distance. These two vectors make and angle. Does that angle somehow correlate to the acceleration/gravity being put forth on the runner?
3) Does the runner in the track have to create more power than the person on flat. The answer would be no if teh answer for (2) is no.
4) Does the runner in the track have to move differently than the runner on the flat in order to run. No if 2 & 3 are no.

Hope this explains my dilema a bit more.

~Matt

>prex

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Dear friends,
i am Giuseppe Scuderi the italian engineer who has invented the centrifugal training track and the hypergravity training track.
I have found these messagges.
I don't know english well i can mail to you the formules on the Hypergravity training track in Italian.
I think it's not difficult to understand it.
I hope that someone can help me to begin new researchs in the world.
It's a new and undiscovery world of research.
In Italy Two university are studing on my idea: University of Rome IUSM and University of mechanical enginering of Catania.
Finally the formules are in internet in this page:
All the people can begin new and interesting study.
ing. Giuseppe Scuderi
 
All your points seem fairly correct.
However you are mixing up physics laws describing the dynamics of a material point (that are simple, sharp and exact) with much less exact concepts like jumping, running, type of movement, work done by a running person, and so on (less exact only because their exact description would require stating an enormous number of parameters, due to the kinematic complexity of the human body).
The only thing that the physics of the material point tells you is that a person moving on a circular track (be it running on a track at rest or standing on a spinning track) will feel an increased gravity applied to its body's center of gravity, and, if the conditions 1 and 2 at the beginning of my preceding post are satisfied, he'll feel and behave (almost exactly) as if he was on a planet with a mass higher than earth mass.
Your example of a runner with double the mass of the reference one is less convincing, as you could be asked to specify whether the additional mass is muscle mass, fat mass or weights he wears on his shoulders. However if you take only the concept and don't bother about such details, then your example of the runner with doubled mass is equivalent to the one of the reference runner on a planet with double the earth's mass.

prex

Online tools for structural design
 
I agree that appling hard mechanical physics to "Body dynamics" is.... less than perfect. However If I can show that the mechanics are similar than I can argue the body dynamics later. In most cases these body dynamics are so unique and variable that true answers generally only come thru testing and research.
I'm still somewhat confused about the acceleration end of the formula. Taking the spinning person in the track. If the force of 2units is created by 2A then if this person jumps in the track don't they have to have 2x the velocity upon leaving the track in order to reach the same heigth as the person on flat land with 2x the mass?
Since that would be true then the legs would have to contract quicker in order to get the body to that velocity. True? or is that incorrect? Mass plays no part in teh trajectory or heigth does it? Only acceleration and velocity.

~Matt
 
If you want both runners do the same work in jumping (hence they must reach the same height), then your statement is incorrect as the initial velocity will be such that the kinetic energy is the same. This means indeed, however, that the starting vertical velocities will be different, as you stated.
But once again you are mixing simple point dynamics (for which the above is true) with dynamics of a running man, and there more complicated behaviours need be considered.
As an example, a good runner has his center of gravity moving substantially at constant height, so that his work (or effort) is not so much related to height changes.
I still can't see your point and think that the only general conclusion you can draw is that the spinning runner, for a suitable pair of radius and velocity, will feel and behave substantially (not exactly but almost) as if he was wearing weights (and this seems to be the main goal of that equipment).
All the rest should be left to the technical judgement of the professional trainer that will follow runner's training.

prex

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I am the inventor of hypergravity training track.
To day i have insert my first research, tomorrow i shall insert in internet the first research of University of Mechanical Engineering of Catania, It was pubblished in Biomedical-Materials Engineering Tokio 99.
There are ten pages written in english.
The research is very interesting to understand the great muscle power that is possible to got with the hypergravity training track.
University of Roma in the third research has calcolated it is 42% more than the muscle power in 100 metres dash.
Excuse my english.
I hope that you understand me.
I hope in all the world the people can begin new researchs on hypergraviy training ( max human power in the run ).
The italian champion 2002 and 2003 on 400 metres under 18 has used this new training method.
Actually also the italian recordman 400 indoor use this method.
Thanks
Scuderi Giuseppe Italian engineer


 
On the track, the runner will move up the track to a steeper section as their speed inreases. As the bank of the track increases the normal force the track applies to the runner increases. At 0 incline the normal force equals the runner's weight. The increased normal force has the same effect as increasing weight.

So, if they are running the same speed and the flat runner is carrying the extra weight such that the runner's weight plus the extra weight equals the normal force of the banked runner there should be no difference in the work out. In reality the experience could be very different because of the biomechanics. The weight belt always applies force to the runner. The normal force only exists when the runner's feet are in contact with the track.

By varying your speed on the banked track you can control the amount of "weight" you perceive yourself carrying. But if you slow down you get less of a work out than if you were carrying a weight.

 
Prex,

My point is this. As my opponent pointed out to me it is quite a different activity to be doing an "explosive" move, physiologically, than it is to do a slow move. As you pointed out the differences in the mechanical world are of little consequence rather 2x the velocity or 2x the mass. The force needed stays the same. Our bodies are much more complex and training for an "explosive" (High velocity) jump is very different than training for a slower high "force" (Higher mass) jump.
The jumper in the rotating track must jump at 2x the velocity in order to reach the same height as the jumper with 2x the mass. That means that the jumper in the track would have to create a velocity of 2X that of the earth bound one in the same distance of muscle motion. This has a habit of recruiting muscles cells and training them in a different pattern. This is also what my opponent was trying to show me by using the "explosive" power lifting method. Unfortunately the scenario he used was incorrect in the sense that the time factor was off. which had me confused.

~Matt
 
You're arguing phisology, not physics, Yes, training for explosive movements differs from training for strength movements.

As for the physics question it gets complicated by the fact that the force you're using to generate "weight" disappears once your feet leave the track.

The first question is which direction is UP. Is up normal to the track or radial to the earth? How high is high? What is the frame of reference for 0 height?

If the frame of reference is the running surface than there is no difference in the acceleration, and therefor the speed, required to reach the same height above the running surface.

For the track runner to generate twice his "weight" he has to be running on a sixty degree incline. In order for him to jump the same height radial to the earth he has to jump twice as high relative to the track. So, relative to the track he needs twice the acceleration and therefore will generate twice the speed.

In fact, running has nothing to do with it. The problem would have the same ballistics if one stood on the ground and jumped away from the center of the earth and the other stood on a 60degree incline (assuming one could) and jumped normal to it. Look at it this way. Leaving from the incline demands that some of the "jump force" be used to generate a horizontal component while the height of the jump only gets credited for the vertical component. To be clear, both "runners" weigh the same when jumping in this static example. The inclined jumper has to wear the weight belt.

I don't know if effective weight as a function of speed offers anything as a training aid. It would depend on what training regimine it replaced. If one never ran with extra weight then this may help. We all know you can go around a banked curve at other than the speed where centripital acceleration overcomes inertia. We do it everytime we take a curve at a speed other than the one posted for the curve. It just means we have to apply acceleration with our feet, just like we do running on the flat.



 
>You're arguing phisology, not physics, Yes, training for >explosive movements differs from training for strength >movements.

Actually were not argueing physiology. We both agree on the physiology as stated above. More explosive(Higher velocity) movement causes different training effect than teh slower lower velocity movement. The argument is whether or not the runner in the track needs to accelerate faster than the one with weight added on the ground.

>As for the physics question it gets complicated by the fact >that the force you're using to generate "weight" disappears >once your feet leave the track.


>The first question is which direction is UP.

Would be differnet in the two differnet scenarios. Up in teh track would be normal to the track on teh ground it would be away from the center of the earth or whatever mass you happen to be standing on.

>How high is high?

High, I think is irrelavant, if they are teh same for both runners


>What is the frame of reference for 0 height?

0 heigth for the earthbound fellow would be ground level. For the person on the track it would be normal to the track. In my attempts at figuring this out I have disreguarded that the track ramps up and have been working with the track as if it is a circle laying down on the ground. This also assumes, falsely that the runner will not be pulled toward the earth, however picture it as if this track is in space.

>If the frame of reference is the running surface than there >is no difference in the acceleration, and therefor the >speed, required to reach the same height above the running >surface.

Here I get very confused.

Case for guy running on ground F=2M*A

Case for guy running in track F=M*2A

Assuming:
1) Gait heigth rise and fall are the same.
2) velocity is the same
3) distance and time per step are the same
Then both runners must rise and fall the same distance in the same time.

If the guy in the track is experiancing 2A then his trajectory must be altered in order to stay in the air as long as the guy on earth. However you bring up an interesting point by pointing out that the acceleration or force causing the 2A is gone as soon as the guy leaves the track. But as soon as he leaves the track his body wants to travel straight, tangent to the point his foot left the track. So he must leave the track at an acceleration that will allow his next step to be at a position that would be the same distance from the last step as it would on Earth. However this step is some point on an arc at "X" distance forward and "Y" distance up. Is it possible that the velocity that he needs is as simple as a trajectory calculation? Or is it even easier a direct velocity "Y" distance/T due to the fact that acceleration won't exist?

Very confused here Sorry.

>For the track runner to generate twice his "weight" he has >to be running on a sixty degree incline. In order for him >to jump the same height radial to the earth he has to jump >twice as high relative to the track. So, relative to the >track he needs twice the acceleration and therefore will >generate twice the speed.

I think you may have misunderstood my problem. I'm trying to figure out the effect on the runner in relationship to the track not the Earth. The heigth of the jump will be similar in both the track and on Earth as this will simulate the rise and fall one experiances during a running gate. I'm assuming the gate inside the track and outside the track are similar.


>I don't know if effective weight as a function of speed >offers anything as a training aid.

Not sure exactly what you meant by "as a function of speed" but running with weight has a different effect than running explosively. Many people do drills such as Polymetrics, "glorified skipping" is one such drill, that concentrates on exagurated large and quick motions. This helps teach muscle fibers to respond concert quicker. Lifting weights has recently been shown to help ones running efforts by recruiting more muscles fibers. Running with a weight vest has long been used, usually succesfully, to replace weight lifting as more sports specific "weight lifting" excercise.

~Matt
 
This is similar to a velodrome that bicycle races are held in.

I believe the physics of it is similar to that of a banked track in racing or a coordinated turn in an aircraft, only with lift being provided by the track.

Because acceleration is involved in a turn, a load factor comes into play. A load factor is the ratio of the lift to the gravitational forces and is represented by n (see website). (1) n = L/W. Also centripetal force Fc = V^2/R where V is velocity and R is radius. The components of lift gives (2) ΣFv = L cos ß –W = 0 or L = W/ cos ß and (3) ΣFh = L sin ß = m*Fa = mV^2/R where ß is bank angle. Placing (1) into (2) and solving for n gives n = 1/ cos ß or sec ß. Fv balances the W of runner while Fh provides turning forces. All forces act through the runner’s centre of gravity.

What this means is when you increase bank angle, the load factor increases. At 60 degrees the load factor is 2g, and because you were at 1g before you started, you have doubled your weight at that angle. If you go to somewhere where g is lower, the load factor is still 2g, so the effect is the same.

For a more detailed explanation see a text on aerodynamics on load factors and turns.

 
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Case for guy running on ground F=2M*A
This equation is more correctly stated as F1=(M1+M2)* G where G is the acceleration of gravity, M1 is the mass of the runner and M2 is the mass of the weight.

Case for guy running in track F=M*2A
This equation is not correct. The correct equation is the vector addition of the centripital acceleration vector and the gravity vector. Using notation from the sketch the vector equation would be F = P + Ac (can't add vector notation in this font)

The track runner is not experiencing an accelerstion of 2A normal to the track but is experiencing a force normal to the track that is equal to the weight of the flat runner and weight.

In order for either runner to jump a force normal to the track that is greater than the force holding him on the track must be applied. The runner's initial normal velocity and therefor the height of the jump depends on how much force in excess of the force holding him on the track he can apply. Notice that I only talk about force, not acceleration. Acceleration is used to calculate the forces. If the forces are the same and the runner's apply the same jumping force then the height of the jump and the initial velocity would be the same.


Case for guy running on ground F=2M*A

Case for guy running in track F=M*2A

Your equations say the forces are the same because 2M*A = M*2A
I know I said your second equation was wrong, and it is. But we agree that the normal force generated on the track would have the same magnitude as if this equation were true.

Your key understanding that an acceleration requires a proportional velocity to overcome is is flawed.


You mention gates in one paragraph. Are you refering to the hurdles used in several olympic events. If you are your assumption that runners jump over these gates is also flawed. They step over the gate. They lower their upper body to raise their hips so they can clear the gate without raising their center of gravity. I think using this track for hurdle training would be impossible. It would be difficult to keep the proper number of steps between hurdles. And you't have to design a hurdle that had the same cross bar height for all radial positions of the runner.
 
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