# Power's relation to actual strength/force.

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#### Solifugae

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The power required to overcome the aerodynamic drag is given by:

P_d = \mathbf{F}_d \cdot \mathbf{v} = {1 \over 2} \rho v^3 A C_d

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.

Is this that fundamental? Applying this to things other than cars, like human sprinters, can we determine someone's actual strength from how fast they run? If an 80 kilo sprinter has a top speed of 27mph, and an untrained 80 kilo person has a top speed of 15mph, does this actually mean that the legs of the sprinter are 5.83200 times stronger practically? Could they then squat that many times greater a weight? That seems a little too extreme in reality, but they definitely produce that many times more power.

I understand that you can have low power and still produce a lot of force through leverage, but in this case, the difference in people's leg length isn't as extreme as in a car's gear ratios, which can create greater torque at a low speed with a low gear and more power for less torque in a high gear. Shouldn't the higher power equal more force linearly when you have only changed power instead of leverage? All other things being the same, 5 times the power equals 5 times the force?

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#### origin

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Solifugae":1qb6mljd said:

The power required to overcome the aerodynamic drag is given by:

P_d = \mathbf{F}_d \cdot \mathbf{v} = {1 \over 2} \rho v^3 A C_d

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.

Is this that fundamental? Applying this to things other than cars, like human sprinters, can we determine someone's actual strength from how fast they run?

That formula is only for aerodyanamic drag. You can only calculate the amount of power needed to overcome wind resistence which is negligable for the typical human. It is true however that for sprinters if the tail wind is high enough the race results cannot be used as a record. But we are talking about 100s of a second for top trained athletes.

If an 80 kilo sprinter has a top speed of 27mph, and an untrained 80 kilo person has a top speed of 15mph, does this actually mean that the legs of the sprinter are 5.83200 times stronger practically? Could they then squat that many times greater a weight? That seems a little too extreme in reality, but they definitely produce that many times more power.

According to your formula the sprinter developes about 8 x the power of the untrained guy, but only for air resistance. Much more power is used to acclerate 80 kilos, not to mention the power needed to keep the 80 kilos balanced, and most importantly the power required to move the legs at a high rate of speed.

Because of gearing in a car, the engine can maintain a realtively constant speed. Unfortunately to run faster your legs must move faster, so our forward velocity is limited by the velocity that we can move out legs. I have seen this inelegantly demonstrated in videos of knuckleheads running down hills.

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#### Solifugae

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According to your formula the sprinter developes about 8 x the power of the untrained guy, but only for air resistance. Much more power is used to acclerate 80 kilos, not to mention the power needed to keep the 80 kilos balanced, and most importantly the power required to move the legs at a high rate of speed.

But how does power relate to strength/force? Does this mean that the sprinter is also 8+ times stronger than the untrained man? This seems too excessive for the relation to be linear, yet at the same time, more power with the same leverage should equal force that increases along with the multiplying of the power, rather than an inequal relationship.

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#### origin

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Solifugae":1ivc1dn4 said:
According to your formula the sprinter developes about 8 x the power of the untrained guy, but only for air resistance. Much more power is used to acclerate 80 kilos, not to mention the power needed to keep the 80 kilos balanced, and most importantly the power required to move the legs at a high rate of speed.

But how does power relate to strength/force? Does this mean that the sprinter is also 8+ times stronger than the untrained man? This seems too excessive for the relation to be linear, yet at the same time, more power with the same leverage should equal force that increases along with the multiplying of the power, rather than an inequal relationship.

Well, I was obviously not very clear in my response so let me try again.

The equation that you posted (aerodynamic drag) has nothing to do with the strength of a sprinter vs an untrained runner. This particular equation tells you nothing usefull about the relationship.

The aerodynamic drag is more of an affect of the higher speed from the sprinter as opposed to a limiting factor. The drag has an affect but it is minimal.

Here as an example taken to the extreme:

Image a weight lifter is dragging a 1000 lb weight across a parking lot at 20 ft/minute and I am walking at 40 ft/minute if we only look at the aerodynamic drag formula you would come to the conclusion that I am expending 8x the power of the weight lifter. Obviously this is absurd.

The problem is the major work being done is to over come the friction between ground and the weight NOT air drag.

It is the same with running; aerodynamic drag is a tiny factor compared to the much larger power needed to move the mass of the legs at high speed.

Hopefully that was clearer. I have a bit of a problem going from brain to paper I guess.

edited to add: The aerodynamic relationship is not linear, if it was linear then doubling the speed would double the drag.

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#### Solifugae

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But your example has greater weight as a factor. Mine does not. While aerodynamic drag may be a smaller factor than friction, everything else being the same it should be the deciding factor. I'm still not understanding why this doesn't indicate the strength distance. Remember, both people weigh exactly 80 kilos, one can run at 15mph, and the other 30mph.

The problem is the major work being done is to over come the friction between ground and the weight NOT air drag.

But isn't the ground friction going to be almost exactly the same for both people if they weigh the same, wear the same shoes, etc.

It is the same with running; aerodynamic drag is a tiny factor compared to the much larger power needed to move the mass of the legs at high speed.

You mean the power needed in accelerating rather than matching the air drag force at the top speed? But since that's even greater, doesn't that imply even more power than 8x the untrained guy that the air resistance implies? How does this not make sprinters ten times stronger than ordinary people?

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#### origin

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My only point is that aerodynamic drag is not the equation to use because it is not the limiting factor. Calculating the power developed from running is not trivial but clearly a professional running at a higher speed is producing much more power than an average 'Joe'.

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#### Shpaget

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Because running involves moving and stopping your legs. You need to accelerate them and than stop them. The faster you run, the faster you need to do that, and that takes a lot of energy.

Cars on the other hand have wheels that rotate constantly and do not require constant acceleration and deceleration, so not much energy is used to keep the rotation going.

Try running on a treadmill and see how fast can you go. Try to find a powered one so you won't need to overcome the friction of the mechanism and there will be no air resistance since you are not actually moving. You'll see that you can't run much faster on the treadmill than really run on the ground.

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#### Saiph

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I'm sure you could find a relationship between the speed, drag and strength between two runners. It's going to be a very loose relationship though, and I doubt it'll be as straightforward as you suggest.

Why? Running fast isn't just about air resistance, but about how efficiently and effectively you use what power you have. The slow runner may be just as strong as the sprinter, but doesn't run right, wastes energy in his motion, etc. This means that any difference in strenght you attribute to your drag calculations, might actually be due to other sources.

So is there a relationship? Certainly, but it's going to be pretty broad and general and the results pretty fuzzy.

Nice idea though!

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