A math question for all you engineers out there..

[naaman]

So... I've been wondering:

How do I calculate the amount of power required for a person to do a push up?

Taking into consideration weight, height, arm-length etc. And, of course, over time, how do I calculate the power involved in a given number of reps in a given number of time?

And, same question, but for sit ups.

Okay: GO!

 

[sahtt]

If only I was qualified.

FYI, you have a lot of variables to quantify if you want to get it fairly accurate. Start with a diagram and figure out all the degrees of travel you want to use as controlled. You could probably just estimate body weight distribution as 'normal'.

I have a pretty good idea of what I'd do for push-ups but sit-ups seems annoying to calculate.

I'll let vega$ come in here and wreck shop then add in any suggestions [doubtful given our history with the air plane saga].

 

[dyee]

Depends how much of your body weight is supported by your arms.

How long is it from the floor when all the way down and all the way back up?

How much time does it take to do a push up?

Power (P) = Energy/time. In this case gravitational PE which is weight x height off the floor then divide this number by time. My recommendation is the see how much time it takes to do 10 push-ups then divide that time by 10and that's the time for 1 push-up. Use two scales under each hand to find out how much of your body weight is being moved. good luck.

 

[JonBoy]

I would treat your body as a pinned support (your feet), rotating about a point. You know your body weight and your center of gravity is the point in your lower back that is 55% of your overall height (measuring from your feet upwards).

Using that, you can get the resultant loads on your toes (pivot point) and your hands.

For instance, I am 190 lbs and 6' (72") tall. My center of gravity is going to be 39.6" from my feet.

Free body diagram is below:

|------------|----------|
Toes 39.6 COG 32.4 Arms

The toes are the pivot point.

Using equations of equilibrium, Ftoes + Farms - Body weight = 0

So, FToes + Farms = 190 lbs

Using a sum of moments about the toes, Farms x 72 - Body weight x 39.6 = 0

Therefore, Farms = 104.5 lbs (ie, your arms are taking 104.5 lbs of your body weight).

One hp = 180 lbs x 181 ft/min = 32580 ft-lbs/min

My arms are probably close to 26" long. Let's assume that I do one complete pushup every 3 seconds. Therefore, I will be moving 104.5 lbs through 26" of travel every 3 seconds (we will assume that you freefall on the downward motion).

26" is 2.1666 ft. We do that 20 times per minute, so total travel will be 43.333'.

So, you'll be moving 104.5 lbs x 43.3 ft/min, or 4528.3 ft-lbs/min.

A ratio of 4528.3/32580 gives 0.14.

Ergo, you will require .14 hp to sustain 20 pushups per minute, in my case. Substitute your own numbers as needed.

 

[usafdarkhorse]

You should also note that that 26" is arm length and not necessarily arm travel.

A standard pushup is where your upper arm makes a 90 degree angle with your forearm ( ie. parallel to the ground ) so your arm travel will actually be roughly half, or however long your upper arm is.

Many runners and athletes can sustain over 1.5hp during the course of their sport.
If you're about 160 lbs., and an average stair height is 1', you would be climbing over 300 stairsteps in a minute, sustained.

Then you would need to take a breather!

 

[BioBanker]

If you did the PU on a force transducer platform, that would tell you.

 

[usafdarkhorse]

If you did the PU on a force transducer platform, that would tell you.

Haha yeah :tongue:

 

[Vega$ NSX]

I'll let vega$ come in here and wreck shop then add in any suggestions [doubtful given our history with the air plane saga].

Haha! Maybe I should take the hint and back out completely! :tongue: :biggrin: :wink:

 

[Vega$ NSX]

JonBoy does a pretty good explaination above. He beat me to the free body diagram (damn lunch!)

If you really wanted to be more technically correct, the work should better be calculated with the formula:

W= Frθ = τφ

Where:
Fr = Rotational force aka torque (τ); note it should be Fr Sinθ, bur if you assume the arms stay perpendicular to the body, then it will always be 1, hence Fr
θ = Degrees of motion
τ = Torque
φ = Degrees formed by the arc of motion

That is because the motion is more of an arc, rather than linear. And if you want to be even more technical then you could use this equation:

W = Integral (τdφ)

so that you account for the change in torque throughout the range of motion.

However, that's splitting hairs and the approximation is close enough. The conclusionary point being that the single largest factor in determining the power is the speed (acceleration) in which the push up is done. The faster you do it the more power it requires. Height, weight, arm length and other factors don't contribute nearly as much as the speed (acceleration) the push up is done. Since that number is not given, the actual answer relies heavily on an assumption. Which drives us engineers nuts becuase there really is no one right answer. :smile: