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Fazor
2009-Sep-25, 05:08 PM
This was just kinda a thought-experiment, to see if I could work it out. I couldn't. So now I'll turn to you all. (It's not that complicated).

Say I weigh X lbs., and I have a standing vertical leap of H inches. All other things equal, if I decrease the weight by 30 lbs. (X-30), how high is my vertical leap now?

Yes, I know there's actually a lot more to it. Changing weight would probably also mean a change in muscle, balance, blah blah blah. To simplify, lets just solve assuming the force generated for acceleration is the same, and use the estimated gravitational constant (9.8 m/s^2?).

Oh, and might as well throw some depressingly close to accurate values in. Initial weight is about 190 lbs., and initial jump height is about 4 inches.

Neverfly
2009-Sep-25, 05:40 PM
This was just kinda a thought-experiment, to see if I could work it out. I couldn't. So now I'll turn to you all. (It's not that complicated).

Say I weigh X lbs., and I have a standing vertical leap of H inches. All other things equal, if I decrease the weight by 30 lbs. (X-30), how high is my vertical leap now?

Yes, I know there's actually a lot more to it. Changing weight would probably also mean a change in muscle, balance, blah blah blah. To simplify, lets just solve assuming the force generated for acceleration is the same, and use the estimated gravitational constant (9.8 m/s^2?).

Oh, and might as well throw some depressingly close to accurate values in. Initial weight is about 190 lbs., and initial jump height is about 4 inches.

Let me get this straight:
You're asking how much force will lift 190lbs to a height of 4 inches and how high that same force will lift 160lbs?

Tog
2009-Sep-25, 05:46 PM
That's the way read it. I have no idea what the answer might be though.

But 4 inches? Really?

Neverfly
2009-Sep-25, 05:55 PM
But 4 inches? Really?

Maybe he is referring to levitation and not jumping.:p

nauthiz
2009-Sep-25, 06:04 PM
Let's say that the actual process of jumping involves a constant force applied over the course of 1/10 second. Furthermore, let the height of the jump is determined by how far your feet get off the ground - that should make it easier, since that means that you're at maximum velocity at that point, so we just need to figure out what speed that is to give a certain jump height, and can use that to figure out the force of the jump.

I'm also going to take the liberty of using 0.1 meters instead of 4 inches and 86 kg in stead of 190 lb. And I'll be doing a lot of rounding for the numerical bits.

Using the ballistic motion equations:
v' = v-gt
y = vt - 1/2gt2

where v' is the velocity at the top of the jump (0) and v is the "liftoff" velocity, first solve the top equation for time:

0 = v - gt
gt = v
t = v/g

Then plug that into the second equation:
y = v(v/g) - 1/2g(v/g)2

Then plug in the height (0.1) meters and solve for v to get the initial velocity of the jump:

0.1 = v2/9.8 - 4.9(v2/96)
0.1 = 0.1v2 - 0.05v2
0.1 = 0.05v2
2 = v2
v = 1.4 m/s

Accelerating from 0 to 1.4 m/s in 0.1 seconds is an acceleration rate of 14 m/s2.

F=ma, so the force of the jump would be (86*14)=1,200 N.

Now if you weigh 30 pounds less, that would be about 73 kilograms. Using the same force, that means the acceleration would now be (1200 / 73) = 16m/s2, so the initial velocity would be 1.6m/s.

Plugging that into the first equation to figure out how long the jump lasts:
0 = 1.6 - 9.8t
9.8t = 1.6
t = 0.16 seconds

And plugging that into the second equation gives us how high the jump is:

y = 1.6(.16) - 4.9(.026)
y - .26 - .13
= 0.13 m

Or about 5 inches.

Neverfly
2009-Sep-25, 06:13 PM
Well, there ya go Fazor. If you lose 30lbs you will see five inches.

IsaacKuo
2009-Sep-25, 06:20 PM
Let's say that the actual process of jumping involves a constant force applied over the course of 1/10 second.
This essentially means assuming that the human legs involved can provide a certain amount of momentum. Force x Time gives us impulse (or momentum).

Another possible assumption would be to assume that the human legs involved can provide a constant force applied over a constant DISTANCE. This is more intuitive, because once the legs are stretched to their maximum length they obviously aren't going to be providing any more force.

Under this assumption, we get a certain amount of kinetic energy. Force x Distance gives us energy.

In this case, the calculations are rather simple. The legs give us some particular kinetic energy; this gets converted into gravitational potential energy at the top of the jump. The formula for this is simply:

E = gH

Where H is the height.

So plugging away, we get:

Gravitational potential energy for 190lbs to a height of 4 inches is 760inch-lbs. Solving for if the weight is only 160lbs, we get a height of 4.75 inches.

Note that this is different from the above calculation because we're starting with different assumptions about what is "constant". Assuming constant momentum capability gives different results than if you assume constant energy capability.

Fazor
2009-Sep-25, 06:27 PM
Maybe he is referring to levitation and not jumping.:p

Nope, sadly that's how high my standing free jump is (two feet together).

Cougar
2009-Sep-25, 08:41 PM
...let the height of the jump is determined by how far your feet get off the ground...

That's not how you do it! :) You stand up next to a wall with a tall chalk board on it. Chalk in hand, you reach up and mark as high as you can reach, with both feet on the ground. Then you coil down into a spring and accelerate your body into the air, throwing your arms up and marking your peak height. Fazor ought to be able to mark 6 inches, even at 190. :doh:

Fazor
2009-Sep-25, 08:45 PM
That's not how you do it! :) You stand up next to a wall with a tall chalk board on it. Chalk in hand, you reach up and mark as high as you can reach, with both feet on the ground. Then you coil down into a spring and accelerate your body into the air, throwing your arms up and marking your peak height. Fazor ought to be able to mark 6 inches, even at 190. :doh:

:) I should point out that I do mean feet-from-ground, keeping legs straight. Didn't want to count pulling your knees up, as that will vary from jump to jump.

Neverfly
2009-Sep-25, 08:52 PM
:) I should point out that I do mean feet-from-ground, keeping legs straight. Didn't want to count pulling your knees up, as that will vary from jump to jump.

How does one jump while keeping his legs straight?

Fazor
2009-Sep-25, 08:56 PM
:) I meant once airborn. More of a 'hop' really.

But for the purposes of the question, had it not already been answered, it could be a robot with a rocket (exhaust nozzle cleverly located, of course). Basically, find force from known weight and "hop" distance versus earth gravity, then adjust weight and use found force to determine next 'hop height'.

ETA: Just for the helluvit, I did a crude test. Basically held my arm straight down with a pencil and tried to move my wrist to mark the apex of my jump., then compare it to my standing position. More like 8 inches. But I'd be happy halving the difference and saying six. I certainly have never been known for my "ups". Youth basketball was ugly. I much preferred soccer and baseball.

Cougar
2009-Sep-25, 09:32 PM
:) I should point out that I do mean feet-from-ground...

You mean measuring from the feet? Tough to measure. Vertical leap using the height of your reach allows your body to do anything to get up there. Yeah, 8 inches. I bet you did 8 inches.

I certainly have never been known for my "ups". Youth basketball was ugly.

Oh, well, there you go. I know this stuff because I was starting forward on my high school varsity squad that went undefeated in league play (10 games). I usually scored in the low double figures, but had one high game of 31 points! This was before the 3-point shot. Went out with the head cheerleader, too. I kid you not.