Comments on: Weight loss and vertical leap
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap/
Comments on Ask MetaFilter post Weight loss and vertical leapThu, 17 Mar 2005 12:26:58 -0800Thu, 17 Mar 2005 12:26:58 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Weight loss and vertical leap
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap
BasicPhysicsFilter: Everything else being equal, how much would my vertical leap improve if I lost 10% of my weight without losing any muscle?post:ask.metafilter.com,2005:site.16448Thu, 17 Mar 2005 12:16:20 -0800callmejayjumpingleapingverticalleapfitnessexerciseweightlossBy: drpynchon
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278734
11.1% higher.. assuming you create the same amount of force or expand the same amount of energy as before...comment:ask.metafilter.com,2005:site.16448-278734Thu, 17 Mar 2005 12:26:58 -0800drpynchonBy: drpynchon
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278739
expend... i mean..comment:ask.metafilter.com,2005:site.16448-278739Thu, 17 Mar 2005 12:31:46 -0800drpynchonBy: Capn
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278742
Dr. P, you must show your work for full credit.comment:ask.metafilter.com,2005:site.16448-278742Thu, 17 Mar 2005 12:37:24 -0800CapnBy: Sonic_Molson
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278748
I figured something like this, but I'm no physics expert:<br>
<br>
F = M * A<br>
100 = 100 *A = 1<br>
100 = 90 * A = 1.1111111<br>
<br>
so 11.1% would be right.comment:ask.metafilter.com,2005:site.16448-278748Thu, 17 Mar 2005 12:44:41 -0800Sonic_MolsonBy: vacapinta
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278751
If the energy is the same, then that means at the height of your leap the new and old potential energy match. This is just MGH, where M-mass, G-grav constant, H-height.<br>
<br>
So, if M1*G*H1=M2*G*H2 just do some arithmetic and you get H2/H1=M1/M2.<br>
<br>
In the Case of 10% weight loss, M2=0.9*M1, so H2/H1=1.11...<br>
<br>
Or, what drpynchon said.comment:ask.metafilter.com,2005:site.16448-278751Thu, 17 Mar 2005 12:46:12 -0800vacapintaBy: Sonic_Molson
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278752
Damn, now that I thought it over; that's all wrong. Your acceleration will improve by 11.1%, but it will definetly not improve your jump height by 11.1%. At least I don't think so.<br>
<br>
I didn't do so well in Physics :-(comment:ask.metafilter.com,2005:site.16448-278752Thu, 17 Mar 2005 12:46:58 -0800Sonic_MolsonBy: Zed_Lopez
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278754
Newton's 2d law: Force = mass * acceleration<br>
<br>
F = ma<br>
<br>
trivially,<br>
<br>
a = F/m<br>
<br>
Distance = .5 * acceleration * time^2, when starting at rest (<a href="http://physics.ucsd.edu/~cdpgrad/speed.html">details</a>)<br>
<br>
d = .5 * a * t^2<br>
<br>
So, original distance:<br>
<br>
d1 = .5 * (F/m) * t^2<br>
<br>
Distance with 90% of the mass and the same force:<br>
<br>
d2 = .5 * (F/.9m) * t^2<br>
<br>
d2/d1 = (.5 * (F/m) * t^2)/(.5 * (F/.9m) * t^2)<br>
<br>
= 1/.9<br>
<br>
= 10/9<br>
<br>
= 1.11<br>
<br>
= 11.1% gain<br>
<br>
You were right the first time, Sonic Moison.comment:ask.metafilter.com,2005:site.16448-278754Thu, 17 Mar 2005 12:50:48 -0800Zed_LopezBy: drpynchon
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278759
You were right the first time Sonic, and you were semi-right the second time. If we assume that his strength is unchanged, neither is the force or energy he exerts. All of these things are conserved. Once you have conservation of energy, the problem is solved as shown above.<br>
<br>
But also, potential energy can be thought of as the path integral of the force over a distance, which is what gets you the simplification of Ep = mgh. The force in question is constant, and while your acceleration will go up 11.1% your mass will fall 10% so F = m1a1 = m2a2 = (0.9m1)(1.11a1) = m1a1.<br>
<br>
You might gain a few inches, which makes sense if you think about it.comment:ask.metafilter.com,2005:site.16448-278759Thu, 17 Mar 2005 12:57:34 -0800drpynchonBy: Doohickie
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278796
Yep, I got 11.1% as well (before peaking into the thread). So apparently I can jump 1/0.8, or 1.25 times (or 25%) higher than at the beginning of the year (since I've lost about 20% of my weight).comment:ask.metafilter.com,2005:site.16448-278796Thu, 17 Mar 2005 13:40:04 -0800DoohickieBy: plexiwatt
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278803
assuming your muscles are equally efficient at pushing two different distributions of weight over your body up in the air, yes.comment:ask.metafilter.com,2005:site.16448-278803Thu, 17 Mar 2005 14:03:50 -0800plexiwattBy: Chuckles
http://ask.metafilter.com/16448/Weight-loss-and-vertical-leap#278874
There is that whole thing about ancient greek standing long jumpers using weights in their hands to improve their jump.<br>
<br>
Then there is the Lance Armstrong vs. Jan Ulrich thing... If your muscles can fire really fast then you want to be light, but if you can't extend as quickly you probably get more energy invested in the jump with a little more weight. A little more energy may or may not mean more height of course. You have to do the math, or preferably the experiment, to know for sure.<br>
<br>
Humph... <a href="http://biomechanics.bio.uci.edu/_html/nh_biomech/long%20jump/longjump.htm">This biomechanics article</a> says that my Armstrong analogy is the same effect as the weights... well sort of anyway...comment:ask.metafilter.com,2005:site.16448-278874Thu, 17 Mar 2005 20:28:05 -0800Chuckles