Only for genius people: jumping stilts on a trampoline.


by rossero-NL
Tags: genius, jumping, people, stilts, trampoline
EnumaElish
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#37
Dec9-05, 04:26 PM
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The law of conservation is there, to be sure. Meaning, energy spent will be identical in the two situations.
emptymaximum
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#38
Dec9-05, 09:05 PM
P: 110
ok
this is what happens:
guy jumps onto trampoline.
trampoline begins to deform.
as the trampoline gets deformed, the springs compress.
when the trampoline is at maximum deformation, the springs are at maximum compression, and the guy has reached a turning point in the motion.
the energy stored in the trampoline is released up, and the energy stored in the springs is released down. They counteract each other.

This is a damped oscillator. Depending on the 'spring constants' you could have it so that the spring/trmpoline deal slows him down to a stop with no return motion (criticaly damped). I said that bit before, and someone else asked about imagining it as well.

you could also have it so that the spring > trampoline, which would increase the deformation of the trampoline in the verticle down past the equilibrium position of the mass of guy + mass of springs alone.

can anyone see anything wrong with the above explanation?
rossero-NL
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#39
Dec10-05, 01:13 AM
P: 15
I think it this way:

You all know that two springs act like one, wit this formule: C_Spring1*C_spring2/(C_spring1+C_spring2)

According to that, they cannot work against eachother.

The damping you are talking about is like the suspension of a bike. It has two chambers with air. One compress with going down, and decompress to go up. And the other chamber has a hole in it, were air is pushed out at high pressure. This one will take all the energy, but can't give it back, because it decompresses already by going down. And because of the chamber one acting like a spring, chamber two is able to suck in air again.

A damper takes more energy than it gives back.

But we're talking aboput ordinairy springs, compressing down, and decompressing up, giving all energy back. So there's no way of damping.

Someone not agreed?

Please tell me.
Cyrus
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#40
Dec10-05, 01:23 AM
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If you want to describe what is happening it goes like this:

(a) He comes down on the trampoline; therefore, he is decellerated as the trampoline starts to sag, and his energy becomes potential energy in the trampoline, and in the springs on the stilts that compress.

(b) The trampoline reaches its maximum deformation, and he comes to a stop.

(c) The spring force of the trampoline reverses his direction and starts to launch him forwards.

(d) [Now here comes the tricky part]
As he is going down, his springs will obviously compress. When his direction of motion changes, his inertia wants to keep him moving down, but the trampoline will force him up. This means his springs will STAY compressed (FOR A WHILE).

As he moves higher and higher, the upward force of the trampoline deminishes. At some point, this force will be equal to the spring force on his stilts. After that point in time, his stilts that were once under compression will now be able to extert a greater downward force on the trampoline that the trampoline exters an upward force on him. So his stilts will cause the trampoiline to locally deform back DOWN, sorta making a cammels hump shape, where his stilts pushed the middle down to make the two humps.

So that small area that his stilts caused to go down, might spring back up as a secondary "boost", but this boost is not not pushing on his compressed springs. It is pusing on decompressed springs, which means it should have no effect. He will get launched more or less the same, but his springs will bounce up and down on the way up due to that secondary push on the uncompressed springs.
rossero-NL
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#41
Dec10-05, 01:37 AM
P: 15
How can it be that the trampoline stops, while the man is still moving down and still has kinetic energy? Because the only thing that can stop the trampoline is when everything is in the turning point. So how can it be?

And how can the stilt build up more energy than the trampoline, without pushing the trampoline down the moment he starts building up? Because again, in my opinion, two different springs pushed onto eachother, cant carry more or less energy then the other.

Can you please explain me?
Cyrus
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#42
Dec10-05, 01:41 AM
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No, it stops when all the energy is stored in both springs, thats what I mean. It stops at the turing point. I said, "The trampoline reaches its maximum deformation, and he comes to a stop" which means all the energy that could go into the trampoline did, and it deformed to its maximum amount.


And how can the stilt build up more energy than the trampoline, without pushing the trampoline down the moment he starts building up?
I dont understand you sorry.


All Im saying is, if that trampoline gives him a LARGE force back up, then its obvious his springs will stay compressed on the way back up. But since the upward force is decreasing with distance moved up, at somepoint the energy in the springs on his stilt will overcome the upward force and decompress while the trampoline contiues to try to launch him up.
emptymaximum
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#43
Dec10-05, 01:45 AM
P: 110
how the springs work is completely irrelevent. pneumatic, hydraulic, rubber bands, whatever. it's what they do that's important, and what they do is damp the upward motion of the trampoline. he'll go less high wearing the things, BECAUSE the damper takes more energy than it gives back.

the springs work by pushing on something solid. if they push on something 'mushy' they'll not very much at all.
have you ever tried jumping while standing on mud?

stand in an inflatable raft in a pool, then try to jump.
are you going to be able to jump higher than when you are standing on solid ground?
are you going to be able to jump at all?
that ought to take care of that now.
/s
Hermite
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#44
Dec10-05, 07:41 AM
P: 23
I agree with cyrusabdollahi and emptymaximum.

rossero you need to reread the above posts and think about what's being said.
alfredblase
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#45
Dec10-05, 02:22 PM
P: 230
Rossero asked for formulas guys =)

hehe as scientists arguing in a thread with so many posts about a classical physics problem im surprised no ones brought out the old maths yet.. :P
emptymaximum
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#46
Dec10-05, 05:52 PM
P: 110
the original post said nothing about formulas.

if you want a formula, it would be the formula for a damped harmonic oscillator, with the restoring force of the stilts being the damping force.

also, it is a good idea to have an understanding of what is going on when you have a formula so's you know what quantities are what.

a lack of a physical understanding of the system will make it difficult to describe the system mathematically, no?
alfredblase
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#47
Dec10-05, 11:03 PM
P: 230
im not so sure, i find it difficult to be sure about which is the right way to visualise the problem. I honestly think that either experimentation or a mathematical proof from first principles is the only way to be 100 percent sure when it comes to a problem that is difficult to visualise, or atleast that many people argue about.
emptymaximum
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#48
Dec11-05, 01:16 AM
P: 110
[tex] m \ddot{x} = -m g + b x - k x [/tex] where +ve x is upward, b is the spring constant of the trampoline, and k is the spring constant of the stilts.
alfredblase
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#49
Dec11-05, 09:10 AM
P: 230
ok i guess u r right. thinking about the problem i guess it would be a damped harmonic oscillator. but if we are visualising the problem incorrectly, (which could be the case) then the above formula wouldnt apply..
emptymaximum
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#50
Dec11-05, 04:09 PM
P: 110
Quote Quote by emptymaximum
also, it is a good idea to have an understanding of what is going on when you have a formula so's you know what quantities are what.

a lack of a physical understanding of the system will make it difficult to describe the system mathematically, no?

so then now you agree with the above statement?
Cyrus
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#51
Dec12-05, 02:06 AM
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Quote Quote by emptymaximum
[tex] m \ddot{x} = -m g + b x - k x [/tex] where +ve x is upward, b is the spring constant of the trampoline, and k is the spring constant of the stilts.
Im afraid its not that simple. The trampoline will not be an ideal spring, but rather some sort of a membrane. It will take the shape of a "well" so to speak, and I doubt the spring constant will be directly linear as you propose, but it could be very nearly the same.


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