# Wall Flips

1. Apr 30, 2007

### Eggroll451

How would you calculate the height needed to do a flip off the wall as described in http://www.wikihow.com/Run-up-a-Wall-and-Flip given the height of the person and the speed.

2. Apr 30, 2007

### Danger

I don't think that it's all so simple. You would also have to factor in the person's weight and leg strength (all leg muscles, including the ankle).

3. May 1, 2007

### Ki Man

height needed? i dont get it.

I agree with danger, there are way more factors in this than that. Angle, power, the course of the flip, stuff like that

4. May 2, 2007

### KingNothing

There are a ton of factors in jumping: muscle fiber recruitment, leg strength, ligament lengths relative to each other, motor skills.

If you are looking for a qualitative approach to this question and simply want it answered, just e-mail the writer of that article or experiment to see what works best.

If you are looking for physical equations to describe this, you probably won't ever find any.

5. May 2, 2007

### Office_Shredder

Staff Emeritus
I think we can simplify the problem by assuming that if you're running at a speed v at the wall, you can efficiently transfer your momentum from horizontal to vertical without any loss. While at first it seems to make no sense, it kind of feels right considering that while you lose energy in changing momentum, you get to add some by taking another step off the wall, so maybe it about balances.

So if you put your launching foot at a height h up the wall, the pivot point of your body is going to be at a height of h once you become horizontal. Note the pivot point of your body will fall as soon as you begin your rotation. It's not really possible to calculate what your angular speed is, so assume your feet move at constant velocity this whole time. The average center of mass is about .55T high, where T is the height of the person (it's about .56 for males, .54 for females), so the distance of your feet (which have constant angular speed) from the pivot point is .55T, where T is how tall you are

Using these assumptions, I would imagine the problem is solvable, or at least much closer to solvable

6. May 2, 2007

### dimensionless

You need to jump to a height h1, with vertical speed v1, horizontal speed v2 and angular velocity w1. You would then have to push off the wall and acquire angular velocity w2, vertical speed v3 and horizontal speed v4. To solve this I imagine you would have to find the person's center of mass as a function of position.

You will also need to double knot your shoelaces, and lift weights/take gymnastics for a year. (In other words, my statements are not to suggest that this should be attempted by anybody.)

7. May 2, 2007

### Danger

Also double-check that nobody greased the wall.

8. May 2, 2007

### KingNothing

How did you beat me to it?

9. May 2, 2007

### aliaze1

yea there are lots of factors involved....

i assume you will need to know a minimum of:

forces exerted by ALL of the muscles of the leg, and their angles since they are not all perpendicular to the ground

angle of the jump in relation to the coordinate plane (the wall and floor)

person's weight and center of mass

velocity b4 the jump

technique of jump

all vector quantities must have their angles known, this is crucial

and any potential losses in the process should be factored in

10. May 2, 2007