Rolling or Cat-Hang to Drop?I was browsing APK earlier and I saw this thread:

http://www.americanparkour.com/smf/index.php/topic,38722.0.htmlMatthew posed the question: To get off a high wall, would it be better to jump and roll off, or cat-hang and drop off without rolling?

Well I thought I could use my nerdy-ness to help. I don’t have something to measure time as accurately or precisely as I would like, so I took a guess-timate on the times and the running speed.

Here’s my made up situation:

A traceur is going for a run in an unfamiliar location when he encounters a 3 meter high (about 10 feet) drop. If he is running at a constant speed of 5m/s, should he

a) Jump off ( at 5m/s) at a 45 degree angle and roll.

OR

b) Cat hang and drop down 2.25 meters (about 2.5 feet less) without rolling?

Assume that a traceur’s average roll takes around 1 second to perform and he exits the roll at a light 2m/s jog.

Assume that, for part b, the traceur absorbs the impact in .25 seconds.

The traceur weighs 77kg (around 170 pounds)

Which situation would have him feel the least average force?

Part b is easier so I’ll do that part first haha!

Through the conservation of mechanical energy, the traceur’s gravitational potential energy (GPE) at the top of the drop (from the cat hang position) should equal his kinetic energy (KE) at the bottom of the drop.

(GPE) mgh = 1⁄2 mv^2 (KE)

Where m is the mass of the traceur, g is the acceleration due to gravity (9.81m/s2), h is the height of the drop, and v is the velocity.

Solving for v gives us:

v = √(2gh)

Substituting numbers for variables gives us the traceur’s speed to be

**-6.64m/s** (the number is negative because the velocity is pointing downwards)

His momentum would be

p = mv

Substituting numbers gives us

**-511 kg*m/s** (again the number is negative because it’s directed downwards)

The average force on the traceur would be:

F = Δp⁄t)

F = (p

_{f} – p

_{i})/t

The subscripts, f and i, represent the final and initial momentums of the traceur

*Substituting numbers gives us 2044 kg*m/s2, or ***2044N.** This is about **460 pounds of force** and close to **3 times the traceur’s weight (2.7x actually).**Time for the fun part xD

The traceur jumps off the drop at a 45 degree angle; that means he’s going higher, so we must find his maximum height.

We must find the y-component (the vertical component) of his velocity; which is fairly easy because he jumps as a 45 degree angle. If you don’t know/don’t remember trigonometry just take my word for it. His vertical speed would be

**3.53m/s**I could use the kinematic equation

v

_{f}^{2} = v

_{i}^{2} -2g∆y

to solve for the additional height, ∆y. v

_{f}^{2} is zero because the vertical velocity at the point we’re solving for would be zero (your vertical velocity at the peak of your jump is zero). This makes solving much easier. We get:

Δy= v

_{i}^{2}/(2g)

Substituting the numbers we find that the traceur added an extra .64m to his height.

Now we must do that whole shehbang with the energy thing to find out his final speed and momentum. But I’ll save you the trouble of reading it all and save me the trouble of writing it all xD

**V = -8.45m/s**

P = -651 kg*m/sWe’ll assume that this downward momentum is completely transferred to forward momentum through “reverse blocking” which was mentioned in the thread.

The final momentum, calculated by multiplying the 2m/s (the traceur exited in a light jog) by his mass is:

**p = 154kg*m/s**Now we use this equation again:

F = (p

_{f} – p

_{i})/t

Except this time the final momentum is 154, the initial is 651, and the time is 1 second.

The average force felt is:

**497N!** That’s a little under a forth of the 2044N felt with the straight drop! And only .66 of the traceur’s weight!

TL;DR?

Conclusion:It’s better to jump off and roll than to cat hang and drop. The math says so.

*BUUUUT!*I did not account for the force it would take to shift the downwards momentum forwards. If somebody isn’t satisfied I’ll break my head trying to find out how to do that xD In fact this whole thing was half assed. Like I said I don’t have correct measurements for time or running speed (although 5m/s seems like a light run) so this could very well have a high percent error. Also, I only considered downwards momentum for the roll. You will have some forwards momentum for the roll and I don’t know how much of a difference that could make to the average force. And the biggest variable of all is us. We need to have a good roll to make sure that momentum is dispersed safely and we don’t hurt ourselves with a 10+ foot drop.

Even so, a roll seems like a better option than a straight drop.

If anybody isn’t satisfied with my math for some reason please tell me if I made a mistake somewhere, even though I’m pretty sure it’s all right.