## Monday, December 29, 2014

### Physics of Bandaloop Dancer

The other day, I was shown a video of one of the most exciting looking things I've seen in some time:

The group of dancers is called Bandaloop and among other things, they dance on the sides of buildings and cliffs by being attached via rappelling ropes. The long ropes keep them from falling, and allow them to make incredible leaps with long hang times.  A video interviewing the founder says some days, from tall buildings, she can get jumps with hangtimes of 9 seconds or more.

My first reaction to the video was "WOW! I wanna do that!".  My second reaction was "What's the physics behind that?  Nine second hangtimes? Really?!"

Since I can't try that anytime too soon, I must instead attempt to describe the physics behind them. Here are the pictures I drew on the side of the napkin:

I didn't have lots of data to go on, so I initially estimated that the Length of the ropes (L) might be approximately 30m, and if harnessed at the center of the body, that puts x at approximately 1m.  At best jump, I estimated approximately 5m out from the building. Since the sin of the angle θ is x/L, the inverse sine of 1/30 and 5/30 suggests that θ ranges between 2 to 10 degrees. At angles this small, sin(θ) and tan(θ) are nearly identical, suggesting that y and L are nearly identical too, and I'll be interchanging them occasionally. This is not true as the rope gets smaller -- so relatively tall buildings and tall cliffs are important.

Next I drew a free body diagram of the forces acting on the dancers as they are away from the building.  There are primarily two forces acting on the dancer -- Weight pulling the dancer down, and tension in the ropes pulling the dancer at angle θ up and in toward the building.  That angled force I broke into components Tx acting in towards the building, and Ty acting to counteract the dancers weight.  Because the ropes are so long, the height of the dancer doesn't change significantly, and since the dancer isn't really moving much vertically, we can say that the forces are balanced vertically.  That is, W = Ty. Since W = mg, this means Ty = mg.

Horizontally, the forces aren't balanced, and so whenever the dancer is in the air, there is a portion of the tension Tx which acts to pull the dancer back in toward the building. This unbalanced force is a net force, and so we can write another equation: Fnet = Tx. Since Fnet = ma, this means Tx = ma.

Finally, Tx and Ty are related to the angle by the tangent relationship, such that tan(θ) = Tx/Ty, and after multiplying, Tx = Ty*tan(θ).

After a few substitutions and a little division, we find a formula for the acceleration inwards toward the building that the dancers feel:
\begin{align*} T_x &=T_ytan(\theta ) \\ \frac{ma}{m} &= \frac{mgtan(\theta)}{m} \\ a &= tan(\theta)*g \end{align*}
It's surprisingly simple and clean -- the acceleration the dancers feel is just a multiple of gravity. Since θ ranges from 2 to 10 degrees, the dancers feel acceleration toward the building ranging between approximately 1/30th to 1/6th that of normal gravity.  For comparisons sake, the acceleration of gravity on the moon is about a 1/6th of that on earth, which according to my rough estimates is about the most that the dancers would feel inward toward the building on their most extreme jumps. Most of the time they are just a meter or two away from the the wall they are feeling much lower attraction toward earth.

Put another way, the dancers feel approximately 1/30th of their "weight" inward toward the wall, so a 120lb dancer might feel only 4 lbs of forces inward. Imagine how easy it would be to jump if you only weighed 4 lbs, but had the strength of a professional dancer!

Which suggests a different approach to this problem. Let's figure out how far out from the building a dancer ought to be able to get, assuming they can jump "off the wall" with as much speed as they can normally jump off the ground.... Assuming that a person can normally jump to a height of 0.5 meters under typical gravitational acceleration of -9.8 m/s/s, we can use the "no time" formula:
\begin{align*} v_f^2 &= v_0^2+2a(\Delta s) \\ 0 &= v_0^2+2(-9.8)(0.5) \\ v_0 &= \sqrt{2(9.8)(0.5)} &\approx 3 m/s \\ \end{align*}
They would have to jump at a velocity of approximately 3 m/s, yielding a hang time of approximately:
\begin{align*} v_f &= v_0-at \\ -3 &= 3-(9.8)t \\ t &= \frac{-6 m/s}{-9.8 m/s^2} &\approx .6 sec \\ \end{align*}
Leaping with the same initial speed (3 m/s) off the side of the building with an acceleration of 1/6th of gravity as these dancers feel would allow them hang times of 6 times as much:
\begin{align*} v_f &= v_0-at \\ -3 &= 3-(\frac{9.8}{6})t \\ t &= \frac{-6 m/s}{\frac{-9.8}{6} m/s^2} &\approx 3.7 sec \\ \end{align*}

Not quite the 9 seconds claimed in the video, but more on that later.  Substituting half of this time (because the maximum height occurs halfway into the trip) into the kinematics equation allows us to calculate the maximum "height" off the buildings this dancer could reach:
\begin{align*} s_f &= s_0+v_0t+\frac{1}{2}at^2 \\ s_f &= 1+3(1.75)+\frac{1}{2}(\frac{-9.8}{6})(1.75)^2 &\approx 4m \\ \end{align*}
That's pretty close to the 5m I estimated from the video.

Now the hangtime of 3.7 seconds is far less then the the claimed 9 seconds of hangtime, and even my closer observation of the video suggests a few jumps were more than 5 seconds long. One way to get more hangtime is to jump with more speed - something that's quite possible with stronger and trained legs. Remember I started with someone able to jump 0.5 meters -- and I'm sure a strong dancer could leap higher.

Another reason the kinematics equations don't give us enough hangtime is that the acceleration is not constantly 1/6th that of gravity -- often it's way less than that even! Smaller accelerations, like those felt close to the building when θ is small, would increase hangtime significantly. Unfortunately, I've forgotten the formulas for how to deal with accelerations that aren't constant -- although I'm sure a few google searches could refresh me.

Either way, the Bandaloopers certainly can experience tremendous "jumps" due to the low horizontal forces they have to fight on their rotated worlds, and simple first-year physics concepts help reveal why.

Now, perhaps someday I'll be able to actually try it for myself.