Comment on xkcd #3152: Skateboard

slippyferret@lemmy.blahaj.zone ⁨5⁩ ⁨weeks⁩ ago

• burntbacon@discuss.tchncs.de ⁨5⁩ ⁨weeks⁩ ago

  I started to answer this, began breaking out the calculus and the kinetics equations, and then said to hell with it. As a really, really simple answer: Average fall rate for a human that knows how to skydive is going to be somewhere around 100-120 mph. I think I average about 118, per the little app I have, but the phone’s sensors probably aren’t really all that great for accurate measurements about it. If a person could sustain two Gs (give or take 40mph/s) for the length of time needed, three seconds, and that your velocity was entirely conserved (I know, I’m throwing out wind resistance here), that would mean a ramp transition distance of ~540 feet (not really accurate, because mph converts to feet per second at 1.47, not 1.5, but easier math here). If it’s a quarter pipe it’s not a perfect circle’s arc (and especially not, because you’d also have to have a consistent normal force with a vertical vector of two G but I don’t actually know their measurements and calculating the angle needed to give a vertical component of 2G at each second is beyond me right now) so I’m just going to peg it as a 90 arc of a circle, which would then have a circumference of 2160 ft, which gives us a diameter of ~687.5696 ft, and a radius (which would be the height of the ramp) of ~343.7whatever ft.

  So to get 600 ft. maybe our intrepid hero calculated his sustainable G force differently, or the angles make for a much longer rampe vertically, or maybe wind resistance and the loss of energy due to friction and other small influences on the wheels and bearings and such would have a bigger bearing on him.

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  • balsoft@lemmy.ml ⁨5⁩ ⁨weeks⁩ ago

    Eww imperial, but interesting maths nonetheless!

    I think your calculations are slightly wrong because you are assuming that the skateboarder will come to a full stop at the end of the quarter-pipe, which is not true - they will keep most of their momentum, it will just be redirected.

    Here’s my take on it

    The world-record quarter-pipe drop (which happened on September 25th, and I think this comic was inspired by it) was on a 60 m high ramp. The skater experienced ~28 m/s and ~4 G. The ramp was far from optimal, with the most curvature at the bottom, so the tightest radius was actually less than 60 m. By a = v² / r ⇒ r = v² / a, we can estimate that the radius of the ramp at the most curved point was actually closer to 20 m (which about lines up with the images online). This doesn’t take into account acceleration due to gravity, so let’s be conservative and say 30 m, which is quite reassuring for already.

    Now let’s calculate what the average acceleration would be throughout our trick in question, assuming a roughly constant speed over the entire quarter-pipe and only the centripetal force acting on the skateboarder. Of course neither will due to air resistance slowing the skateboarder down, and gravitational force increasing the contact force as time goes on, but those two are counteracting. Contact force increasing with gravity should win out (because air resistance depends on the velocity and so we get exponential decay or so, while contact force due to gravity will increase sinusoidaly or so), but I don’t think by much.

    Assuming a circular quarter-pipe, the centripetal acceleration is |a| = v² / r. With a terminal velocity of ~50 m/s and a radius of 600 ft ≈ 180 m, we get |a| ≈ 14 m/s², or just 1.5 g (at least at the top of the quarter-pipe), which is really easy to pull off. Even if cueball wouldn’t slow down at all, that would mean 2.5 g at the bottom, which is still doable. Adding a perpendicular air resistance acceleration (which at least at the top will be equal to 1 g for obvious reasons) to both, we get a range from 1.5 g to 2.7 g, which is much less than the world record and so I deem doable.

    Now let’s try being more precise.

    Since the contact force is always perpendicular to the velocity vector, it will not change the speed; so, let’s try thinking in the reference frame of the rotating object.

    So, the only changes in speed will come from the relationship between drag and gravity.

    Let’s assume that acceleration due to drag is just C * |v|, i.e. some constant times the speed.

    Let’s say x is the distance traveled along the ramp; then angle traveled α = x / (R * τ/4) radians.

    Acceleration due to gravity then becomes

    In that case we get acceleration due to gravity = g (cos α) = g cos (x / (R * τ/4))

    Finally,

    x’’ = g (cos (x / (R * τ/4))) - C x’

    where x₀ = 0, x’₀ = 50 m/s, x’'₀ = 0, R = 180 m, C = 1g / (50 m/s) ≈ 0.2 s⁻¹.

    I don’t know if there’s an analytical solution for this, but WolframAlpha refuses to find one. If you manage to wrangle this around and find out, let me know!

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    • burntbacon@discuss.tchncs.de ⁨5⁩ ⁨weeks⁩ ago

      Eww imperial

      He started it!

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