January 27, 2009 10:40 AM   Subscribe

As a purely hypothetical question, if a five-pound cat were to fall from an arbitrarily tall point, how long would it take him to reach terminal velocity? How fast would he be traveling at that point? Part II: If, say, someone was standing at the initial point, how long would it take the cat to drop completely out of sight? What is the longest distance an unaided human eye can perceive an armload-sized object, in other words? Assume the area is lit at about a 60-Watt bulb in a uniform manner.

No cats were or will be harmed in the answering of this question. I'm just very bad at math and am unable to come up with any kind of sensible answer to a physics question posed by an acquaintance. I'm hoping someone in the hivemind will be better able to explain physics to me.
posted by Scattercat to Science & Nature (12 answers total) 1 user marked this as a favorite
This describes what may be the only study precisely related to cat terminal velocity, which was estimated to be 60 mph.
posted by Zed at 10:58 AM on January 27, 2009

As a point of reference, the rule of thumb when skydiving is that the first thousand feet of freefall takes ten seconds, and every thousand feet after that takes five seconds, so terminal velocity is reached somewhere in the first ten seconds, and it's around two-hundred ft/sec. This assumes a stable, spread eagle body position. Assuming the cat was flailing and/or otherwise unstable it might fall a little faster, but probably not much.

Tell him to open the 'chute a little early the first few times; if he experiences "ground rush" it's too late.
posted by dinger at 11:10 AM on January 27, 2009

Terminal velocity of a cat is around 28m/s.

It takes about 2.9s to reach that speed.

Again from Wikipedia, the angular resolution of the human eye is around 30-60cm at 1km distance (and 30-60cm is about the size of a cat). So at 1km the cat should be roughly on the point of being too distant to see.

A cat would take about 14 seconds to fall 1km in a vacuum.

Accounting for terminal velocity is slightly more complicated, so I'll leave that to someone else.
Ditto the 60W bulb.
posted by le morte de bea arthur at 11:12 AM on January 27, 2009

Okay, so...

*whips out Calculator*

A cat's terminal velocity is somewhere around one and a half feet per second? (Assuming the 60 MPH figure has any validity.) Would they still reach it in the first ten seconds? (Acceleration from gravity is constant, isn't it?)
posted by Scattercat at 11:14 AM on January 27, 2009

Nevermind, my math was off. I typed things wrong. (This is why I'm asking you guys!)
posted by Scattercat at 11:14 AM on January 27, 2009

Okay, so thirty seconds of falling would likely leave the cat too far away to see, if I understand aright? How does falling in air compare to falling in a vacuum?
posted by Scattercat at 11:16 AM on January 27, 2009

Terminal velocity is not related to the weight of an object. It's the speed where the air resistance balances the downward pull of gravity. That means that there is not a set terminal velocity - a fat, hairy cat with it's legs stretched out will fall slower than a skinny hairless cat in a high dive pose.

I seem to remember that gravity 'pulls' (accelerates) at 9.8 m/s per second. So each second, the object gets 9.8 meters per second faster. I also seem to think that a human sky-diver falls at around 150 mph.

150 mph is about 67 m/s.
So about 6.8 seconds.
For cat with the air resistance of a human.

Part II - no idea.
posted by Xhris at 11:21 AM on January 27, 2009

You can't just use v=at here, though, as the cat's air resistance isn't constant while it flips into position and uses its flying squirrel style kung fu. (I'd expect only a small fraction of a second's difference from la morta de bea arthur's 2.9s, though, not anything your observer could notice.)
posted by Zed at 12:20 PM on January 27, 2009

Well, the physics answer is that, in an ideal world, the cat never reaches terminal velocity. All we can say is that as time goes on, the cat gets closer and closer to terminal velocity.

What we can calculate is how long the cat will take to reach 99% of terminal velocity (or whatever percentage of terminal velocity you care to use). Wikipedia has the derivation of the formula to use (click on show next to Derivation of the solution for the velocity v as a function of time t).

We can use t = arctanh(αv)/αg
From earlier in the derivation, α = (k/mg)1/2
We know that when the cat is at terminal velocity (vt = 28m/s per le morte de bea arthur), the net force on the cat is zero, so the force due to gravity is equal to the force due to air resistance: mg = kvt2
Therefore α = 1/vt

So we now have t = vtarctanh(v/vt)/g
We can then calculate the time to reach 99% of terminal velocity, which is 7.6s.

An exact answer to part II would require more calculus and so is left as an exercise for the reader. Very roughly, the cat travels about 150m in those first 7.6s and then continues to fall at a speed near 28m/s, so it takes the cat roughly 38s to fall 1km, which is (again, per le morte de bea arthur) about as far as you will be able to see it.
posted by ssg at 12:22 PM on January 27, 2009

man, I spent like thirty minutes deriving an answer to this, and then ssg tells me it's on Wikipedia??? crap.

Here are a couple random additions that might still be useful, though. By solving the equation of motion, you end up finding that the velocity as a function of time goes like


where t_drag is sqrt(2m/(rho*Cd*A*g)), with m the mass of the cat, rho the density of air, Cd the drag coefficient, A some measure of area, and g the acceleration due to gravity. V_t is the terminal velocity (which you can find by taking the net force=0), ~sqrt(2*m*g/(rho*Cd*A)). The drag coefficient Cd will vary depending on whether the cat is curled up into a ball or splayed out or whatever, but for now let's just say it's a number of order one.

For air, rho is about 1.2 kg per m^3 at sea level. (Oddly, I checked _that_ part on wikipedia...) Let's say the cat has an area of 0.05 square meters, and take Cd=1. This gives us a terminal velocity of 27.1 m/s, which is about what people have quoted above. The characteristic timescale is t_drag=2.8 s -- so after one 2.8 s interval has elapsed, the cat's trucking along at 76% of terminal velocity; after 2 such intervals it's at 96%, after 3 it's at 99.5%.

The distance traveled is

x~ (2m/(rho*Cd*A))*ln(cosh(t/t_drag))

which is basically v_term*t_drag*(an increasing function of t, which asymptotically approaches t). So after 2.8 seconds, the cat's dropped about 32 meters -- i.e. about half as much as it would've fallen had it been going at terminal velocity the whole time. For large times, you can essentially just take x=v_terminal times the time.

My sense is that formulating an exact answer to part 2 is probably so dependent on what you assume that you're better off guesstimating. Do you think you can still see a cat 32 meters away? How about at 64? etc, etc.
posted by chalkbored at 1:38 PM on January 27, 2009

Thank you everyone for your help! I hope you are glad to know that

A) You have assisted me in a completely pointless endeavor, but one which I enjoyed and

B) I have no idea what ssg or chalkboard even said, but it sure SOUNDED cool.

posted by Scattercat at 4:31 PM on January 27, 2009

Assume the area is lit at about a 60-Watt bulb in a uniform manner.
FWIW 1 km is probably reasonable for being able to see the cat under good illumination--say under decent sunlight.

However if a 60-watt bulb is your ONLY source of illumination, that isn't much.

But it's not easy to figure out how far exactly you could see because it depends on much else than the source of illumination, including the color of the cat, the surroundings, how dark adapted your eyes area, etc.
posted by flug at 5:20 PM on January 27, 2009

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