i should go for the KH24 in your case. But in my case I will safe a lot of money and buy a stronger one because I have already bent my KH crank ( after 3 months:( ) So if you do a few drops. Not very big ones, you bent you crank. I have jumped 4 thimes of a 5 foot skate ramp. So it isn’t strong enough.
Ferko
A five foot drop to flat is pretty big. If you don’t roll out properly on a drop that big you’re going to be bending or breaking cranks. I’d bet you could even break a Profile crank if you do repeated 5 foot drops to flat with poor technique.
How smoothly can you land a 3 foot drop to flat? Can you roll it out every time? Do you land with a thud and a double bounce? Are you able to land lightly and collapse your body to absorb the impact?
Technique has a lot to do with how well your unicycle will survive big drops, especially to flat. If you’re a big guy and you land big drops like a ton of bricks, you’re going to be breaking or bending almost any crank.
I can do 3 foot drops to flat fairly smoothly. I still sometimes mess up and land poorly, but most of the time I land them smoothly. I don’t land 4 foot drops as well. I’ll land and sometimes get a double bounce because I don’t roll out and/or collapse my body properly to absorb the impact. Until I can get 4 foot drops as smooth as my 3 foot drops I’m not going to do anything higher.
KH cranks can handle 5 foot drops. Kris uses KH cranks now and he does big drops. He just lands all of his drops very smoothly. If you’re 180+ pounds and you land a 5 foot drop poorly, yeah, you probably can bend the crank.
Adding just a foot to the drop height adds a lot of additional energy that needs to be absorbed. It’s an exponential increase in energy. Doubling the height does a lot more than double the amount of energy. I’m too lazy right now to go get my physics book to find the exact equation, but I know it’s exponential.
Anyways, if you’re bending cranks like the KH cranks it would be a good idea to get real smooth at smaller drops before going for the bigger drops. The alternative is to get a Profile hub and Profile Dirtjumper cranks and hope for the best. But even Dirtjumper cranks can break.
Re: best Muni under $500?
“john_childs” <john_childs@NoEmail.Message.Poster.at.Unicyclist.com> writes:
> Adding just a foot to the drop height adds a lot of additional energy
> that needs to be absorbed. It’s an exponential increase in energy.
> Doubling the height does a lot more than double the amount of energy.
> I’m too lazy right now to go get my physics book to find the exact
> equation, but I know it’s exponential.
I’m not surprised that adding height feels dramatically harder.
There’s only so much energy your body can absorb, only so far you can
roll out without losing balance, plus everything happens faster as you
hit harder.
Still, the amount of energy goes up linearly with height. The
equation is E = mgh, where m is mass, g is gravitational acceleration,
and h is the height dropped.
Ken
I just calculated this out, i dunno whether i helps your arguement or what. its pretty interesting tho
DROP HEIGHT: SPEED WHEN YOU HIT THE GROUND:
1 foot 3.9 mph
2 feet 5.4 mph
3 feet 6.7 mph
4 feet 7.8 mph
5 feet 8.7 mph
6 feet 9.5 mph
and if you’re kris holm,
100 feet 39 mph
does all that sound right? check my math if you think im crazy.
-grant
Re: Re: best Muni under $500?
That’s for potential energy. For a drop we’re interested in the kinetic energy and the impulse and the momentum. Kinetic energy increases with the square of the velocity. That’s where the exponential component would come in.
Ugh, It’s been a long time since I’ve done any physics. I hope I’ve got that right. I might have to dust off my physics book.
im taking physics at the moment and i think john is right.
kinetic energy = 1/2 (mass)(velocity)^2
thus, velocity is being squared, so it is in fact exponential.
looks like you dont have to worry about getting out your physics book after all, john 
-grant
Ooh, threads on physics. Time for one of my essays then…
Kinetic energy is proportioanl to the ssquare of the speed (KE = (1/2)mv^2 ) But speed doesn’t increase linearly with drop height. Notice from tennisgh32’s numbers that speed from 100’ is only 10 times higher than 1’, where it would be 100 times if it was linear.
Potential energy is PE = mgh ( or mass x gravity x height as words) so it’s linear with height increase. When you drop, all that potential enegry is converted into kinetic energy, so your kinetic energy when you hit the floor will increase linearly with drop height. Because you accelerate as you fall, for the last foot you’re travelling faster, so while you’re still accelerating (at 9.8m per second per second) the actual time for which you are accelerating is shorter, so you don’t speed up as much.
If you start from 0 vertical speed, v^2 = 2as, so velocity is the square root of accelearation x height (ie distance travelled) x 2. Which makes sense, 'cause the kinetic energy is the square of a square root, so it’ll be linear.
Just as a vague nod to the topic, that means that weight is as important as height, so you should get the onza. Or have a smaller lunch before a ride.
John
P.S. If anyone’s doing GCSE or A-level physics this year, I hope you paid attention to that. You never know when questions on unicycling will turn up. That, and ping pong balls.
Looks like I should dust off my physics book. I used to know all that stuff pretty well. It’s just been a long time since I’ve used any of it. At least my vague recollection of kinetic energy was right. Unfortunately my vague recollection of velocity due to drop height was not right.
Knowing how drop height and mass affects the jolt (impulse) when you land is a good thing to know. It’s linear.
Re: best Muni under $500?
“tennisgh22” <tennisgh22@NoEmail.Message.Poster.at.Unicyclist.com> writes:
> kinetic energy = 1/2 (mass)(velocity)^2
Quite. However, the kinetic energy is converted from potential
energy, and is therefore bounded by PE (by the First Law of
Thermodynamics).
> thus, velocity is being squared, so it is in fact exponential.
No. Velocity is proportional to the square root of distance, so it
(energy) is in fact linear (with distance).
But this is a roundabout way to analyze kinetic energy of a falling
object. Step back and ask yourself what kinetic energy is. A very
relevant answer is the capacity to do work, e.g. exert a force over
distance. But exerting a force, m*g, over a distance, h, is exactly
how that kinetic energy was acquired.
> looks like you dont have to worry about getting out your physics book
> after all, john
You might want to have another look at yours