best Muni under $500?



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

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eenwieler wrote:
> *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.

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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.

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johnhimsworth wrote:
> *Ooh, threads on physics. Time for one of my essays
> then... *

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.

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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

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capital punishment for stupidity, but why don't we just take the safety
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Ken Cline wrote:
> * 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
>
> *

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.

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"john_childs" <[email protected]> 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
 
"tennisgh22" <[email protected]> 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 :)