Author Topic: Is there a record for High Toss? (29.55m)  (Read 35729 times)

mrpink

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Is there a record for High Toss? (29.55m)
« Reply #20 on: May 04, 2005, 08:14:00 PM »
Some confusion and lots of formulas... so Ill just add some more...
More about gravity and falling objects can be found in wikipedia:
http://en.wikipedia.org/wiki/Gravity#Newton.27s_law_of_universal_gravitation

The formula I found there for calculating the distance traveled for a freely falling object falling t seconds is
d=1/2*g*t^2

Formula for calculating diabolothrowheight becomes
h=g/2(time/2)^2   (here t is the total airtime)
aproximating g=10 we can simplify and get

h=5/4*t^2 alternativley

h=1.25*t*t (wich is verry easy to remember)

t is still the total airtime upp and down.


Lemony you forgot  the factor 1/2 in your formula so your throw is sadly enough just 20 m...

JGherkin

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Is there a record for High Toss? (29.55m)
« Reply #21 on: May 05, 2005, 07:42:57 PM »
Regarding Seans confusion over what xfirebladex said, Sean is assuming acceleration on the way up as a negative acceleration, the decelleration as it travels, whereas xfirebladex is assuming the acceleration from the throw, the acceleration of the diabolo from the start of the throw through to when you release it..........I could be wrong though :)

I swear I've just been doing this in physics getting ready for my exam in a couple of weeks but its all way above my head...o dear  :(

Sean

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Is there a record for High Toss? (29.55m)
« Reply #22 on: May 05, 2005, 09:31:10 PM »
Quote from: JGherkin
xfirebladex is assuming the acceleration from the throw, the acceleration of the diabolo from the start of the throw through to when you release it

That could be, but knowing the acceleration during the throw isn't necessary to find out the height of the throw in this situation. All you need is the time from release back to the same point. Sure, you could also calculate the peak height if you knew the acceleration applied to the diabolo and the quantity of time that that acceleration was applied for... but measuring that acceleration accurately wouldn't be easy, not to mention that the acceleration applied by throwing a diabolo is unlikely to be constant. Then we get into "jerk" (or "jolt" for some of you UKers) and so on... timing the throw is much simpler. :)

Regarding a point made earlier about not knowing when it will reach its peak:
The diabolo will reach its peak at exactly half the total time and be caught with the same speed as it was thrown with (assuming no air friction).

Chiok

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Is there a record for High Toss? (29.55m)
« Reply #23 on: May 07, 2005, 04:59:12 PM »
Quote from: Sean
Then we get into "jerk" (or "jolt" for some of you UKers) and so on... timing the throw is much simpler. :)
Regarding a point made earlier about not knowing when it will reach its peak:
The diabolo will reach its peak at exactly half the total time and be caught with the same speed as it was thrown with (assuming no air friction).

Oh right, I thought the speed at which it went up wouldn't be the same as you're giving it force to send it up.  I thought it would be travelling faster.  Possibly not, I'm really not sure (about it going up.)

I'm in the UK and I thought they were called "jerks" too, and then rate of jerks is snaps, rate or snaps is crackles, rate of crackles is pops.
I kid you not.

Chiok
www.gravityvomit.co.uk - Gravity pulls down, we throw up.
University of Bath Juggling and Circus Skills

kragen

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Is there a record for High Toss? (29.55m)
« Reply #24 on: May 08, 2005, 02:26:00 PM »
ok, even though several people have the answer, all the explanations are fairly difficult to follow, so here is my attempt :P

Ignoring air resistance and all the other complicated stuf, the height of the diabolo with time is a parabola:

Code: [Select]
h = v t - 1/2 g t^2

where h is the height at time t, and v is the initial velocity that the diabolo was thown at.

Code: [Select]
h = t ( v - 1/2 g t )

This equals 0 at two points, when t=0, i.e. when the diabolo is thrown, and when v = 0.5 g t, when it hits the ground again. We know the time at this point (because we carefully timed it :P) so lets call it time T, meaning we know v.

Code: [Select]
v = 1/2 g T

The maximum height is at time T / 2, so substituting v, and t = T / 2 into the eq-n for h...

Code: [Select]
H = 1/2 g T ( 1/2 T ) - 1/2 g ( 1/2 T )^2

H = g / 8 * T^2


Where g = 9.8 (approximately) is the gravity constant for earth, H is the maximum height the diabolo reaches, and T is the time it took to go up and come back down again :)

Sorry, that was probably more long-winded than neccessary, but I was having problems following the other explanations, and im on the 2nd year of a maths degree :P

Here is a pretty picture:

Justin
Gravity Vomit - Gravity pulls down, we throw up.

mrpink

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Is there a record for High Toss? (29.55m)
« Reply #25 on: May 08, 2005, 06:19:56 PM »
So now we have a formula that is easy for annybody to use.'
How about some experimenting?
It would be great if we could have some actual record atempts here as well.
What is the highest throw you can do? (and is it sientifistically documented?).
Mabe we wont find the highest throw ever done but it culd be fun annyway...

Matt_H

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high throws
« Reply #26 on: May 09, 2005, 11:53:15 AM »
Well i jumped on this one a bit late but Kragen's analysis is correct, all that is needed are the equations of motion, as for negative sign for gravity (-9.81) this just denotes the direction of the acceleration so when using the formula

D=ut + 1/2at^2

the distance will come out negative as the diabolo is traveling down and the time, t is half the total time of flight.

regards

Dr Matt_h

p.s. i do have a Ph.D in Physics.

seán_

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Is there a record for High Toss? (29.55m)
« Reply #27 on: May 09, 2005, 01:12:42 PM »
Define your terms slacker.

Do they give Phd's out with cereal boxes these days? No wonder we have three doctors at Hullabaloo juggling club.

how about nailing the friction issue and the initial variance in starting height?

Matt_H

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high throws
« Reply #28 on: May 09, 2005, 01:15:56 PM »
Oi listen here ya cheeky git, as i said Kragens analysis is correct so the terms  in the tiny formula i put in are defined in his post ...... or did you not read that  :wink:

if you really want me to do the friction analysis i will but you may regret it !



regards

Dr Matt_h

seán_

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Is there a record for High Toss? (29.55m)
« Reply #29 on: May 09, 2005, 01:29:57 PM »
you're the one who is allways banging on about the lementable state of undergrads work so of course I read it (and looked at the great graph), but I was expecting a full analysis from you.

Any body with more nouse than me willing to rattle of a ready reckoner table so we can easily work out heights from times?

Matt_H

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high throws
« Reply #30 on: May 09, 2005, 01:54:36 PM »
what do you need a ready reckoner table for ?

distance in meters = (10 x (t x t))/2
 t is time in seconds, use the total time from throw to it hitting you in the head divided by 2.

so a throw lasting 2 seconds means t =1 so thats 10/2 = 5 metres,
4 seconds means t = 2 so thats (10 x (2 x 2))/2 = 20 metres,

now Sean_ i'm sure you can do a little maths in ya head   :wink:

Later

Dr Matt_H

seán_

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Is there a record for High Toss? (29.55m)
« Reply #31 on: May 09, 2005, 02:16:38 PM »
look, its fingers for counting or fingers for holding the handsticks! One or the other.

I cant do your height fudge (rewriting newtonian physics) AND work out the full fluid mechanics analysis (not a mere laminar one) without taking my shoes of either.

F= 1/2rC Au^2 = 1/2 prC r^2u (C being the drag coeficient, 0.42 for a smooth sphere, so another guestimate required there)

I'm going to stick to my photographic memory* to picture a heights throw and compare it to buildings of known height for the moment.

*more of an amatuer photographic memory actually. Some of the images are out of focus, a lot of peoples heads are out of shot and there is this whole black section from 17 - 21 years old when I left the lens cap on

Matt_H

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high throws
« Reply #32 on: May 09, 2005, 02:20:54 PM »
O i see laminar flows ey, maybe you would like to explain them to the rest of the group  :wink:

or maybe you'll wait till i explain it to you over messanger then paste it in hey Sean_  :wink:

lol


regards

Dr Matt_h

seán_

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Is there a record for High Toss? (29.55m)
« Reply #33 on: May 09, 2005, 02:30:41 PM »
I could but that would involve effort on my part. I might as well point them to this
http://academic.reed.edu/physics/courses/Phys100/Lab%20Manuals/Air%20Resistance/Air.Resistance.pdf
as it has pretty pictures and everything (and it saves you having to write it all out in messenger longhand ;) )

mrpink

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Is there a record for High Toss? (29.55m)
« Reply #34 on: May 09, 2005, 02:57:11 PM »
Wow fluid dynamics... soon we will have people talking about chaostheory and  silly butterflies making bad wether...
Dont get me wrong, I really think this is quiet interesting.
The diabolo is spinning pretty quickly. How does this affect the air resistance?
If I want to throw high shuld i have lots of spinn or barley enough to keep it stable?

Matt_H

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high throws
« Reply #35 on: May 09, 2005, 03:03:58 PM »
if you consider just the rotation/spin of the diabolo in the air then the diabolo will slow down due to air resistance so its just the same frictional problem but applied to the spin rather than the diabolo falling. Either way the difference it makes in the hieght calculations is small enough to be ignored.
The other consideration is that the diabolo is launched from a standing position and not the floor so your measurement of the time as being 1/2 the total time in flight (from throw to landing) is not strictly true and is therfore the largest source of error .

now about chaos theory ...........................

regards

Dr Matt_H

seán_

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Is there a record for High Toss? (29.55m)
« Reply #36 on: May 09, 2005, 03:06:15 PM »
I have mulled over that in recent days, the spin rate would come in somewhere and in all honesty I reckon Dr Matt's the man to answer then but unless it had a great bearing on results I would say the two points to remember with spin for high throws are

1. low spin = les stable and less forgiving when you launch it
2. high spin = more stable and running further after it if you drop it :)

As you can see, Best left to Dr. Matt_H (whom I failed to credit for previous link. Will the accusations of academic mallpractise ever cease?)

Matt_H

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high throws
« Reply #37 on: May 09, 2005, 03:08:25 PM »
Fair point Sean, with the diabolo spinning faster it is more likely to not to tumble and hence a more stable path will be followed by the diabolo.

regards

Dr Matt_H

seán_

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Is there a record for High Toss? (29.55m)
« Reply #38 on: May 09, 2005, 03:17:00 PM »
but on a deeper level, would the rotatinal forces affect the actual diabolos flight, Say in relation to air friction, lift or some kind of freaky antigravity deal?

to further show my lack of understanding when it comes to physics, I do have time for the theory that you can achive both levitation and perpetual motion by strapping a piece of buttered toast (butter side up) to the back of a cat then let go of the cat

Matt_H

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high throw
« Reply #39 on: May 09, 2005, 03:19:03 PM »
No

 

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