Is there a record for High Toss? (29.55m)

[Chiok]

After some googling on bogleg and other sources, I wondered whether there was an actually record for the highest throw of a diabolo by an individual.  A guy I know got some arlequinos (or piccolinos, not sure) for christmas, and we've just been chucking them high into the air.  Tried it in our sports hall and it ricocheted off the roof (what other times does an object go that high in a hall?)  And being outdoors now, it just gets thrown higher.

Is there a record for this?  Possibly with people standing on different levels of a hotel or something.

Chiok

[norbi]

well i'd say the record is 'ridiculously high' as there is no good way to measure the height, unless somehow you could attatch a devise to the diabolo that measured the speed it was going, and you could count the airtime, then voila distance = speed x time

But barring that, no way to tell really.

[The Void]

Was Newton crucified in vain?

If your physics teacher was alive today, Norbi, she'd turn in her grave.

The Void
..................
Hyperbolic

[seán_]

or refererence the height against a known height say a building
I know norbi was having some fun with how many pirouettes skips etc beneath a high throw at sheffield. Remis 7 skips and a forward roll a few years back in Leeds is a hard one to beat though.

[norbi]

what? 2mph for 2 hours = 4 miles*, right?

*for example

[seán_]

got to take gravity into consideration Norbi

[Alex]

All you have to do is count the time its in the air.  You can figure out how high it went from that.

[seán_]

oops I forgot decelleration due to gravity is a constant. (still easier to measure against a known height though )

Should get Matt_H onto the case, he has a PhD in Physics he should have the required sum type stuff to hand

[mrpink]

If we dont take the airfriction into consideration and apply newtons laws of gravity and a little algebra and the aproximate value 10 for the gravityconstant (its about 9.82 really) the formula to calculate the height of a diabolothrow in terms of airtime becomes

h=1.25*t*t

h is the height in meters
t is the total airtime in seconds

a 4 second throw wuld be aproximatley  1.25*4*4=20 m high
and a 5 second throw 1.25*5*5=30 m
and a 6 second throw 1.25*6*6= 45m



Another way to messure the height of a throw wld be to put an observer (possibly a camera) at a known distance say 20 meters from the thrower.
He wuld then mesure te angle he observes the diabolo in when it peaks.
A little trigionometery is all that is neded to get the result then...

[Ifrit]

i remember seeing something on Tv about this guy who did have the record. I think it was about the 5 second mark. but i cant get more than 3 and a half on a good day. However i think his diabolo was abit on the small size.

[el_grimley]

You can also work it out if you happen to know the shutter spead on the camera and you can work it out by the length of the motion blur. But as far as high goes I would say uber high.


g is for geek

[Twilite]

Displacement = 0.5 * acceleration * time * time

I do believe, where acceleration (decelleration in this case) is 9.81, gravity.

I'm here instead of revising for my physics exams, so I hope that's right!

[Andy]

just remember you have to half the time to get the time it was at the top, and also that the acceleration due to gravity is negative. You also need to know the velocity at the start if this is the method - using
distance = - acc due to gravity * time * time + velocity at start * time

it would be much easier to do the trig and angle method mentioned above somewhere using a clinometer i think its called...

[Alex]

I just worked this out.  In this case, t (time) would be half of the recorded time, so it would be just the time going up:

Given:
Vf^2 = Vi^2 + at
x = .5at^2 + Vit

Work:
Vf^2 = Vi^2 + at

Vf = 0

a = -9.8

9.8t = Vi^2

Vi = root (9.8t)


x = .5at^2 + Vit

x = -4.9 t^2 + t * root (9.8t)


I believe that should be right.  If anyone spots a mistake, please point it out.  Also, it could probably be simplified a bit more, but I'm kinda lazy.

[Yoda]

Well, I just found the guy who can do the highest throw ever, and I doubt anyone can beat him!

Open the DSSS and type z0000000000000000000000000000000000 in the high pattern. I counted 35 seconds.
If we calculate the height, with gravity as 10m/s^2
V=v+at
0=v-10*35
v=350

S=v*t+1/2a*t^2
S=12250+6125
S=18375m

So, if newton was correct (unfortunately he wasn't) the guy threw it up 18,375 Kilometers high.


Can anyone beat him???

[Andy]

Alex - its Vf^2 = Vi^2 2ax
Yoda - You forgot to half the time - the velocity when the time equals 35 is the opposite to the starting velocity ie 2v = 350 so his throw would be only just over 12km(!) should point out you can change his speed - it would appear he can throw higher....

[mrpink]

I did a high throw last summer , it still hasn't come down...
But since its kind of stuck on top of a roof I dont think it beats the 12k throw...

[lemony]

well i could be wrong but i rather thought the equation was just

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(time/2)^2 * a = H

(think in terms of units, time is in s, accelleration is in M/s^2 therefore the result must be in metres)

which means that given that i can do just over a 4 second throw (timed by a mate yesterday) then i can throw:

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(4/2) = 2

2^2 = 4

4*10 = 40 Metres high (taking gravity as 10M/s^2)

that seems vaguely feasible to me... but correct me if i'm being dense

[Chiok]

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(4/2) = 2

2^2 = 4

4*10 = 40 Metres high (taking gravity as 10M/s^2)

Well, you square the time first and then divide by 2, so actually it's 80m IF it took 4 seconds to come down from the top of it's fall (i.e at rest).  If you started timing from when it started coming down, then it would have been an 78.48m throw if assumed no air resistance and only under the effect of gravity.

Of course it's hard to say when it starts coming down, and also the acceleration of it going up isn't the same as it coming down because you threw it.  I'm still in favour of the tall building with windows and standing on the right floor.

Chiok

[Sean]

the acceleration of it going up isn't the same as it coming down because you threw it.

Excuse me? :-s

Ignoring air friction, a projectile will have a constant acceleration towards the Earth of 9.81 m/s*s.

Accounting for air friction, the downward acceleration will be greater going up than coming down.