| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
06 Sep 2005 04:15:46 AM |
| Object: |
Herc's Juggler Problem - Not bad at all |
|-|erc wrote in
http://groups.google.com/group/sci.math/browse_frm/thread/afc2097a83bf687f/60414891f39430b8#60414891f39430b8
:
: > a juggler is juggling 3 balls in a standard pattern.
: > how fast does a bird have to fly in a horizontal straight
: > line to collide with all 3 balls.
: >
: > Jugglers hands are approx. 1 meter apart, balls reach
: > 1 meter higher, balls and bird are relatively small.
Extra assumptions for the dimwitted :
the projectiles are point sources, they have no wings,
no rubber, no wehl and tensor attractors.
how fast CAN a bird fly... (maximum speed) and collide
with all 3 balls, constant speed.
the balls move back and forth on the same parabolic curve,
(this is the easy version since noone will program a
simulation of real juggling with extra lateral motion)
(the hands are still and use 100% perpertual rebounce
with no losses)
standard pattern : the balls are hitting the hands at
evenly spaced intervals
Ah, that looks like a reasonable problem now, in fact quite
a good one, even if it is a bit laborious to state.
If I haven't messed up the GCSE physics and geometry, the
formula for the maximum bird speed is surprisingly simple,
mainly it seems due to the fact that if balls are thrown
from alternate hands at equal intervals at the maximum
rate then when two consecutive balls cross in flight is
exactly when the next ball is thrown!
Assume the balls travel in a parabolic path of horizontal
range (hand separation) 2D and maximum height H, and from
being thrown reach the peak of the parabola in time T.
[N.B - There are no vector dot-products in this post.
The dots indicate products of scalars!]
At time t after being thrown, the vertical speed of a ball,
v_t, is given by v_t = v_0 - g.t, where g is the acceleration
due to gravity. Thus T = v_0/g (the time where v_t = 0).
Also, assuming unit mass balls, equating the vertical part
of a ball's kinetic energy with its potential energy at the
peak of flight gives: 2.g.H = v_0^2 [ = g^2.T^2 ], so that
2.H = g.T^2
Likewise, at time t after being thrown the height, h_t, of
a ball is given by h_t = v_0.t - g.t^2/2 [ = g.(T.t - t^2/2) ]
As Herc, argues, assuming the bird travels at a uniform
speed, it must meet two of the balls together on one side
of the parabola.
Only consecutive balls are relevant, since alternate balls
(being thrown from the same hand) never meet, and the third
next ball is the same one being thrown from the opposite
hand.
Equating two h_t expressions for consecutive balls thrown
at times t = 0 and t = 2T/3 shows that these balls have
the same height when t = 4T/3, i.e. 2T/3 after the second
ball was thrown (which happens to be when the _next_ ball
is thrown!)
But as T < 2T/3, the first ball has passed the peak when
the second ball is thrown, and thus where their heights
are equal is where they meet, and the height, h, where
this occurs is given by h = g.(T.4T/3 - (4T/3)^2/2),
i.e. h = 4.g.T^2/9
The bird flies along a horizontal chord, length 2d, of
the parabola, and by definition of a parabola we have:
d^2/D^2 = (H - h)/H.
The bird's speed v, must be such that it crosses this
distance in the time, 2T/3, it takes for the third ball
to rise to height h, i.e.:
v = 2d/(2T/3)
= 3d/T
= 3D.SQRT((H - h)/H)/T
= 3D.SQRT(1/9)/T (using H, h = g.T^2/2, 4.g.T^2/9)
= D.SQRT(g/2H) (using T = SQRT(2H/g))
So for a bird in uniform horizontal flight the maximum
speed it can intercept all three balls is D.SQRT(g/2H).
(As a sanity check, this is dimensionally consistent.)
Cheers
John R Ramsden
.
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| User: "|-|erc" |
|
| Title: Re: Herc's Juggler Problem - Not bad at all |
07 Sep 2005 07:21:28 AM |
|
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<john_ramsden@sagitta-ps.com> wrote in message
:
: |-|erc wrote in
: http://groups.google.com/group/sci.math/browse_frm/thread/afc2097a83bf687f/60414891f39430b8#60414891f39430b8
: :
: >
: > : > a juggler is juggling 3 balls in a standard pattern.
: > : > how fast does a bird have to fly in a horizontal straight
: > : > line to collide with all 3 balls.
: > : >
: > : > Jugglers hands are approx. 1 meter apart, balls reach
: > : > 1 meter higher, balls and bird are relatively small.
: >
: > Extra assumptions for the dimwitted :
: > the projectiles are point sources, they have no wings,
: > no rubber, no wehl and tensor attractors.
: > how fast CAN a bird fly... (maximum speed) and collide
: > with all 3 balls, constant speed.
: > the balls move back and forth on the same parabolic curve,
: > (this is the easy version since noone will program a
: > simulation of real juggling with extra lateral motion)
: > (the hands are still and use 100% perpertual rebounce
: > with no losses)
: > standard pattern : the balls are hitting the hands at
: > evenly spaced intervals
:
: Ah, that looks like a reasonable problem now, in fact quite
: a good one, even if it is a bit laborious to state.
:
: If I haven't messed up the GCSE physics and geometry, the
: formula for the maximum bird speed is surprisingly simple,
: mainly it seems due to the fact that if balls are thrown
: from alternate hands at equal intervals at the maximum
: rate then when two consecutive balls cross in flight is
: exactly when the next ball is thrown!
:
: Assume the balls travel in a parabolic path of horizontal
: range (hand separation) 2D and maximum height H, and from
: being thrown reach the peak of the parabola in time T.
:
: [N.B - There are no vector dot-products in this post.
: The dots indicate products of scalars!]
:
: At time t after being thrown, the vertical speed of a ball,
: v_t, is given by v_t = v_0 - g.t, where g is the acceleration
: due to gravity. Thus T = v_0/g (the time where v_t = 0).
:
: Also, assuming unit mass balls, equating the vertical part
: of a ball's kinetic energy with its potential energy at the
: peak of flight gives: 2.g.H = v_0^2 [ = g^2.T^2 ], so that
: 2.H = g.T^2
:
: Likewise, at time t after being thrown the height, h_t, of
: a ball is given by h_t = v_0.t - g.t^2/2 [ = g.(T.t - t^2/2) ]
:
: As Herc, argues, assuming the bird travels at a uniform
: speed, it must meet two of the balls together on one side
: of the parabola.
:
: Only consecutive balls are relevant, since alternate balls
: (being thrown from the same hand) never meet, and the third
: next ball is the same one being thrown from the opposite
: hand.
:
: Equating two h_t expressions for consecutive balls thrown
: at times t = 0 and t = 2T/3 shows that these balls have
: the same height when t = 4T/3, i.e. 2T/3 after the second
: ball was thrown (which happens to be when the _next_ ball
: is thrown!)
:
: But as T < 2T/3, the first ball has passed the peak when
: the second ball is thrown, and thus where their heights
: are equal is where they meet, and the height, h, where
: this occurs is given by h = g.(T.4T/3 - (4T/3)^2/2),
: i.e. h = 4.g.T^2/9
:
: The bird flies along a horizontal chord, length 2d, of
: the parabola, and by definition of a parabola we have:
: d^2/D^2 = (H - h)/H.
:
: The bird's speed v, must be such that it crosses this
: distance in the time, 2T/3, it takes for the third ball
: to rise to height h, i.e.:
:
: v = 2d/(2T/3)
:
: = 3d/T
:
: = 3D.SQRT((H - h)/H)/T
:
: = 3D.SQRT(1/9)/T (using H, h = g.T^2/2, 4.g.T^2/9)
:
: = D.SQRT(g/2H) (using T = SQRT(2H/g))
:
: So for a bird in uniform horizontal flight the maximum
: speed it can intercept all three balls is D.SQRT(g/2H).
:
: (As a sanity check, this is dimensionally consistent.)
:
:
: Cheers
:
: John R Ramsden
:
thanks, guess I have to check if you're correct.
the ball goes up 1 meter and falls 1 meter in even time intervals.
Fall 1 meter.
1 = 1/2 10 t^2
1 = 5 t^2
t = sqrt(0.2)
t = 0.447
time in air = 2*t = 0.894
time for ball to return = 1.789 UP DOWN UP DOWN
CONSIDER ONE HAND
time between balls = 1.789 / 3 = 0.596
how high does the ball get half way between throws
t2 = .0.596 / 2
= 0.298
total drop = initial drop + remaining drop
0.447 = 0.149 + 0.298
s = 1/2 * 10 * 0.149^2
= 0.111
Height of ball at impact = 0.888m They do seem to hover up there!
At that moment, half way between throws with one hand, the other hand has a ball at 0,
since the hands are alternating, left throw, right throw, left throw right throw....
Takes another 0.298s for 3rd ball to reach 0.888m high.
Bird travels less than a meter in 0.298 seconds, about 1 meter / second.
John's answer is a few times quicker than mine, perhaps the distance is smaller than my estimated 30cm.
Herc
.
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| User: "" |
|
| Title: Re: Herc's Juggler Problem - Not bad at all |
07 Sep 2005 08:55:15 AM |
|
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|-|erc wrote:
<john_ramsden@sagitta-ps.com> wrote in message
:
: |-|erc wrote in
: http://groups.google.com/group/sci.math/browse_frm/thread/afc2097a83bf687f/60414891f39430b8#60414891f39430b8
:
: [...]
:
: Likewise, at time t after being thrown the height, h_t, of
: a ball is given by h_t = v_0.t - g.t^2/2 [ = g.(T.t - t^2/2) ]
:
: As Herc, argues, assuming the bird travels at a uniform
: speed, it must meet two of the balls together on one side
: of the parabola.
:
: Only consecutive balls are relevant, since alternate balls
: (being thrown from the same hand) never meet, and the third
: next ball is the same one being thrown from the opposite
: hand.
:
: Equating two h_t expressions for consecutive balls thrown
: at times t = 0 and t = 2T/3
Explicitly this equation (cancelling g from both sides) is:
T.t - t^2/2 = T.(t - 2T/3) - (t - 2T/3)^2
in which the LHS is the height of the first ball, and the RHS
is that of the second, thrown 2T/3 seconds later.
: shows that these balls have
: the same height when t = 4T/3, i.e. 2T/3 after the second
: ball was thrown (which happens to be when the _next_ ball
: is thrown!)
:
: [...]
:
: So for a bird in uniform horizontal flight the maximum
: speed it can intercept all three balls is D.SQRT(g/2H).
[...]
John's answer is a few times quicker than mine, perhaps
the distance is smaller than my estimated 30cm.
Assuming D = 0.5 m (half the hand separation), H = 1 m,
and g = 9.8 m/s^2, I get a bird speed of 1.1 m/s.
.
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| User: "|-|erc" |
|
| Title: Re: Herc's Juggler Problem - Not bad at all |
07 Sep 2005 05:07:40 PM |
|
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<john_ramsden@sagitta-ps.com> wrote in message :
: |-|erc wrote:
: >
: > <john_ramsden@sagitta-ps.com> wrote in message
: > :
: > : |-|erc wrote in
: > : http://groups.google.com/group/sci.math/browse_frm/thread/afc2097a83bf687f/60414891f39430b8#60414891f39430b8
: > :
: > : [...]
: > :
: > : Likewise, at time t after being thrown the height, h_t, of
: > : a ball is given by h_t = v_0.t - g.t^2/2 [ = g.(T.t - t^2/2) ]
: > :
: > : As Herc, argues, assuming the bird travels at a uniform
: > : speed, it must meet two of the balls together on one side
: > : of the parabola.
: > :
: > : Only consecutive balls are relevant, since alternate balls
: > : (being thrown from the same hand) never meet, and the third
: > : next ball is the same one being thrown from the opposite
: > : hand.
: > :
: > : Equating two h_t expressions for consecutive balls thrown
: > : at times t = 0 and t = 2T/3
:
: Explicitly this equation (cancelling g from both sides) is:
:
: T.t - t^2/2 = T.(t - 2T/3) - (t - 2T/3)^2
:
: in which the LHS is the height of the first ball, and the RHS
: is that of the second, thrown 2T/3 seconds later.
:
: > : shows that these balls have
: > : the same height when t = 4T/3, i.e. 2T/3 after the second
: > : ball was thrown (which happens to be when the _next_ ball
: > : is thrown!)
: > :
: > : [...]
: > :
: > : So for a bird in uniform horizontal flight the maximum
: > : speed it can intercept all three balls is D.SQRT(g/2H).
:
: > [...]
: >
: > John's answer is a few times quicker than mine, perhaps
: > the distance is smaller than my estimated 30cm.
:
: Assuming D = 0.5 m (half the hand separation), H = 1 m,
: and g = 9.8 m/s^2, I get a bird speed of 1.1 m/s.
:
I approximated the distance but got 1m/s, working looks similar in both our approaches so that's 1 point.
Which newsgroup?
Your question if you want to put up with continuous inane complaints, random guesses, phalic jokes, death threats, abuse, taunting
and belittling, motivation analysis, killfiling, ISP complaints, mob tactics and moronic pedanticism.
Herc
--
100% of bums prefer the streets to Centrelink
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| User: "" |
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| Title: Re: Herc's Juggler Problem - Not bad at all |
08 Sep 2005 03:12:56 AM |
|
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|-|erc wrote:
I approximated the distance but got 1m/s, working looks
similar in both our approaches so that's 1 point.
Which newsgroup?
sci.math, if it makes any difference.
.
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| User: "|-|erc" |
|
| Title: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all |
08 Sep 2005 04:12:32 AM |
|
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<john_ramsden@sagitta-ps.com> wrote in message :
: |-|erc wrote:
: >
: > I approximated the distance but got 1m/s, working looks
: > similar in both our approaches so that's 1 point.
: > Which newsgroup?
:
: sci.math, if it makes any difference.
After Q2
sci.math 2
uk.mensa 1
rec.mensa 0
alt.physics 0
sci.physics -1
people lose half a point for a wrong answer and you gain that half a point
1 point to the 1st correct answer and they get the serve.
give more clues the more tries they have
Your serve!
Herc
--
100% of bums prefer the streets to Centrelink
.
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| User: "Autymn D. C." |
|
| Title: Re: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all |
08 Sep 2005 06:21:09 AM |
|
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|-|erc wrote:
<john_ramsden@sagitta-ps.com> wrote in message :
: |-|erc wrote:
: >
: > I approximated the distance but got 1m/s, working looks
: > similar in both our approaches so that's 1 point.
: > Which newsgroup?
:
: sci.math, if it makes any difference.
After Q2
sci.math 2
uk.mensa 1
rec.mensa 0
alt.physics 0
sci.physics -1
Hey arsehole, you said at equal intervals and I solved the case for
bosonic juggling. The balls start at extrema of the biparabola and
reach the same height at once. The math guy went the hard route and
solved for fermionic juggling. My answer is faster than his. My bird
flies infinitely fast! Fix those scores!
-Aut
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| User: "" |
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| Title: Re: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all |
08 Sep 2005 07:52:50 AM |
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Autymn D. C. wrote:
|-|erc wrote:
[...]
After Q2
sci.math 2
uk.mensa 1
rec.mensa 0
alt.physics 0
sci.physics -1
Hey arsehole, you said at equal intervals and I solved the
case for bosonic juggling. The balls start at extrema of
the biparabola and reach the same height at once. The math
guy went the hard route and solved for fermionic juggling.
Debbie, what _are_ you talking about? A competent 14 year old
could solve this problem, once the conditions were explained.
It's not rocket science (well actually - er, no never mind).
I hope you don't mind people calling you Debbie (or Debs?);
but I feel uncomfortable using the name Autymn as it seems
like I'm misspelling "autumn", and you know how much you
hate spelling mistakes!
Cheers
John R Ramsden
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| User: "Autymn D. C." |
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| Title: Re: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all |
11 Sep 2005 01:04:28 AM |
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wrote:
Debbie, what _are_ you talking about? A competent 14 year old
could solve this problem, once the conditions were explained.
It's not rocket science (well actually - er, no never mind).
One doesn't learn kinematics at that age, or how to juggle. But if you
define "competent" to mean one who does, then one could solve anything
learning anything else. My juggling was better nonetheless.
I hope you don't mind people calling you Debbie (or Debs?);
but I feel uncomfortable using the name Autymn as it seems
like I'm misspelling "autumn", and you know how much you
hate spelling mistakes!
It's a name, not a word.
-Aut
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| User: "Jeremy Boden" |
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| Title: Re: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all |
17 Sep 2005 04:56:52 PM |
|
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In message <1126418668.311918.149710@g14g2000cwa.googlegroups.com>,
Autymn D. C. <lysdexia@sbcglobal.net> writes
john_ramsden@sagitta-ps.com wrote:
Debbie, what _are_ you talking about? A competent 14 year old
could solve this problem, once the conditions were explained.
It's not rocket science (well actually - er, no never mind).
One doesn't learn kinematics at that age, or how to juggle. But if you
define "competent" to mean one who does, then one could solve anything
learning anything else. My juggling was better nonetheless.
I hope you don't mind people calling you Debbie (or Debs?);
but I feel uncomfortable using the name Autymn as it seems
like I'm misspelling "autumn", and you know how much you
hate spelling mistakes!
It's a name, not a word.
-Aut
But a name *is* a word or words.
--
Jeremy Boden
.
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| User: "Autymn D. C." |
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| Title: Re: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all |
18 Sep 2005 06:01:28 AM |
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Jeremy Boden wrote:
But a name *is* a word or words.
not always
.
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| User: "" |
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| Title: Re: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all |
08 Sep 2005 06:25:58 AM |
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Hey arsehole, you said at equal intervals and I solved the case for
bosonic juggling. The balls start at extrema of the biparabola and
reach the same height at once. The math guy went the hard route and
solved for fermionic juggling. My answer is faster than his. My bird
flies infinitely fast! Fix those scores!
***************************
Pecans are an excellent source of polyunsaturated oil.
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