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 Post subject: More travel fun
PostPosted: Wed, 17 Aug 2011 03:37:07 UTC 
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A[U@u].........................k..........................[V@v,W@w]B
Straight path AB, length k.
U (speed = u) is at A, V (speed = v) and W (speed = w) are at B.
All leave at same time.

U meets V at point C, picks up V, turns around, heads towards A, drops V
off at point D, turns around and heads towards B; V continues towards A.

U meets W at point E, picks up W, turns around, heads towards A:
they arrive at A at same time as V.

When V and W were "aboard" U, speed was u.
No loss of time at pick-ups and turn-arounds!

BC = a, AD = b and BE = c.

What is k in terms of a,b,c ?

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 Post subject: Re: More travel fun
PostPosted: Wed, 17 Aug 2011 06:19:25 UTC 
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Denis wrote:
A[U@u].........................k..........................[V@v,W@w]B
Straight path AB, length k.
U (speed = u) is at A, V (speed = v) and W (speed = w) are at B.
All leave at same time.

U meets V at point C, picks up V, turns around, heads towards A, drops V
off at point D, turns around and heads towards B; V continues towards A.

U meets W at point E, picks up W, turns around, heads towards A:
they arrive at A at same time as V.

When V and W were "aboard" U, speed was u.
No loss of time at pick-ups and turn-arounds!

BC = a, AD = b and BE = c.

What is k in terms of a,b,c ?


Assuming D is between A and B?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: More travel fun
PostPosted: Wed, 17 Aug 2011 09:42:16 UTC 
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outermeasure wrote:
Assuming D is between A and B?

Yes, of course...sorry...plus E also.

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 Post subject: Re: More travel fun
PostPosted: Wed, 17 Aug 2011 09:51:30 UTC 
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Spoiler:
From the start to when U,V meet at C, U travelled distance k-a at speed u and V travelled distance a at speed v.

From drop off at D to all meet at A, U travelled distance DE+EA=(k-b-c)+(k-c)=2k-b-2c at speed u and V travelled distance b at speed v.

Hence \dfrac{k-a}{a}=\dfrac{2k-b-2c}{b}, which rearranges to k=\dfrac{2ac}{2a-b}.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: More travel fun
PostPosted: Wed, 17 Aug 2011 20:32:30 UTC 
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Yep; matches mine; as example:
a = 40, b = 20 and c = 90 makes k = 120 ; (speeds u:v:w = 10:5:6)

If the given a was AC instead of BC (thus BC = k - a), then we have a slightly different:
a / (k - a) = (2k - b - 2c) / b , leading to a quite more complicated:
k = [SQRT(d^2 - 8e) + d] / 4 where d = 2a + b + 2c and e = 2ac

I'm unable to make that simpler; can you?

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