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SHOOTING POOL By: Alvaro Francisco Manuel Western Oregon University
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Summary We have a (3,4) pool table. Starting position was at the lower left corner. Five bounces before landing in the lower right corner pocket.
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PROBLEM STATEMENT Our task is to figure out a general formula to predict the number of bounces before landing in a corner pocket for any size pool table As well as, predicting which corner pocket.
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ASSUMPTION
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NOTATION We will refer to our pool table with two dimensions as (n,m) n corresponds to the vertical dimension (y-axis) and m corresponds to the horizontal dimension (x-axis) Lower Left: LL Lower Right: LR Upper Left: UL Upper Right: UR T(n,m) is defined to be the size of pool table. B(n,m) is defined to be the number of bounces P(n,m) is defined to be the corner pocket the ball lands in. gcd(n,m) is defined to be the Greatest Common Divisor of n and m. lcm(n,m) is defined to be the Least Common Multiple of n and m.
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NOTATION 4 3 For example, we have T(3,4). The starting position is at the LL corner pocket. B(3,4) = 5. P(3,4) = LR (Lower Right)
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MY STRATEGY The plan was to work on many examples starting with small table size.
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TABLES Dimensions# of BouncesLocation 1x10UR 1x21LR 1x32UR 1x43LR 1x54UR 1x65LR 2x11UL 2x20UR 2x33UL 2x41LR 2x55UL 2x62UR 2x77UL 2x83LR 2x99UL 2x104UR 3x12UR 3x23LR 3x30UR 3x45LR 3x56UR 3x61LR 3x78UR 3x89LR 3x92UR 3x1011LR 4x13UL 4x21UL 4x35UL 4x40UR 4x57UL 4x63UL 4x79UL Dimensions# of BouncesLocation 5x47LR 5x50UR 5x69LR 5x710UR 6x711UL 6x85LR 6x93UL 6x106UR 6x1115UL 7x813LR 7x914UR 7x1015LR 8x23UL 8x39UL 8x410UL 8x511UL 9x411LR 9x512UR 9x63LR 9x714UR 10x66UR 10x715UL 10x87LR 10x917UL 11x312UR 11x413LR 11x514UR 11x615LR
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PATTERNS I FOUND Any T(n × m) will have the same number of bounces as the reverse table i.e. T(m × n). For example, look at T(4,5 ) and T(5,4), they have the same number of bounces. If you rotate a T(4, 5) 90˚ we get a T(5,4) so clearly the number of bounces does not change, but the corner pockets is now in a different relative position. The corner pockets will be opposites of each other. For example, P(4,5) = UL and P(5,4) = LR. Look at T(1,2), T(2,4), and T(4,8), and notice B(1,2) = B(2,4) = B(4,8) = 3. In general, it seemed that T(n,m) = T(kn,km)
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FORMULAS Theorem 5.1 If the gcd(n,m)= k, then B(n,m) = B (nk, mk) and P(n,m) = P (nk,mk). Theorem 5.2 If the gcd(n,m) = 1 then B(n,m) = n + m − 2. Theorem 5.3 Let T(n,m) be a given table with gcd(n, m) = 1. Then P(n,m) = UR n and m odd LR n odd and m even UL n even and m odd
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EXAMPLES Solve a T(10,11), T(15,20), and T(448,320).
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EXAMPLES T(10,11) gcd(10,11) = 1. We use our Them 5.1 B(10,11) = 10+11-2 = 19. Since 10 is even and 11 is odd, then P(10,11) = UL T(15,20) gcd(15,20) = 5, we use Thm. 5.1 to get T(3,4). Thm. 5.1/.2 B(3,4)= B(15,20)=5 Since 3 is odd and 4 is even then P(3,4) = P(15,20) = LL T(448,320) gcd(448,320)= 64. By Thm 5.1 we get T(7,5). By Thm. 5.1/.2 B(7,5)=B(448,320)=10 Since both 7 and 5 are odd then P(7,5)=P(448,320)= UR.
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ANY QUESTIONS
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Our starting position is always going to at the lower left corner. The velocity of the cue ball is the same as it travels around the pool table until it goes into a corner pocket. We are going to hit the cue ball at a 45˚ angle. The angle of incidence equals the angle of reflection.
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