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Math Help - Determining nmber of bricks I need for garden edging

  1. #1
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    Determining nmber of bricks I need for garden edging

    Gudday all
    I am doing some circular edging in my garden and want to determine how many bricks I would need (at least get close).
    The brick edge is 76mm, the radius of the curve is 5,500mm. After rifling through my Uni text books I felt that using part of the method of exhaustion might give me the answer.
    The equation is l = 2rsin(pi/n) where l = 76, r =5500 and n = no. of bricks (I hope that this is correct.)
    Plugging in and simplying I get 6.91^10-3 = sin (pi/n).

    Question: How do I rearrange to solve for n?

    I have looked at this for a long time but Uni was such a long, long time ago.
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  2. #2
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    Quote Originally Posted by Tigers1 View Post
    Gudday all
    I am doing some circular edging in my garden and want to determine how many bricks I would need (at least get close).
    The brick edge is 76mm, the radius of the curve is 5,500mm. After rifling through my Uni text books I felt that using part of the method of exhaustion might give me the answer.
    The equation is l = 2rsin(pi/n) where l = 76, r =5500 and n = no. of bricks (I hope that this is correct.)
    Plugging in and simplying I get 6.91^10-3 = sin (pi/n).

    Question: How do I rearrange to solve for n?

    I have looked at this for a long time but Uni was such a long, long time ago.
    Dear Tigers1,

    I=2r\sin\left(\frac{\pi}{n}\right)

    Substituting your value we get,

    76=11000\sin\left(\frac{\pi}{n}\right)

    \sin\left(\frac{\pi}{n}\right)=\frac{76}{11000}

    \frac{\pi}{n}=m\pi+(-1)^m\sin^{-1}\frac{76}{11000}~where~m\in Z

    \displaystyle n=\frac{\pi}{m\pi+(-1)^m\sin^{-1}\frac{76}{11000}}~where~m\in Z

    m=1; n= 1.002204096797

    m=2; n= 0.4994507915636

    m=3; n= 0.3335778736504

    Hope this is helpful.
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  3. #3
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    Quote Originally Posted by Tigers1 View Post
    Gudday all
    I am doing some circular edging in my garden and want to determine how many bricks I would need (at least get close).
    The brick edge is 76mm, the radius of the curve is 5,500mm. After rifling through my Uni text books I felt that using part of the method of exhaustion might give me the answer.
    The equation is l = 2rsin(pi/n) where l = 76, r =5500 and n = no. of bricks (I hope that this is correct.)
    Plugging in and simplying I get 6.91^10-3 = sin (pi/n).

    Question: How do I rearrange to solve for n?

    I have looked at this for a long time but Uni was such a long, long time ago.
    Hi Tigers1,

    An easy approximation is to compute the circumference of the circle and divide by the length of a brick. This is not an exact answer, because the bricks are not curved, but I bet it will give you a result good enough for gardening.

    The circumference is

    2 \pi r \approx 34,600 \; mm

    and

    \frac{34,600 \; mm}{76 \; mm/brick} \approx 455 \; bricks.
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  4. #4
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    Quote Originally Posted by Sudharaka View Post
    Dear Tigers1,

    I=2r\sin\left(\frac{\pi}{n}\right)

    Substituting your value we get,

    76=11000\sin\left(\frac{\pi}{n}\right)

    \sin\left(\frac{\pi}{n}\right)=\frac{76}{11000}

    \frac{\pi}{n}=m\pi+(-1)^m\sin^{-1}\frac{76}{11000}~where~m\in Z

    \displaystyle n=\frac{\pi}{m\pi+(-1)^m\sin^{-1}\frac{76}{11000}}~where~m\in Z

    m=1; n= 1.002204096797

    m=2; n= 0.4994507915636

    m=3; n= 0.3335778736504

    Hope this is helpful.
    Sudharaka forgot to set his calculator to radian mode

    int the equation I=2r\sin\left(\frac{\pi}{n}\right) , the expression \frac{\pi}{n} would be a very small number in gardening. I wouldn't imagine your garden requires less than say 15 to 20 bricks. When x is very small, y= sin(x) can be very well approximated by y = x. so your equation can be well approximated simply by I=2r\frac{\pi}{n} which basicly gives awkard's estimation anyway
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