# Thread: Determining nmber of bricks I need for garden edging

1. ## 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.

2. Originally Posted by Tigers1
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,

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

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

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

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

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

m=1; n= 1.002204096797

m=2; n= 0.4994507915636

m=3; n= 0.3335778736504

3. Originally Posted by Tigers1
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

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

and

$\displaystyle \frac{34,600 \; mm}{76 \; mm/brick} \approx 455 \; bricks$.

4. Originally Posted by Sudharaka
Dear Tigers1,

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

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

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

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

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

m=1; n= 1.002204096797

m=2; n= 0.4994507915636

m=3; n= 0.3335778736504

int the equation $\displaystyle I=2r\sin\left(\frac{\pi}{n}\right)$ , the expression $\displaystyle \frac{\pi}{n}$ 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 $\displaystyle I=2r\frac{\pi}{n}$ which basicly gives awkard's estimation anyway