I have a pipe 5000mm diamater and a known length of string 59996mm long, how would i figure out 'x' for when the string wraps around the pipe coming of the tangent and up to a center point at the top, see attached sketch to better understand.
I have a pipe 5000mm diamater and a known length of string 59996mm long, how would i figure out 'x' for when the string wraps around the pipe coming of the tangent and up to a center point at the top, see attached sketch to better understand.
Hello, jambanks!
Using the numbers given in your diagram,
. . the problem is quite easy.
I have a pipe 5000 mm diameter and a known length of string 59996 mm long.
How would i figure out $\displaystyle x$ in the diagram?Code:A * - /|\ : / | \ : / | \ : 20,974 / | \ : / | \ : / * * * \ x /* | *\ : B * | * : * * | * * : * | * : * 2500 * | * * : * * * - * O * * * * * * * * * *
We see that $\displaystyle x \,=\,AO$
In right triangle $\displaystyle ABO\!:\;\;AO^2 \:=\:AB^2 + BO^2$
Hence: .$\displaystyle AO^2 \:=\:20974^2 + 2500^2 \:=\:446,\!158,\!676$
Therefore: .$\displaystyle x \:=\:AO \:=\:\sqrt{446,\!158,\!676} \:\approx\:21,\!122.5 $