1. ## Height to tangent

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.

2. 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: .$\displaystyle x \:=\:AO \:=\:\sqrt{446,\!158,\!676} \:\approx\:21,\!122.5$

3. Thank You for the reply. I never really made it clear in my first message but you wouldn't know the 20974 value, only the complete value of the lines 59996, you you would have to figure out ao first