#### Ingafgrinn Macabre

##### Can't get enough of FH

- Joined
- Jan 4, 2004

- Messages
- 3,155

- Thread starter
- #1

I have this problem I cannot seem to find the answer for, and I was hoping some of you could help me with this. It's about three cables of different sizes flush to eachother, fitted in a round tube, which is also flush to the three cables.

[NOTE:] d and r are the diameters and radials of the cables, D and R is the diameter and radial of the tube

so these three cables of different sizes (d1, d2, d3) in a tube, what will the absolute minimum (inner) size of the tube be to fit all the three cables. The cables will always be set in sized order, so r1 >= r2 >= r3.

when the diameter of the third cable is small enough, (image 1) the tube diameter D can be d1+d2. I've got the formula for finding that borderline of when that is possible.(image 2)

but now the difficult part, is to come up with a set of formulae to calculate the size of the tube when this is not the case (image 3).

The lines from the centerpoint of the tube to the centerpoint of the cables I have named r1', r2', r3'. with these names:

r3' - r2' = r2 - r3

r2' - r1' = r1 - r2

r3' - r1' = r1 - r3

r1' + r1 = r2' + r2 = r3' + r3 = R

Also the three triangles from the tube centerpoint to two tangent points have two equal edges.

I was thinking this could be solved by previous formulae substituted into area- or heightformulae from the three triangles from the centerpoint of the tube to the centerpoints of the cables, but I'm not that good in simplifying the formulae, and it's getting extremely messy on my piece of paper here

btw: Heron's areaformula for triangles where only the sidelength's are known:

Area = SQRT(s(s-a)(s-b)(s-c)),

where s=(a+b+c)/2 or perimeter/2.

if anyone knows the answer to this, or feels intrigued by trying, I would be in your eternal debt

Thanks in advance

[NOTE:] d and r are the diameters and radials of the cables, D and R is the diameter and radial of the tube

so these three cables of different sizes (d1, d2, d3) in a tube, what will the absolute minimum (inner) size of the tube be to fit all the three cables. The cables will always be set in sized order, so r1 >= r2 >= r3.

when the diameter of the third cable is small enough, (image 1) the tube diameter D can be d1+d2. I've got the formula for finding that borderline of when that is possible.(image 2)

but now the difficult part, is to come up with a set of formulae to calculate the size of the tube when this is not the case (image 3).

The lines from the centerpoint of the tube to the centerpoint of the cables I have named r1', r2', r3'. with these names:

r3' - r2' = r2 - r3

r2' - r1' = r1 - r2

r3' - r1' = r1 - r3

r1' + r1 = r2' + r2 = r3' + r3 = R

Also the three triangles from the tube centerpoint to two tangent points have two equal edges.

I was thinking this could be solved by previous formulae substituted into area- or heightformulae from the three triangles from the centerpoint of the tube to the centerpoints of the cables, but I'm not that good in simplifying the formulae, and it's getting extremely messy on my piece of paper here

btw: Heron's areaformula for triangles where only the sidelength's are known:

Area = SQRT(s(s-a)(s-b)(s-c)),

where s=(a+b+c)/2 or perimeter/2.

if anyone knows the answer to this, or feels intrigued by trying, I would be in your eternal debt

Thanks in advance