How to calculate rope length around sheave with obstacle?

Hello
This may look like a school assignment but it's not. We're trying to load test a wire sheave in real life. Need to find minimum length of the wire sling to clear the obstacle.

I found a set of equations now that seemed right but I couldn't solve them. Used solver so I think I have a solution.
Still, it would be awesome to get the proper equations solved.

See attached.

You shouldn't need to use solver.

See attachment to see how I named the proportional triangles
Because they are proportional triangles
DE/AD = BC/AB
BA = x
DE = r
BC = L/2
BD = h
AD = h + x
r/(h+x) = (L/2)/x
Solving for x
x = (L/2)h(1-L/2)

Then distance AE by Pythagrian theory
= SQRT of (h+x)squared + r squared

Total wire = 2 times distance AE + (Pi * r)

I came up with this, Looks quite a bit like ChemistB's solution I think, but the answer is undoubtedly a bit short, as I am missing some arc wrap I'm sure

My result should be slightly off as it assumes the tangent creates an angle ADE which is a right triangle. Such an angle would slightly intersect the circle and would not be a true tangent. SHG's calculations are more accurate except, L2 should be L2/2. Yes?

In the picture, I changed the definition of L2 to the obstacle half-height, consistent with the apex half-angle.

Chemist, could you read the equation? I'm not sure how it presents to people without MathType.

I have no problems but we'll see if the OP can read it.

I didn't know this equation.
d2 = r (180 + 2theta)
Good one.

Well, you always go halfway (180 degrees) around the sheave, plus theta (by similar triangles) additional at top and bottom.

Thanks for the reply guys! Yup, I can read the equations, no problems.
As far as I can see the solutions from Chemist B and Dredwolf produce different results but very similar. I have my own simplified solution that I think also produces a very similar but different result... Part of me wants to go to the bottom with that but what I really wanted was the full exact solution.

shg, your solution looks excellent and is what I was looking for. But were is P on the drawing? d1 seems to occupy the space where I would expect to find P. Also the text description of P and d1 are very similar.

In true Excelforum poster tradition I will now add more demands. The wire thickness needs to be included in the equations.
Some real life numbers are 74 mm for wire thickness, 1470mm for L1 and 450mm for L2 (shg denominations).

For the record the test was performed this afternoon, stretching the wire to a healthy 136 tonnes.

Regardless of that these equations are still of great interest to me, could easily happen again.
By the way, why are some of the equations in blue?

P is the distance from the sheave center to the apex. d1 is the distance from the apex to the sheave tangent.

why are some of the equations in blue?

They are the formulas used in the worksheet.

The wire thickness needs to be included in the equations.

Increasing the sheave radius by the wire radius would get close. The trig would be messy for an exact solution.

Yikes, a mistake

I had a tangent in the first formula that should have been a sine. Corrected, I think, in the attachment.

There's a quadratic root UDF because I didn't want to solve it symbolically.

Thank god! I thought I was loosing my mind trying to figure it out
This looks more like what I came up with when I tried on my own. While in theory this is within reach for me I just don't trust myself with handling lumps of calculus like this, hence I reached for Excel Solver and then for Excel Forum.
Excellent job shg!

To add the wire thickness to the equation I added the wire diameter/2 to the sheave radius.
To L2 I added wire diameter/2/cos(theta) which would add a whole lot of trouble to the equations, definitely not worth the effort.
Allow iterations? Yes please! Problem solved as accurate as I imagine is possible.

CAD, couple of lines and a circle = 15.91642971

Good one Kevin. I've seen that mentioned before but I didn't think about it now. I tried in Draftsight and it works fine and doesn't take long. I'll keep it in mind. Very useful for double checking if you are making an mathematical model of something in Excel.
However, now that I have my Excel model ready it is even faster.

It actually took me a minute to figure out why everything was error when I opened it again. Of course the circular reference had crashed it.
Added an IFERROR to make it start properly. Funny, I never thought that was an issue before...

I'm not entirely happy with finishing off a nice analytical solution with a circular reference but for practical use I'm still satisfied with it.

Originally Posted by Kevin UK

CAD, couple of lines and a circle = 15.91642971

That affirms the calculation in the workbook, thank you.