Any math-wizzkids around? need some help :)

Ingafgrinn Macabre

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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 ;)
 

tris-

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obviously the tube is bigger than the wires so i think thats all YOU need to know.

thank you and good night.
 

Natswoo

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I'm not sure how, and it would be impossible for me to explain how I did it, but i worked it out using multiple symultaneous equations.... though the answers are a bit funkeh... as in A = 1.9342552726E41 or A = 1.29342552726E-14 ........ /confused

tbh i wouldn't listen to me, cos that looks seriously wrong

Btw pic 3 cannot work as the two largest wires will touch the sides, but the smallest could be "floating"

Eurg will ask my son when he gets back from uni next wednesday.. he's doing Physics... he should know...

my brain is hurting :mad:
 

Natswoo

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The theory cannot work unless 2 of the cables are of the same diameter.............
 

Ormorof

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mad people doing maths, i got 5+3 wrong in my maths GCSE (i put 7, though i still managed to get a C rawr :D )
 

Natswoo

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Ye, for all the wires to touch the tube, and all touch each other two of them have to be the same diameter... rest shall be solved by my son... i hope
 

tris-

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look i told u, all u need to know is the tube is bigger than the damn wires.
 

Ingafgrinn Macabre

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The wires do not need to be of the same size, though when the third wire gets small enough it will not touch the tube anymore as seen in image 1 and image 2. I do have the formula to see when that is the case, but it's not always the case. It should be solveable since there is only one answer possible, I just do not know how to :) (though I do know that the centerpoint of the tube will always stay within the triangle created by interconnecting the centerpoints of the wires)

Anyway, Natswoo, thanks a million for trying it, and if you or your son could solve it, I'd be in 7th heaven ;)
 

Natswoo

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Ingafgrinn Macabre said:
The wires do not need to be of the same size, though when the third wire gets small enough it will not touch the tube anymore as seen in image 1 and image 2. I do have the formula to see when that is the case, but it's not always the case. It should be solveable since there is only one answer possible, I just do not know how to :) (though I do know that the centerpoint of the tube will always stay within the triangle created by interconnecting the centerpoints of the wires)

Anyway, Natswoo, thanks a million for trying it, and if you or your son could solve it, I'd be in 7th heaven ;)
I'll see what i can do... 25 years ago i might have been able to do it.... nerf vast amounts of beer and doing an un maths based job.
 

Lamont

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I may be completely wrong with this... but I'll risk embarassing myself anyway :drink:

IF I understand the question correctly -

Assuming that all 3 wires touch each other and also touch the outer tube, is it not safe to say that the diameter of all 3 wires must be equal? I'll assume your question doesn't include proving this.. ;)

Now assuming that r1 = r2 = r3, then it's relatively straight forward to work out the internal angles and thus the diameter of the tube...

Joining the centre of each internal wire to each of the other wire centre points forms an equilateral triangle with each side length 2r. The distance from the centre point of this triangle (a point in the 'open space' formed between the 3 circular wires) to any corner of this triangle (i.e. the centre point of a wire) PLUS the radius of the wire (r) equals the minimum radius of the tube.

Using pythagoras, the distance from absolute centre of tube to centre of a wire = r / cos30 = 1.1547r.

Therefore, the minimum radius of the tube itself is 1.1547r + r = 2.1547r

Diameter is of course twice this, i.e. 4.3094r
 

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Ingafgrinn Macabre

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if r1=r2=r3 then indeed the angles between the Cable centerpoints and the tube centerpoint is three times 120deg, that ain't too hard to calculate indeed since the triangle between the three cable centerpoints is a regular triangle with all sides being d1=d2=d3 and the tube centerpoint is exactly in the center of this triangle, but the problem is exactly when that is not the case so r1 is equal or bigger then r2 which is equal or bigger then r3, so I need a way to calculate the radial of the tube, but I might have an idea on how to solve it, I don't know for sure yet though if it works... if it does, I'll let you guys know ;)
 

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