/ Geometry help please!

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Deadeye - on 13 Aug 2017
This is going to be something of a challenge to describe.
I'm making a cyclone separator and want to work out how to draw the line for cutting the hole for the side inlet. A similar product is here (but out of sock and rather expensive): http://www.cyclonecentral.co.uk/Store/

I have a 1000mm square sheet of 2mm polycarbbonate. I'm going to cut a quarter circle radius 250mm from one corner and then a 1000mm radius quarter circle arc parallel.
When rolled up this will give me a truncated cone with radius 250mm at the top and 125mm at the bottom.

With the cone vertical and widest at top, I want to insert a 100mm side pipe with centre 80mm from the top and edge just touching the inside of the cone when inserted to the diameter line. To make the join easier, I don't want the outside edge to be effectively a tangent, but set in perhaps 3cm from the tangent.

This will require a horizontal teardrop-shaped hole in the cone. The challenge is to draw it onto the unfurled net of the cone.

Is there a tool? Or a method to graph paper it?
aln - on 13 Aug 2017
In reply to Deadeye:

You wanna put a banging donk on that.
keith-ratcliffe on 13 Aug 2017
In reply to aln:
A true RadMac response !

Timmd on 14 Aug 2017
In reply to Deadeye:

There must be a programme which could calculate/draw the template for you.
Jim C - on 14 Aug 2017
In reply to Deadeye:
If I'm not too confused this sounds like the tests they used to set me for my Higher Engineering Drawing ( many years ago before computers were invented.)

We would draw it in plan and elevation , then divide the circle into equal divisions and number them and project it off to gain the true lengths ( identically numbered) we occasionaly took the developments and cut out the shapes on stiff card and made models . it was called interpenetration of solids Can't do it off the cuff now, and never had to use a CAD programme.

Lots of worked examples on You Tube.

Development of the true shapes is on section 4
Post edited at 03:37
Deadeye - on 14 Aug 2017
In reply to Jim C:

That's a beautiful book! Thank you.
marsbar - on 14 Aug 2017
In reply to Deadeye:

Personally I think I'd go for a 1:10 scale model made of thin cardboard covered in graph paper and a piece of 1cm diameter dowelling and a sharp knife and then unfold it and scale it up.

I suspect it is possible mathematically but I'm reminded of the story of Edison asking someone to find the capacity of a lightbulb, wondering where they have got to and filling it with water.
marsbar - on 14 Aug 2017
In reply to Jim C:

Wow that's amazing. I never did anything more than very basic drawings, I wish I had.
jrck2 - on 14 Aug 2017


1. Imagine a Cartesian coordinate system with the origin located where the point of the cone would be. Have the z-axis parallel with the centre-line of the side pipe.

2. The centre-line of the side pipe is located at x0,y0, and the side pipe has radius r0. The side pipe has an equation something like: (x-x0)^2 + (y-y0)^2 - r0^2 = 0.

3. If the cone widens at a rate k, then the equation for the surface of the cone is x^2 + z^2 - (ky)^2 = 0.

4. Solve these two equations to give you another equation for the line where the two parts intersect. Should be something like: x0^2 - 2x*x0 - z^2 +(k^2 -1)y^2 - 2y*y0 + y^2 - r0^2 = 0.

5. Create a coordinate system based on the flat sheet, with coordinates r and a. These are related to the Cartesian coordinate system by y = r/(1+k^2)^0.5, x=r cos(a) and z=r sin(a). r is the radial coordinate on the flat sheet, and a is an angular coordinate.

6. Now express the intersect in terms of r and a. Something like: x0^2 - 2x0r cos(a) - r^2 sin^2(a) + (k^2 - 1)r^2 / (1+k^2) - 2y0r/(1+k^2)^0.5 + r^2 /(1+k^2) - r0 = 0 will be the equation of the line you need to cut on your flat sheet.

7. Remember the cone is rotationally symmetric, so you can shift coordinate a by any amount a0, to avoid cutting on the join. Also, the above equations should have 2 solutions, as I haven't specified a finite length side pipe (ie. don't cut 2 holes).

On reflection, the methods suggested by others might be better...
Post edited at 10:31
Deadeye - on 14 Aug 2017
In reply to jrck2:

Thanks. My cone has sin(tip angle/2)=1/4
What does that make k?
jrck2 - on 14 Aug 2017
In reply to Deadeye:
I think from the numbers in your OP the radius of the small end of the cone will be 62.5mm, not 125mm. This is consistent with sin(tip angle/2)=1/4.

k = tan(arcsin(1/4)) = 0.2582.

As others have suggested, definitely check with cardboard first!

Also, just noticed the final term in the equation in point 6. of my previous post should be r0^2, not r0.
Post edited at 14:31
Deadeye - on 14 Aug 2017
In reply to jrck2:

Thanks - and you're right, I gave the diameter.

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