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#1 Re: Help Me ! » Expanding Torus Problem » 2009-09-02 09:39:45

PDS

Thank you Jane - this has given me something to work with on vacation.

I have had a quick look at your Cardano link and as I am only seeking an indication of the percentage reduction in the ring size will pursue the graphical option.

I did iterate a close solution using a spreadsheet (about 11% reduction in this instance) but was seeking a ‘quick and simple’ routine, where I could ‘just’ input a few numbers and obtain an answer, oh well.

Thanks again

(Puzzled, Dumb and Stupid)

#2 Re: Help Me ! » Expanding Torus Problem » 2009-08-31 06:37:03

PDS

Hello Soroban

Thank you for the prompt reply – however I have obviously not explained my problem correctly.

Initially the diameter (D) of the hole in the Torus is 3, this is equal to (R – r) x 2 [or (2.25 – 0.75) x 2]. Producing a Volume of 24.98

Using this volume, what is r when diameter D expands from 3 to 4.5?

#3 Help Me ! » Expanding Torus Problem » 2009-08-30 23:11:20

PDS
Replies: 4

Hi Guys

The volume of a Torus is V = (2 pi R)(pi r2) – where ‘R’ is the distance from the axis of rotation to the centre of the ring and ‘r’ is the radius of the ring.

My little problem is that I know the inside diameter (the hole of the doughnut) and the diameter of the ring and can therefore find the volume – BUT what is the diameter of the ring if I keep the same volume for the torus but increase the size of the hole in the centre?

If you need numbers the hole diameter is 3 and the ring diameter is 1.5, but what would the ring diameter if the hole was 4.5?

Is there a 'simple' equation I can use?

This is similar to my brain ---> dizzy

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