a little diversion

Seems demonstrably untrue. We commonly talk about revolutions per minute of a motor shaft.

Rod, yes that was the "original thread I was referring to" ..... and there should have been a winkie at the end of my post (a weak attempt at humour) - my apologies.

This comes with no explanation, but is worth a look : **LINK**

https://youtu.be/Ot-xEGQrfhA

MichaelG.

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Edit: __ Now watch this one: https://youtu.be/Xh5coE5wJ2I

https://youtu.be/Xh5coE5wJ2I

Edited By Michael Gilligan on 12/10/2021 00:19:53

Hi,

It's funny how a problem like this can get stuck in your head. I think I have finally sussed it out and for what it's worth this is what I have come up with.

If we refer to Macolms drawing there are two forms of rotation going on here.

The first of these is the rotation of the small wheel around the axis of the large wheel. If the 2 circles have smooth circumferences and the small circle is rotated from the 12 o'clock position to the 3 o'clock position around the centre of the large circle without rotating about its own axis (rolling), the A point on the small circle will change from the 6 o'clock to the 9 o'clock position which gives an apparent clockwise rotation about the axis of the small circle of 90 degrees.

The second form of rotation is the small circle rotating about its own axis (rolling). If we assume the large circle to be a 120 tooth gear and the small circle to be a 40 tooth gear (3:1) then rolling the small gear from 12 to 3 o'clock will involve 120/4 = 30 teeth meshing which will cause the small gear to rotate by 30/40 X 360 = 270 degrees.

The total apparent rotation of the small circle is therefore 90 + 270 = 360 degrees which gives a total of 4 apparent rotations per full traverse of the large circle. This is made up of 3 rotations about the axis of the small wheel and 1 rotation about the axis of the large wheel.

Interestingly, this problem doesn’t arise when the large circle is rotated about the small circle. While playing with a 28 tooth gear and a 56 tooth gear, I could see that the small gear apparently rotated 3 times when circumnavigating the large gear instead of the intuitively expected twice.

When the large gear was rolled around the small gear however, things went as expected, with the large gear rotating 180 degrees for the first circuit and 360 degrees to complete the second.

I think this contradicts the “mathematical solution” presented in the video Michael linked to:

**LINK**

Regards,

Alan C.

Do remember that this was a GCSE level question, not a degree level thesis discussion. Half (or more) GCSE students are poor at reading the questions properly, let alone answering the question actually asked.

.

Do they have GCSE in the United States ?

… or are you saying it is an equivalent level ?

MichaelG.

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https://www.princetonreview.com/college/sat-information

Edited By Michael Gilligan on 12/10/2021 08:51:58

Hi Alan, the maths works just the same for your 28 & 56 gears, the ratio being two to one, 1+2=3, so three turns for the 56 round the 28 or visa-versa. It works just the same for gears that don't have a whole number for the ratio, e.g. a 100T & 30T gear has a ratio of 3.333> to 1, therefore it will take four and one third revs for the 30T to orbit the 100T gear once and 13 revs of the small gear to get it back to it's starting state.

Starting state.

First orbit @ 4 - 1 / 3 revs.

Second orbit @ 8 - 2 / 3 revs.

The third orbit takes it back to the starting state, i.e. 3 x 4.333> revs = 13 revs in total.

Regards Nick.

Edited By Nicholas Farr on 12/10/2021 10:01:19

Ref. __ **LINK**

https://www.nytimes.com/1982/05/25/us/error-found-in-sat-question.html

MichaelG.