the whole concept of an open forum.
Yesterday I made a post asking about the calculation of divisions when creating a vernier scale.
http://bbs.homeshopmachinist.net/showthread.php?t=29789
I made the post thinking I'd get a few ideas, pick up some theory and with any luck I'd learn something. A side benefit to these open discussions is that others can learn a little something while the thread is active or search through the archives later on. It's really a wonderful system.
While this thread was active I received a PM from member DJC informing me I was incorrect. Here is the PM:
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> Let's say my cross feed dial has 100 divisions total. If I
> was to add a vernier to the system I'd space the divisions
> so that 10 marks on the vernier would fit within 9 marks on
> the dial. That would mean my if my dial has 100 divisions,
> my vernier would need to have it's circumference divided by
> 110 (only needing 10 marks).
I think you need to think again. If 10 marks on the vernier is 9 marks on the dial, 100 marks on the vernier is 90 marks on the dial and 110 marks on the vernier is 99 marks on the dial.
The spacing between each mark on the vernier is 90% of that between the spacing on the main dial. The spacing between the main dial marks is 1/100 of the circumference of the dial, so you need 90% of this i.e. 9/10 x 1/100 = 9/1000. Hence you have to divide your circumference by 1000 and mark every ninth division.
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I responded back with this:
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http://www.tpub.com/math1/7e.htm was the document that made it click for me. Perhaps I'm not reading it right.
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Followed by this reply from DJC:
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> Perhaps I'm not reading it right.
Well, you are correct in this assertion at least. There is nothing wrong in the link you posted.
It is your post that says dividing the circle into 110 parts will achieve a vernier scale that is incorrect.
Look at the link you posted. When the zero on the vernier scale lines up with the zero on the main scale, the one on the vernier scale is 9/10 of the way along the gap between zero and one on the main scale. If the gap on the main scale is 1/100 (of the circumference), the gap in the vernier scale is 9/10 of this, or 9/10 multiplied by 1/100 which equals 9/1000 as already posted.
If you don't believe me, go and buy George Thomas' "Model Engineer's Workshop Manual" ISBN 9781857610000 and read page 178. For your education, I have scanned the paragraph and it's at http://www.desolation-row.demon.co.uk/vernier.jpg
Now kindly post a correction to your erroneous post on the public forum, so nobody else reads it and makes the same mistake. If you don't correct your own work, I will.
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Giving his point some thought I think he's right, and I'm glad he pointed it out to me.
I only wish he would have participated in the thread while it was active. I also would like to point out to DJC that my "Work" was not work at all, it was a question looking for an answer. I posted what I thought was a solution but was making a mistake in my thinking. Even then, I was questioning my solutions correctness.
Lastly I'd like to point out that threatening people as you did in your closing statement does no good. In fact it simply makes you look like an A55Hole.
Anyway, thanks for the correction DJC and I hope you take a more active part in the forum, it seems like you have something to offer.



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