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March 17th, 2017, 12:45 AM   #1
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Can anyone help me with a problem I'm having working out an arc

Hi I'm having problems drawing an arc of drawn a bit of a basic sketch if worked out I need my arc to have a rad of 5646mm but can anyone tell me what the height of line A will be and even shown me a formula please

David
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March 17th, 2017, 04:42 AM   #2
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I don't know what accuracy you require or if you know any trigonometry.

The simplest way of setting this out is to use the formula that the deviation of a circular curve from the straight (tangent) is almost exactly the square of the length measured along the tangent divided by twice the radius.

So if you set up a tangent, parallel to the ground, at the midpoint, which is also the lowest point,
Then you can mark off any point on the curve by adding the offset to 1220.
You have to work from the centre for this.

I note that by the time you reach the extremity the calculated offset is 296, but you show 310.
If this small error is unacceptable (it is only this much at the extremity) then we will need to look for a more accurate, but more complicated, trigonometric formula.
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 March 17th, 2017, 02:45 PM #3 Newbie   Joined: Jul 2016 From: Ringwood Posts: 7 Thanks: 0 Thanks for the help just struggling to get the formula working on my calculator lol
 March 17th, 2017, 06:35 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,722 Thanks: 1375
March 17th, 2017, 07:51 PM   #5
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Quote:
 Originally Posted by studiot . . . the deviation of a circular curve from the straight (tangent) is almost exactly the square of the length measured along the tangent divided by twice the radius.
For radius r and length along the tangent x, where x $\small\leqslant$ r, the deviation (by Pythagoras) = r - √(r² - x²), which is 304.8001... for r = 5646 and x = 1830.

 March 17th, 2017, 09:33 PM #6 Newbie   Joined: Jul 2016 From: Ringwood Posts: 7 Thanks: 0 Thanks skipjack a formula I can read, thankyou so much Last edited by Nutcase1; March 17th, 2017 at 09:35 PM.
March 18th, 2017, 01:43 AM   #7
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In quoting me and then posting this

Quote:
 Originally Posted by skipjack For radius r and length along the tangent x, where x $\small\leqslant$ r, the deviation (by Pythagoras) = r - √(r² - x²), which is 304.8001... for r = 5646 and x = 1830.
Are you disagreeing with my formula or supplying one of the many more accurate and more difficult formulae I mentioned?

 March 18th, 2017, 06:23 AM #8 Global Moderator   Joined: Dec 2006 Posts: 18,691 Thanks: 1523 Trigonometry isn't needed. Any exact formula could be simplified to the exact formula I gave, which can be expanded to justify the approximate formula you gave.
March 18th, 2017, 07:49 AM   #9
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Quote:
 Originally Posted by skipjack Trigonometry isn't needed. Any exact formula could be simplified to the exact formula I gave, which can be expanded to justify the approximate formula you gave.
So are you saying that surveyors, mining and civil engineers were using the wrong formula for the last couple of hundred years?

 March 18th, 2017, 11:52 PM #10 Global Moderator   Joined: Dec 2006 Posts: 18,691 Thanks: 1523 That depends on what you mean by "wrong". r - √(r² - x²) = r((x/r)²/2 + (x/r)$^4$/8 + (x/r)$^6$/16 + . . . ), so the approximate formula may be close enough to the exact value if x/r is small, but the exact formula is quite easy to evaluate.

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