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July 29th, 2015, 08:16 AM  #1 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 32 Thanks: 0  Arc Length given Chord Length & Height
I got a formula for arc length from this page https://en.wikipedia.org/wiki/Circular_segment s = Arcsin(c / (h + c^2 / 4h)) X (h + c^2 / 4h) given c = 163' & h = 7' I do s = Arcsin (163 / (7 + 163^2 / 4 X 7)) X (7 + 163^2 / 4 X 7) s = Arcsin (163 / 955.89) X (955.89) s = Arcsin (0.1705) X (955.89) s = 9.818 X 955.89 s = 9,385' This is obviously not correct. What am I doing wrong? 
July 29th, 2015, 12:02 PM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
Can you not do this one from first principles more easily than finding some longwinded formula?
Last edited by skipjack; July 29th, 2015 at 02:11 PM. 
July 29th, 2015, 01:17 PM  #3 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 32 Thanks: 0 
Thanks studiot for your solution. Now I guess my question is one of curiosity on my part. Is the published formula correct? If it is correct what am I doing wrong, as I get an incorrect answer when I solve for arc length using it. 
July 29th, 2015, 02:41 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324  
July 30th, 2015, 06:26 AM  #5  
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 32 Thanks: 0 
Thanks skipjack Redoing: s = Arcsin (163 / (7 + 163^2 / 4 X 7)) X (7 + 163^2 / 4 X 7) s = Arcsin (163 / 955.89) X (955.89) s = Arcsin (0.1705) X (955.89) s = Arcsin (0.1705 X pi / 180) X (955.89) s = Arcsin (0.002976 rad) X (955.89) s = 0.1705 X 955.89 s = 163.00024063' Still don't get 163.83' as does studiot Quote:
7' is best I can measure from engineering drawing.  
July 30th, 2015, 08:29 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324 
arcsin(0.1705212) = 0.1713586 (radians) approximately, and 0.1713586 × 955.893 = 163.8 approximately. 
July 30th, 2015, 09:49 AM  #7 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 32 Thanks: 0 
s = Arcsin (163 / (7 + 163^2 / 4 X 7)) X (7 + 163^2 / 4 X 7) s = Arcsin (163 / 955.89) X (955.89) s = Arcsin (0.1705)radians X (955.89) s = 9.8169radians X pi /180 X 955.89 s = 0.17134degrees X 955.89 s = 163.78' If I carry full decimal places in the above steps I get s = 163.800456979' Thanks skipjack 
July 30th, 2015, 10:49 AM  #8 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
Since this thread is obviously still live here are some useful formulae. These are much better than the clumsy arcsin one from Wiki. First Engineers used to use versed sine (versine) tables before calculators. This easily gives the maximum rise from the chord if the radius and angle subtended to the centre are known. A great many more offsets from the chord than just the centre one will be needed. These can be calculated as in the first formula, after calculating the max offset. An alternative are the offsets from the tangent (or any sideways displaced straight line) as shown in the second diagram. 
July 30th, 2015, 12:01 PM  #9 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 32 Thanks: 0 
Thanks studiot Your additional formulae would be very useful to me. I can't decipher the "?" below h = ho?  { R  sqrt(R^2  x^2) } 
July 30th, 2015, 02:08 PM  #10 
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
c/2 is the half chord length. ho is the maximmum height of the arc above the chord and is at the centre of the chord and arc. Distance x is measured from the centre. h is the height of the arc above any point at distance x from the centre and is equal to ho minus the expression (square root) in brackets. It arises thus: The arc is symmetrical about the centre so if you divide the chord into an even number of sections as in the diagram you calculate two offsets at a time, one each side of the centre. The geometry of the formula is shown in the diagram. What you are effectively doing is repeatedly using the centre height formula across a smaller and smaller chord as shown by the dashed line and subtracting it from the main centre height. Before computers drawing offices had what were known as 'railway curves' which were shaped pieces of wood or plastic formed to arcs of standard curvature, for drawing arcs of very large circle. Surveyors used the formulae above for setting out such large curves. I would hate all this old knowledge to be lost just because we can plot it out by computer these days. Last edited by studiot; July 30th, 2015 at 02:14 PM. 

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