Calculation of Radius

gardenshed

gardenshed

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How do I calculate the radius of curvature of a piece of mainsheet track?
I know the curved distance from end to end (2.0m) and the distance that each end drops from the centre (10cm)

I need to know this as the proposed supplier of the replacement track recommends a minimum radius of 6m and I can't work out what if my track has a greater or smaller radius than this.
 
"Bulges are something that women have (mostly to please the opposite sex it seems) and something that guys try to get by placing socks in strategic places. At least until they get older. Which is the time they tend to develop bulges in not so strategic places. In other words: bulges are all about curvature."


This site has some useful background, and suggested equations.
http://www.afralisp.net/lisp/Bulges1.htm
 
I make the radius about 4.96m if your 2m is measured around the curve of 5.05m if your 2m is measured in a straight line.
 
The two equations you need to solve are R(Theta)=1, and cosine(theta)=(R-10/100)/R, where theta is half the angle subtented and 1 is half the distance along the track. Rearrange in terms of Theta, subtract and you are left with an equation in R. Easy!....Alternatively (and much quicker) get out a piece of paper and a school compass and ruler and draw a 2m line to scale and then try drawing different radii until you get the 0.1m drop and then scale it off. In an ideal world the two methods would match exactly, although the paper one will be reasonably accurate. You will need to mess around abit until the track distance along the curve scales 2m.
 
For the curved chord solve:
Angle in Radians: Theta = 1/R
And: R*cos(Theta) = R - 0.1

For the straight it is just Pythagoras:
1*1 + (R - 0.1)*(R-0.1) = R*R

But I do have a few bits of software here to solve the equations for me!

Either way it's less than 6 metres. /forums/images/graemlins/frown.gif
 
T square of half the arch length = the product of the two part of the diameter of circle. radius near enough 5m. Such a small deviation from straight that fixings will pull it into curve.
 
nearly got it with my algebra and geometry but not quite.
3 years of maths at university and I still couldn't work it out myself.
many thanks to all of you who supplied answers.
I'll re-measure the track to see whether or not I've overestimated the curvature.
 
G'day GardenShed,

If you are planning to replace the track it might save you a lot of extra work if you simply send your old track, this way you can be sure the holes will not be in the wrong places resulting a lot of extra work, filling and re drilling.

Avagoodweekend......
 
You could get a piece of string 6m long, tethered at one end, chalk at the other, draw a 6m arc and compare it.
would save all that maths...makes my ears bleed.
 
good idea but I am reliably told that even 30 years ago, all tracks were drilled with 100mm centres for their fastenings.................that's why I intend to go for "sliding bolt" track, i.e. where the bolts are slid inside the extrusion and don't have pre-drilled holes! This way the bolts can be aligned with the existing holes without any bother.
The track is fully supported all along its length on a 20mm thick aluminium beam so no worries there.
 
simple is best. like your idea and may well use it to double check the calcs.

Ma heid blew up earlier this evening with the thetas, radians and cosines

'mazing how clever some of the folks on this forum are. Easlier than googling for an answer.
 
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