How to find arc length of this frustum..?

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Nimisha
Nimisha 2015-3-20
评论： Nimisha 2015-3-20
I had this frustum. I want to find the arc length of it. I had smaller side radius, larger side radius, cone height and no. of turns.
How to find the arc length..?
Is there any equation or Code.? Please help.
Will this attached equation will applicable..?
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Nimisha
Nimisha 2015-3-20
owww.. it may be clicked two times..
John D'Errico
John D'Errico 2015-3-20
As for the question of whether that equation applies, without re-deriving it, how would we know? You don't give a source for the expression, nor do we know what the parameters in it mean. Anyway, that question is not a MATLAB question, so off-topic on a site for questions about MATLAB.

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John D'Errico
John D'Errico 2015-3-20
I'd just generate a sufficient number of points on the curve, then call my arclength tool (download it from the file exchange to use this tool.)
Use the 'spline' method for a fair accuracy. As a wild guess, you should get 8 significant digits from 100 points on the curve (Depending on the number of wraps, etc.)
If you wanted a more accurate solution, then it is quite doable with numerical integration. There are several methods that would suffice to yield essentially full precision, depending on the effort you are willing to make. (Without your relationship for that curve and the set of parameters that define it, I'll not go any further.)
Finally, with that relationship in hand, it MIGHT be possible to generate an analytical solution for the arc length.
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John D'Errico
John D'Errico 2015-3-20
What can't you understand? Read the help for arclength.
Did you do as I suggested? Did you generate a sequence of points along that curve?
What is the formula for that curve? Telling me some basic measurements about it is not useful. This is your curve, not mine. While I COULD spend some time to develop a parametric representation of the curve, if you don't have a mathematical representation for that curve, how did you expect to compute an arc length?
ARGH!
If theta varies from 0 to 5*2*pi, this represents 5 turns of a periodic function.
A representation, in cylindrical coordinates for this curve would reasonably be...
theta = linspace(0,5*2*pi,100);
z = 150*theta/(10*pi);
r = 19 + (35 - 19)*theta/(10*pi);
Converting to cylindrical coordinates, we would have
x = r.*cos(theta);
y = r.*sin(theta);
plot3(x,y,z,'o')
grid on
box on
Nimisha
Nimisha 2015-3-20
Thank you very much, John..
I solved it.

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