Helix Arc Length

[ArPEn]

Would someone mind demonstrating how to calculate the arc length of a helix, or show how to integrate over a 3D vector on the HP Prime?

I am attempting to calculate the length of wire required to wrap around a cylinder, and need to mathematically determine length to calculate resistance.

Parameters:
Height: 10 inches
Revolutions: 35
Diameter: 0.54 inches
Pitch: 0.29 inches

A=∫(from 0 to 1)? √((x′(t))^2+(y′(t))^2+(z′(t))^2) dt

(Taken from https://math.stackexchange.com/questions...of-a-helix)

If anyone is able to direct me how to integrate over a 3D vector function I can probably figure out the correct formula and/or inputs. I just can't figure out how to input the equation properly.

Thanks kindly!

[ijabbott]

Hint: a cylinder is flat if you unroll it. So for a single turn, the length is SQRT((2*PI*R)^2 + L^2), where R is the radius of the cylinder (helix) and L is the longitudinal separation (or "lead") between turns. Multiply by the number of turns N for the total length.

EDIT: Since you mentioned the diameter, you can of course replace 2*PI*R with PI*D in the above. Or if 'D' is the inner diameter of the helix, and 'W' is the diameter of the wire, you can use PI*(D+W)

[Nigel (UK)]

...or, if you want to do it by integration, parameterise the curve as
\[x=r\cos\theta\qquad y=r\sin\theta\qquad z={l\over2\pi}\theta\]
where \(r\) is the radius and \(l\) is the distance between one turn and the next. The square root in your integral is
\[\sqrt{r^2+{l^2\over 4\pi^2}}\]
which is a constant. Integrate over \(\theta\) from \(\theta=0\) to \(2\pi\) and you get the length of one turn; then multiply by the number of turns.

Nigel (UK)

[ArPEn]

Would you be able to demonstrate how to enter this into the calculator? I am unable to determine how to actually enter these into the calculator.