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I'm having a tough time searching for the answer to this as most of the results are in the realm of mechanical engineering and appear to be unrelated, I apologize if this is a really obvious question.

Say there is a circular arc in $2$ dimensions covering $90$ degrees at a radius of $21$. I know the length of the arc would be $\frac{21}{2}\pi$ or about $32.99$, but what if it were then stretched through the third dimension by some number $x$? How do you calculate the screw thread length?

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Take an old paper towel (or toilet paper) tube. You see the seam that runs in a helix around the tube? What is the length of that helix? Split the tube along this seam and unroll the tube. the paper unrolls into a parallelogram. What is the length of the edge?

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Pythagoras to the rescue.

The horizontal distance is $2\pi\cdot r\cdot R$ where one $r$ is the radius of the tube and the other $R$ is the number of revolutions.

And the vertical distance is the height of the tube.

$l= \sqrt { (2\pi\cdot r\cdot R)^2 + h^2} $

Your screw thread problem is nearly identical.

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As an intuitively convincing method, you could consider straightening out the arc into a line of length $\frac{21}2\pi$ and then stretching it along another axis by length x and using Pythagorean theorem to calculate its new length: $\sqrt{(\frac{21}2\pi)^2+x^2}$.

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If the screw surface can be modelled like the curved surface of a cylinder, then by flattening it, the helical screw thread becomes a diagonal line.

The length of that quarter circle screw thread can be obtained by Pythagoras's theorem, with one side of the triangle being the $(2\pi/4)\cdot21\approx 32.99$ you have calculated, and the other side being how deep the screw would go if it is turned by $90^\circ$.

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To define terms, let's say we have a helix with radius R, height H, and number of turn N (not necessarily a whole number).

The length of a small piece of the helix is $$ds = \sqrt{dx^2+dy^2+dz^2}$$ Since the 2D projection of this curve is a circle, we have $$dx^2+dy^2 = R^2d\theta ^2$$ Where $\theta$ is the angle of this small segment. We need to relate $dz$ to $d\theta$. We know that the total angle is $2\pi N$ and the total height is H, and that there is a linear relationship between z and $\theta$, so $$dz = \frac{H}{2\pi N} d\theta$$ So we then have $$ds = \sqrt{\left(R^2+\left(\frac{H}{2\pi N}\right)^2\right)}d\theta$$ Ans the arc length is $$s = \int_{0}^{\theta_{TOT}}\sqrt{\left(R^2+\left(\frac{H}{2\pi N}\right)^2\right)}d\theta = \sqrt{R^2+\left(\frac{H}{2\pi N}\right)^2}\theta_{TOT}$$ Where $\theta_{TOT}$ is the total angle of the helix (so if it winds around 4 times, it would be $8\pi$)

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  • $\begingroup$ Shouldn't it be $dx^2 + dy^2 = r^2d\theta^2$? Also, 4 windings would be $8\pi$. $\endgroup$ – Mark H Jun 2 '17 at 23:26
  • $\begingroup$ Also, the expression for $dz$ needs to be squared under the square root. $\endgroup$ – Mark H Jun 2 '17 at 23:27
  • $\begingroup$ Yes to both questions $\endgroup$ – idontseethepoint Jun 2 '17 at 23:35
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A screw thread is simply a helix. The parametric equations are, for example,

$$ x=a\cos t\\ y=a\sin t\\ z=c~t $$

Now, for any parameterized space curve, the differential arc length is given by

$$ds=\sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dy}{dt} \right)^2+\left(\frac{dz}{dt} \right)^2}~dt$$

For the case of the helix described above

$$ds=\sqrt{a^2+c^2}~dt$$

And, therefore the length of the spiral is $s=\sqrt{a^2+c^2}~t$, where $t$ is the length of the planiform curve, which, in this case, is given by $t=2\pi n$, where $n$ is the number of turns. Putting it all together we get

$$s=\sqrt{a^2+c^2}~2\pi n$$

For example, a unit helix (i.e., $a=c=1$) has a length 0f $s=\sqrt{2}~2\pi n$. I have verified these results numerically for arbitrary (random) $a,c$ and $n$.

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