# Area of a helical surface

1. Mar 5, 2015

### BALU E GEORGE

Hello.
I would like to know the derivation of surface area of the helical plate, of single turn with a pitch p, diameter D, attached to a shaft of d. It will look like a circle in plan
Also, I would like to get the equation relating the torque resisted by the surface, if it is made to enter a material of shear strength s or frictional coefficient μ.
Thankyou

2. Mar 5, 2015

### haruspex

The plate shown appears to have been produced by splitting a flat annular disk. If so, the area will be the same as when it was flat. The plan view (as it is, mounted on the shaft) would not be a complete disk. There would be a small radial gap at the split, due to the strain from the shaft.
If your question really concerns a helical flange (like a screw thread) extending over multiple turns then the area will be a bit larger.

3. Mar 5, 2015

### BALU E GEORGE

Yes sir,
My question concerns the helical flange. But it is a sungle turn. So the area will be a bit larger, as compared to a circular disc and there is no radial gap at the split. This area will increase as th pitch increases. So i was looking for a relation connecting the pitch and the radius to the area.

4. Mar 6, 2015

### haruspex

The pictured flange cannot be continued into an arbitrary number of turns. If you look at either straight edge and project it back through the shaft you will see that it meets the flange on the other side. If it were truly just one turn out of a continuing thread, like a spiral staircase, that would not happen. Instead, the line would be perpendicular to the axis of the shaft. Its continuation through the shaft would come out on the other side half way between two passes of the flange/thread.
This is why I remain convinced that it has been formed from a flat washer, split, then forced onto the shaft. The radial gap I infer would be very small. It might be hard to detect it by eye because of parallax. Whether there's a measurable gap or not, the difference between the actual area and its original area before being mounted on the shaft would be minute. Just treat it as flat.

5. Mar 9, 2015

### LCKurtz

If you want the surface area formula for a theoretical helicoid of the form$$\vec r(t) = \langle r\cos t, r\sin t , kt\rangle~~0\le t \le a,~~0\le r\le b$$it is easy enough$$A =\int_0^a\int_0^b \sqrt{k^2+r^2}~drdt$$
[Edit:] Corrected typo.

Last edited: Mar 10, 2015
6. Mar 9, 2015

### haruspex

Sure, but the pictured flange is clearly not like that.
Also, I guess you mean $A =\int_0^a\int_{r_0}^{r_1} \sqrt{k^2+r^2}~drdt$

7. Mar 10, 2015

Yes.