Helix
You have a cylindrical tube, 9 inches long, with a 4 inch outside circumference. Ten turns of a wire are helically wrapped around its total length and secured at the ends.
How long is the wire?
How long is the wire?
posted by Zaux at
10:10 AM
9 Comments:
you can imagine the wire as the diagonal of a rectangle with one side as long as the tube is high and the other side as long as the tubes circumference * 10
so simply use Pythagoras' theorem:
heigh^2+(circumference * 10)^2= wire^2
so the lengh of the wire is roughly 90.089 inches
oh, i mixed up the lengh and the circumference, the right answer is exactly 41
Good question and good explanation guys.
regards, Curtis
Hi Carc ...
41 is correct
Roll the cylindrical surface and wire onto a plane. The element
(9"). the repeated circumference
(10"*4)= 40", and the wire(L) now form a right triangle.
Using Pythagorean theorem,
L^2 = (40)^2 + (9)^2
= 1600 + 81
L = 41" of wire
I agree, that was a nice explanation.
I looked at it as 4 inches circumference, 10 times, plus 1 tenth inch each time round to make the 10 turns to the top of the cylinder, 41 inches.
Then I saw the other explanations, and realised the error of my ways and maths, and have duly hung my head in shame....
Karl ...
very clever analysis
Zaux,
This comment is not about this post. I have not even read it. I just got home, and posted the solution to the Two 2 inch intersecting cylinders.
You are so good at posting I can't keep up.
Ragknot ...
it will be easy to keep up henceforth ...
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