# Helix / Spring Volume

#### Bunion

Hello, I need some help learning how to calculate the volume of a Spring. The only resource i can find is on wikipedia, but i am having trouble understanding it.

Please see : Spring (mathematics) - Wikipedia, the free encyclopedia This is the forumula they give.

where
R is the distance from the center of the tube to the center of the helix,
r is the radius of the tube,
P is the speed of the movement along the z axis.
Ok, the helix will be 400 wide, so R = 200 units.
The radius of the tube will be 30, so r = 30 units.

Now this is where im stuck, they ask for P for the speed it moves upwards / downwards. How should i get this? For example if i want my spring to have 2 complete revolutions and have a total height of 500 units, how should i know what P is?

I also need to know n, which is not explained very well what n is on the page.

Many thanks if you can help me understand.

#### galactus

MHF Hall of Honor
Your helix has radius 200 and the tubing as radius 30. Correct?. This conflicts with R which says they're both 200.

In this event R+r=230. That's what I'll use.

Assume you're wrapping a tube of radius 30 around a helix of radius 200,

This forms a helix. A helix can be represented by $$\displaystyle x=acos(t), \;\ y=asin(t), \;\ z=ct$$

a in this case is R+r=230. The radius of the helix + radius of tubing.

Since you want two revolutions when z=500 units, you have $$\displaystyle c=\frac{500}{4\pi}=\frac{125}{\pi}$$

You have the helix: $$\displaystyle (230cos(t))i+(230sin(t))j+(\frac{125}{\pi}t)k$$

It's length in making two revs in a distance of 500 units is given by

$$\displaystyle 2\pi\sqrt{230^{2}+(\frac{125}{\pi})^{2}}\approx{1466.6}$$

Now, this is a cylinder of radius 30 with height 1466.6

$$\displaystyle {\pi}900(1466.6)=4,146,713.8 \;\$$

Check my calculations out.

• topsquark and Bunion

#### Bunion

Wow thanks for taking the time to explain that to me more clearly.

Hopefully your message will help others aswell if they find this through a search engine.

#### jaeminshin11

This is so useful, thank you so much.

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