# Car racer seeks help

• Oct 8th 2006, 12:28 AM
Chippy
Car racer seeks help
Hi All,

A bit of an odd one this, but I am getting desperate and just stumbled across your very knowledgable forum, I wonder if you could help me? I race a car here in the UK and I'm designing some new parts to make us go faster, however I have come undone with my Math!

I need to divide a circle of 60mm diameter into 3 parts of equal area using two horizontal lines, so I need to know how far apart to space the lines, hope this makes sense, please see enclosed diagram if it doesn't! Please note, on the diagram it says volume this is because it is a tube, what I really mean for the purposes of this is equal Areas by correct spacings for the lines.

Although the theory and equations behind it would be useful, what I really need is an answer so I can get the order in for the parts!

Many thanks in advance for your help,

Chip.

http://i103.photobucket.com/albums/m..._II/IR2001.jpg
• Oct 8th 2006, 04:22 AM
CaptainBlack
Quote:

Originally Posted by Chippy
Hi All,

A bit of an odd one this, but I am getting desperate and just stumbled across your very knowledgable forum, I wonder if you could help me? I race a car here in the UK and I'm designing some new parts to make us go faster, however I have come undone with my Math!

I need to divide a circle of 60mm diameter into 3 parts of equal area using two horizontal lines, so I need to know how far apart to space the lines, hope this makes sense, please see enclosed diagram if it doesn't! Please note, on the diagram it says volume this is because it is a tube, what I really mean for the purposes of this is equal Areas by correct spacings for the lines.

Although the theory and equations behind it would be useful, what I really need is an answer so I can get the order in for the parts!

Many thanks in advance for your help,

Chip.

See the attached figure:

The area of the shaded segment is:

A=(1/2) r^2 [theta-sin(theta)]

(this is a standard result). Now we want this to be 1/3 of the area
of the circle, or:

A=(1/3) pi*r^2,

so we have:

(1/2) r^2 [theta-sin(theta)] = (1/3) pi*r^2,

or after a bit of simplification:

theta - sin(theta) - (2/3)*pi = 0.

Now this is a mixed algebraic-transcendental equation so really
has to be solved numerically. Doing so gives:

theta ~= 0.264932.

Finally we can see from the diagram that:

cos(theta/2) = h/r,

so:

h = r*cos(theta/2) ~= 0.264932 r.

Finally we may observe that your B = 2h, and A = (2r-B)/2, and as
your r= 30 mm, we have:

h ~= 7.94796 mm,

B ~= 15.8959 mm

A ~= 22.052 mm.

RonL
• Oct 8th 2006, 05:42 AM
Chippy
Thankyou very much! It's been an education and a privilege.
• Oct 8th 2006, 06:55 AM
CaptainBlack
Quote:

Originally Posted by Chippy
So you are saying that:

A = 22.052 mm.

B = 15.8959 mm

C = 22.052 mm.

And you are 100% certain?

I'm reasonably confident in the values, but errors can always occur.

I've done a construction on squared paper, and the areas are equal
to within the limits of accuracy of that method (which is of a few percentage
points).

Also Monte-Carlo integration of the areas finds them equal to about 1
part in 1000.

If there are any glaring errors one of the other regulars here will probably
correct them within about 24 hours.

RonL