A Math Question

[The Helper]

There is a formula to calculate the weight per foot of steel pipe.

OD - this is the Outside Diameter of the Pipe
Wall - this is the Thickness of the Pipe
WTPerFt - this is the Weight per foot of the pipe

The formula for calculating weight is

WtPerFt = ([od]-[wall])*[wall]*10.68

The above formula is good when OD and Wall are known. The problem is trying to
modify this formula if we don't know Wall and we have WtPerFt and OD. Then it goes
into the Quadratic formula, or at least I think it does and it gets hard.

Can anyone work out a formula from this to get the Wall size if WtPerFt and OD is given.

Here are some correct values if you want test the answer.

OD     Wall   WtPerFt
---    ----   --------
6.625  .250   17.02
10.75  .365   40.48
12.75  .375   49.56

Thanks!

 

[Bronxernijn]

This should do it!

 

[The Helper]

Thank you very much! I will not have a chance to try it out until next week but cool!

 

[Slapshot136]

http://www.wolframalpha.com/input/?i=f+%3D+%28[o]-[w]%29*[w]*10.68+solve+for+w

 

[The Helper]

So which one is right?

 

[Dave312]

Both are correct. The Wolfram Aplha solution has factored out the decimal points to use whole numbers only so that is why it looks different.