Very little information to go on here. I wish I could read minds but unfortunately I can't.

I am assuming that you mean "How do you caculate the length of belt in a roll?" and not how much belt is around a pulley.

I use these formulas:

When thickness of the belt is known:

L = ((D*D)-(C*C)) / (15.3*t)

L = Length; D = Outside Diameter (inches); C = Inside Diameter (inches); t = belt thickness (inches)

When wraps are known:

L = ((D+C)*3.14*n) / 24

n = Number of wraps in roll ■

Dear Mr.RK2007,

Imagine you are looking the edge of the unwound straightened belt.

You will be seeing a very long rectangle.

The area ( A ) of the rectangle is

= Length of the belt ( L ) x thickness of the belt ( t ) ------- ( 1 )

Now you have started rolling the belt and making a drum .

Now again you are looking from the side.

You measure Bigger diameter ( D ) of the drum & smaller diameter ( d ) where there is no belt.

The area ( A ) remains same

= Area of the Bigger diameter circle - Area of the smaller diameter circle ----- ( 2 )

Equation ( 1 ) = Equation ( 2 )

So Length of the belt in the belt drum = ( Area ) / ( Thickness of the belt )

Word of caution : "The drum should be properly wound. It should not be loose & do not have many gaps in between the wraps."

Regards, ■

Originally Posted by Gary Blenkhorn

Very little information to go on here. I wish I could read minds but unfortunately I can't.

I am assuming that you mean "How do you caculate the length of belt in a roll?" and not how much belt is around a pulley.

I use these formulas:

When thickness of the belt is known:

L = ((D*D)-(C*C)) / (15.3*t)

L = Length; D = Outside Diameter (inches); C = Inside Diameter (inches); t = belt thickness (inches)

When wraps are known:

L = ((D+C)*3.14*n) / 24

n = Number of wraps in roll

Dear Mr.Gary Blenkhorn,

Help us to understand the derivation of the above formulas.

Thanks & regards, ■

L = ((Daverage)*3.14*n

Daverage = (D+C)/2

L = ((D+C)*3.14*n) / 2

( Instead of L = ((D+C)*3.14*n) / 24 )

Example:

C=1 m

t = 0.005 m

n = 50 wraps

D = 1+2*n*0.005 = 1.5 m

L = L = ((1.5+1)*3.14*50) / 2 = 196 m

The formula:

L = ((D*D)-(C*C)*3.14) / (48*t)

should be:

L = ((D*D)-(C*C)*3.14) / (4*t)

example:

L = ((1.5*1.5)-(1*1)*3.14) / (4*0.005) = 196 m

Success

Teus ■

Dear sganesh,

Lateral area of the coil is:

Area = 3.14/4*(D*D - C*C) = L*t

Resulting in:

L = 3.14/4*(D*D - C*C) /t

Mathematics can be simple.

Have a nice day

Teus ■

The 24 is used to convert the roll measured in inches into feet for the length of the roll. Thus 2*12 = 24. I should have stated that L = Length in Feet

Imperial is sometimes more confusing than metric. But that is what I learned in school instead of metric.

The 15.3 is dirived by 4/3.14(Pi)*12 (inches per foot).

I think in imperial and not thinking about our Metric friends. Sorry for any confusion. ■

Dear Gary,

Are you also driving on the left side in a metric country?

Greetings

Teus ■

Originally Posted by Teus Tuinenburg

Dear Gary,

Are you also driving on the left side in a metric country?

Greetings

Teus

Teus

The country may have gone metric but us old school guys didn't change. Unless it is a government project all designs are done in imperial. Maybe when us old guys are all dead and gone that may change.

Funny thing is all building materials are still supplied in imperial. If you framed a house to the stud spacings as per the building code you would not be able to find a piece of plywood that would fit because they are sold as 48" x 96". Oh you could buy it in metric sizes but it would cost about 25% more.

And no we do not drive on the left side, we drive on the RIGHT side. The correct side. haha

Cheers,

Gary ■