TIP: Formula for rubber drive band path length

[N_Cook]

ie the stretched length where it is too restricted to pass a thread around
to measure. Or to measure without taking apart a worker, to obtain a spare,
before having to take apart to replace. Using callipers, pair of rods and
ruler or whatever to take measurements of diameters and so radii and gap,
templates, whatever intrusions will allow.

2 pulleys , small diameter "a" and big diameter b and gap between the 2
pulleys c.
So distance between pulley centres is a +b +c
Straight belt section between the two pulley rims (doubley tangential) is x
N is the angle, in degrees, between the line-of-centres and the b radius to
the tangent point of the large pulley. Also the angle between this extended
line and the radius a of the small pulley.

Parallelogram with parallel sides a, b and the other two sides x and (a+b+c)
and two right angles.
So reducing the parallelogram by lenghth "a" down to a right angle triangle
of lengths x, b, (a+b+c) and N between b and (a+b+c)

x = sqrt [(a+b+c)^2 - (b-a)^2]

and N= cos(-1) [ (b-a)/(a+b+c)]

Then band length = 2x + 2*Pi*a (2N/360) + 2*Pi*b[1 - (2N/360)]

Confirmed practically in one case anyway

--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/

 

[Joe]

I think that your math is correct, but I worked out the formula using
radians, not degrees, so I would have to convert my formulas to the degree
form before I could say that yours is correct.

There is however one small quibble. In the first part of your
explanation, you state that "a" and "b" are are diameters, and of course
you meant that these are the radii, as you correctly state later on.

This brings back memories. The total length of a belt around two pulleys
of different diameters was a trig problem in my 11th grade algebra-trig
class. Only two of us thought we had the correct solution, and I, the
11th grader, got it right. The other guy, a 12th grader, got the
tangential lengths correct, but screwed up the lengths that are wrapped
around the pulleys.

I got to present my solution at the board in front of the whole class.

--- Joe


In article <h17k5f$283$1@news.eternal-september.org>, "N_Cook"
<diverse@tcp.co.uk> wrote:

ie the stretched length where it is too restricted to pass a thread around
to measure. Or to measure without taking apart a worker, to obtain a spare,
before having to take apart to replace. Using callipers, pair of rods and
ruler or whatever to take measurements of diameters and so radii and gap,
templates, whatever intrusions will allow.

2 pulleys , small diameter "a" and big diameter b and gap between the 2
pulleys c.
So distance between pulley centres is a +b +c
Straight belt section between the two pulley rims (doubley tangential) is x
N is the angle, in degrees, between the line-of-centres and the b radius to
the tangent point of the large pulley. Also the angle between this extended
line and the radius a of the small pulley.

Parallelogram with parallel sides a, b and the other two sides x and (a+b+c)
and two right angles.
So reducing the parallelogram by lenghth "a" down to a right angle triangle
of lengths x, b, (a+b+c) and N between b and (a+b+c)

x = sqrt [(a+b+c)^2 - (b-a)^2]

and N= cos(-1) [ (b-a)/(a+b+c)]

Then band length = 2x + 2*Pi*a (2N/360) + 2*Pi*b[1 - (2N/360)]

Confirmed practically in one case anyway

--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/

 

[N_Cook]

confirmed, my error , a+b+c beween centres is correct, a and b are radii.
I edited , incompletely, from diameters to radii, also the word radius gauge
templates should have been inserted after the word ruler