Re: Math problem
- From: "Don Strenczewilk" <nospam@xxxxxxxxxx>
- Date: Fri, 31 Aug 2007 13:03:42 -0400
Hmm, I can think of a few:
1. The physical pressure put on the roller will slightly decrease the radius
and cause distortion.
2. The point where the rubber contacts the paper will be slightly flattened.
This distortion amount will be affected by the pressure too.
3. The motion of turning it will very slightly expand the rubber / radius.
Is one of these along the lines of what you're thinking?
"Russ" <nobody@xxxxxxxxxxxxxxxx> wrote in message
news:46d8448b$1@xxxxxxxxxxxxxxxxxxxxxxxxx
Actually, I was just trying to give you the gist of why it was happening
to show you where to start, not a formula. I am kind of lazy that way!
Just for grins, can someone determine why even with the formula which
gives an exact result you will no get a perfect result.
Russ
"Don Strenczewilk" <nospam@xxxxxxxxxx> wrote in message
news:46d83cb5@xxxxxxxxxxxxxxxxxxxxxxxxx
Thanks very much to all of you. Uwe was right on, Russ was halfway there,
and a little tweaking made the results jibe with a formula a supplier
gave to me (which is what I was trying to verify). Although Hopeful
Skeptic's solution is a good one, there is little hope of getting it
implemented here. In case anyone wants to verify the other two methods:
program DistortionFactor;
{$APPTYPE CONSOLE}
uses
SysUtils;
function Uwe( thickness, outercircumference : extended ) : extended;
var
r : extended; // radius of the roller
begin
r := (outercircumference / pi / 2) - thickness;
result := r / (r+thickness);
end;
function Russ( thickness, outercircumference : extended ) : extended;
var
k : extended; // elongation lenth (difference in length from top to
bottom of the strip)
begin
k := 2 * pi * thickness;
result := 1 - (k / outercircumference);
end;
const
T = 0.107; // thickness
O = 15; // outer circumference
begin
writeln( Uwe( T, O ) );
writeln( Russ( T, O ) );
readln;
end.
.
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