What is wrong with my instnataneous torque formulas?

[Mark944turbo]

I thought I had these formulas working and I messed around and lost it

If work per cycle = integral of Pressure * dVolume, the most instantaneous work should get done where piston velocity is highest, because dV will be biggest, if you assume constant pressure. So doing that out, my engine has the highest piston speed at 77 degrees ATDC, and it all works out, torque will be highest at 77 degrees if pressure in the cylinder is constant.

Next formula: instantaneous torque on crankshaft by 1 piston will be equal to connecting rod force in the direction of the cranks tangent, times the crank radius. So from a formula right off of this site :

http://www.mayfco.com/rods.htm

sin(x)*(1-((r/l)sin(x))^2)^1/2 + cos(x)*...sin(x)*(r/l)
M(x) = F * r* -----------------------------------------------------
(1-((r/l)sin(x))^2)^1/2

where M is torque
F is piston force
r is crank radius
x is degrees
r/l = crank/rod ratio


And, when I do this out, the highest torque happens around 66 degrees ATDC for my rod ratio, assuming constant pressure. I dont see why these 2 formulas disagree.

I dont expect anyone to take the time to fully work this out, but maybe I am missing a concept?

 

[EngJW]

I think the formulas are for two different things (1) work which is force times distance, and (2)torque which is force times moment arm. So you can't compare them: distance that piston moves is not equal to moment arm.

Pressure is not constant. Maximum pressure occurs near tdc where the moment arm is smallest, and it falls off rapidly to the point where the moment arm is greatest.

Without studying your equation in detail, it looks like the classic equation for reciprocating motion found in any engine text. "F" in the equation is a variable.

 

[ivymike]

so essentially you're saying that
dW/dV = P (derivative of work w/re volume is pressure)
and then you're jumping ahead to say that
dW/dt is greatest when dV/dt is greatest if P is constant

(dW/dt) = (dW/dV)(dV/dt) = P*(dV/dt)

and you know that dV/dt is proportional to piston velocity,
dV/dt = A*(ds/dt)

so dW/dt = P*A*(ds/dt) ... seems right so far

...but at the flywheel you're saying that
dW/dt = T(t)*(dq/dt) (instantaneous torque times crankshaft angular velocity)

if dq/dt is constant, then dW/dt = T(t)*crkvel

so T(t)*crkvel = P*A*(ds/dt)

if you have a constant crank velocity, and a constant cylinder pressure, then you can say that
T(t)/(ds/dt) = constant

but your equation for T(t) gives a peak at a different time than your equation for (ds/dt) does... which means that one of your two equations is wrong.