Need help-integration? I'm embarassed how much my math skills have slipped.?

Question

I'm trying to determine an expression for average torque during the power stroke from a one-cylinder engine with a standard crankshaft. 

Let the crankshaft angle when the piston is all the way down be 0 degrees, and when the piston is all the way up, let it be 180. Let the connecting rod be long so the angle between it and vertical can be ignored.

An expression for the instantaneous torque at 90 degrees (when the piston is halfway from bottom to top) would be 
T = F*d 
where F is the force and d is the distance from the crankshaft centerline to the bearing where the connecting rod and crankshaft mate (the lever arm).

The expression
T = F*d*sin(theta)
should be valid while theta varies from 0 to 180, the 1/2 revolution during which power is available. )Probably F would not be a constant but let's assume it is.)

My problem is how to get the average torque over the 180 degrees of the power stroke. Please, I really need to learn how, more than I need the answer.

Thanks

miamidot · Accepted Answer

You are presuming the force F of combustion is constant throughout the cycle. 

For convenience, replace the notation theta with u, the angular displacement in radians.

Integrate the total work W done over u=0 to u=π .
Average Torque = W/(π)

Why can't we say the average Torque is W/(180)?
That is because the foundations of angular mathematics have determined that 2πr is the circumference of a circle. 180 or 360 are merely arbitrary angular representation that bears no relation to the dimension of a circle.

For the case where we presume F is constant throughout the cycle.
Work done W = integral of Fdsin(u)du
W = ∫ Fdsin(u)du
 = Fd(1-cos(π))
 = 2Fd

Simple answer: Average torque = W/π = 2Fd/π.

However, in reality, the explosive force is not constant.

Let us presume that the explosive Force diminishes to k% at end of the cycle. The most efficient engine would have k=0%, but some hotrodders would want their engine produce max torque rather than fuel efficiency. For simplicity, let's say k is 0%, and that force at angle u is f(u).

But u is an angular function of piston displacement e.

It would be smart to have the engine time the max explosive force when the cam is exactly at 90 degree position, i.e. at π/2.
We also have to presume that the explosion started at 0 degrees and that the explosive force f is a function of u.
f(u) = Fdsin(u), where F is peak force.

In this manner, we don't have to worry about calculating f based on e, and then worry about e as a function of u.

Therefore, the Torque at angle u
T(u) = f(u)dsin(u)
 = Fdsin(u)sin(u)
 = Fdsin^2(u)

Ignoring residual torque of previous cycle,
Work done W = Integral of T(u)du
W = ∫ Fdsin^2(u)du
 = Fd(u/2 - sin(2u)/4)
 = Fd(π/2 - sin(2π)/4)
 = Fdπ/2

Average torque = W/π = Fd/2.
Remember that this F(peak force) is different from the previous F (constant force).

Since I am looking at this problem only from the view of calculus without intimate knowledge of the engine involved, it must be pretty inaccurate. Does the manufacturer of the engine provide or can you tune the engine:
- what is the residual torque from the previous cycle?
- at what angle is the spark ignited?
- what is the curvature of torque across the cycle?
- is it a diesel engine or a plug ignition engine?
- what is the residual force at the end of cycle?

For a proof for integral of sin^2u refer to
http://calc101.com/special_1.html

For a list of trigonometric integral solutions:
http://mathworld.wolfram.com/IndefiniteIntegral.html

mith · Answer

I'm sorry I don't know the answer, but to help restore some of that lost knowledge look at you diet. The single biggest element that causes brain damage is too much Mercury. Their are of course many others. Go to the doctor for a better answer to this one.

With anything of course, it will take time.

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Need help-integration? I'm embarassed how much my math skills have slipped.?

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