# Need help with a math problem...

First of all, I am not asking for help with homework - I aced all three semesters of calculus when I took it 120 years ago. It’s just that I haven’t used it since then.

Here’s the problem I’ve been pondering: describing the velocity and acceleration of a piston in a car. Using fairly standard notation, I’ve got a triangle with sides a, b and c; with angles A, B and C opposite the respective sides. “a” represents the rod length, “b” half of the crankshaft stroke, and “c” the position of the piston wrist pin. Angle “A” is the angle of the crankshaft.

The law of cosines says:

a^2 = b^2 + c^2 - 2bc*cosA

Rearranging terms:

0 = c^2 + c*(-2bcosA) + (b^2 - a^2)

c = ((-(-2bcosA)±((-2bcosA)^2 - 41(b^2 - a^2))^.5)/(2*1)

(provided I got all of the parentheses right), or:

c = bcosA ± (a^2 - b^2 + b^2cosA*cosA)^.5

Taking the derivative with respect to time (indicated by “’”):

c’ = -bsinAA’ ± (.5*(a^2 - b^2 + b^2cosAcosA)^-.5)b^22cosA-sinA*A’

or:

c’ = -bsinAA’ ± b^2*(a^2 - b^2 + b^2cosAcosA)^-.5)cosA-sinA*A’

(Here’s where I get doubtful):

c" = -b*(sinAA" + cosAA’A’) ± b^2((-.5*(a^2 - b^2 + b^2cosAcosA)^-1.5)b^22cosA-sinAA’)cosA-sinAA’ + (a^2 - b^2 + b^2cosAcosA)^-.5)-sinA-sinAA’A’ + (a^2 - b^2 + b^2cosAcosA)^-.5)cosA-cosAA’A’ + (a^2 - b^2 + b^2cosAcosA)^-.5)cosA-sinA*A"

I’m only interested in constant RPM, so A" = 0:

c" = -bcosAA’A’ ± b^2((-.5*(a^2 - b^2 + b^2cosAcosA)^-1.5)b^22cosAcosAsinAsinAA’A’ + ((a^2 - b^2 + b^2cosAcosA)^-.5)sinAsinAA’A’ + ((a^2 - b^2 + b^2cosAcosA)^-.5)cosA-cosA*A’*A’)
Well, did I get it right?

I may have seen a more unnecessarily complicated and pointless solution for SHM in my time, but I doubt it.

Errr, what’s SHM?

Is this something I could easily have looked up if I knew where to look?

Simple harmonic motion.

The equations for the piston’s position, velocity and acceleration are all of the form:

A cost + φ),

where

ω = angular velocity of the crank,
t = time,
and A and φ are constants that depend on the physical dimensions of the system.

If the crankshaft is turning at a constant angular speed, then the piston will not execute simple harmonic motion. I think.

As far as I can tell, you got it right. That trig could really stand to be cleaned up a bit. But not bad. If you’re describing a physical system like a piston, you don’t need to carry the ± through. You can just pick positive.

However, may I ask what the heck you’re trying to do? Simply get an expression for c’’ as a function of A and A’?

Achernar is correct, as usual.

An exact translation from uniform circular motion to simple harmonic motion in a reciprocating engine requires a sliding con-rod, rather than the typical arrangement.

For most engines SHM is only an approximation, which introduces errors when the con-rod length is small compared to the stroke.

Let θ = ωt = angular position of crank.

Piston position = a + b - b cos(θ) - a sin(cos[sup]-1[/sup](b sin(θ)/a))

= a + b - b cos(θ) - √(1 - (b sin(θ)/a)[sup]2[/sup]))

Crap.

Rearrange that last line as:

= a + b - b cos(θ) - √(1 - (b/a)[sup]2[/sup]sin[sup]2/sup)

Still doesn’t look good. Try:

= a + b - b cosθ - √(1 - (b/a)[sup]2[/sup]sin[sup]2[/sup]θ)

That’s better.

Of course, if you’re going to derive velocity and acceleration from that equation, you have to replace θ with ωt, and differentiate with respect to t.

Well, I really don’t remember any of the tricks from college days, so I tried to get it by brute force. I’d welcome any “cleanup” suggestions.

Just trying to be rigorous, so I wouldn’t do something stupid and botch it all up.

Exactly.

The reason for all this is a comment on a car discussion group to which I belong - something like “5mm extra rod length wouldn’t make any difference to engine performance”. Depending on the factory specs, it sure as heck could. My fapidly fading recollection is that race cars usually have a rod/stroke ratio of around 2.0:1. Thought I’d run the equations through Excel and see what the difference actually would be.

Thanks for the response.