rod length / engine stroke

This makes people think that at a 1.5:1 ratio , the rod is only 50% longer then the crankshaft.

The problem with that is the crank throw is only

*half*the engine stroke.

therefore it should be stated as:

rod length / crank throw

now this same 1.5:1 r/s ratio is 3:1. Sounds a little better doesn't it?

now lets get into angles. The "popular internet argument" is that a lower ratio radically increases rod angles, therefore pushing the piston into the cylinder wall.

Lets compare two popular engines.

first one being the k20, second k24, since it seems most people here are interested in the two.

the k20 has a 3.39 stroke, 1.7" crank throw.

the rod length is 5.985"

the rod to crank throw ratio is 3.52:1

now lets use some high school trig to figure it out. remember SOH CAH TOA?

lets use it to figure out the most extreme rod angle, at 3 oclock and 9 oclock.

the formula to use is cos(theta) = adjacent/hypoteneuse(spelling anyone?)

cosine is simply a button on your calculator(using a computer, hit "View" then

click "scientific".

theta is what we are trying to figure out(it will be the rod angle in degrees)

adjacent is the crank throw(NOT THE STROKE)

hyp is the rod length, center to center obviously

lets plug the #'s

cos(theta) = 1.695 / 5.473

cos(theta) = .3097 blah blah blah...

now "check" the inverse on the calculator and click cos, we have figured out a rod angle of 71.958 degrees.

now lets figure the k24

the stroke is 3.9" crank throw 1.95"

rod length is 5.985"

cos(theta) = 1.95 / 5.985"

cos(theta) = .3258

theta = 70.985

a whole degree.

just for fun lets toss in the f20, with its "awesome rod angle"

I couldn't find exact measurements, but i did find the r/s online at 1.82. We can just double that and inverse it. 1 / 3.64 = 2.74

theta is 74 degrees.

There you have it, 3 degrees between the f20 and the "terrible" k24.