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What Is The R/s Ratio?

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Old 02-02-2003, 10:33 PM   #1
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Think about what R/S means for a second. Rod length / Stroke length. All it says is that ideally your rods are 1.75 times the length of you stroke. Now if you stop and consider why that ratio is so "good". The longer your rods are compared to you stroke the smaller the angles are between vertical and rod position as the pistons travels up and down. Envision a cross section of an inline reciprocating piston engine. As the piston starts to climb from BDC the rod forms an angle with vertical at the wrist pin. If you have taken any physics this is pretty basic. Think of a force diagram illustrating the forces acting on the wrist pin ( for simplicity sake assume that the piston and wristpin are one unit). The larger the angle between vertical and the rod, the larger the sine of that angle, and the larger the horizontal component of the force applied to the wrist pin (meaning more side loading on the piston and cylinder wall causing ovaling of the cylinders, increased piston wear, etc.).

1.75:1 is "perfect" because it is a good comprimise. The steeper rod angles make for more torque and longer rods put less pressure on the sidewalls. 1.75:1 is the optimization of these two factors (so you get the most torque with the least sidewall loading). Most engines run a an RS ratio of less than 1.75:1.
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Old 02-04-2003, 12:21 AM   #2
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Just for reference... stock rod/stroke ratios are as follows, rounded to the nearest thousandth:

B16A / B16B = 134.36 / 77.0 = 1.745
B17A = ? / 81.4 = ?
B18A / B18B / B20B / B20Z = 137.0 / 89.0 = 1.539
B18C = 137.9 / 87.2 = 1.581
B20A = ? / 95 = ?

If you want more precise ratios, calculate them yourself. If anyone can give proof of what the B17A and B20A rod lengths are, please PM me and let me know. You can use these numbers to make your own crank/rod combos to see what kind of ratios you'll get with different builds.
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