snips

Rolf,
I followed your math but I don't understand why the R/1 of .28 is added
at TDC and subtracted at BDC???  It seems like it should be added at BDC
as well. Also, I didn't see where rod length is factored, with a short
rod accelerating faster from TDC and BDC than a long rod (the benefits
of this is always a hot debate around car guys; it does seem that the
longer rod would result in lower side loads on the piston).

Ken Powell

Hi Ken and Bob,

Ken the R/1 should read R/L where L is the length of the connecting rod. I know a small ?l? can easily be confused with a 1.
The connecting rod length in this example is about 9 ? long, perhaps a bit less. Hence R/L = 2.5 / 9 = .278. This length is just an assumption. They are generally in that neighborhood.

By this ratio R/L the length of the connecting rod is accounted for. The impact on acceleration is exactly that of its value. I think 28 % is not small.

Now, how to explain the addition and subtraction of this value?

Assuming the piston is halfway down. Draw a circle from the centre of the wrist pin with the radius of the connecting rod length. This circle passes through the centre of the crankshaft and cuts the
radius of the crank throw above the centre line of the crankshaft.

This is where the piston is near its maximum velocity. The maximum speed of the piston occurs when the crank radius and the connecting rod form an angle of 90 degrees. This is nearby, at about 15.5
degrees slope upwards from a horizontal line through the centre of the crankshaft.

It follows that from this maximum velocity, where acceleration is zero, the rotation angle of the crankshaft, and therewith the time, is shorter around the top for for TDC than it is for BDC.
Consequently the acceleration is larger at TDC and smaller at BDC. Or said differently, acceleration is larger at TDC and smaller at BDC than the acceleration of a rotating body, here the crankpin.
The formula for the crank pin acceleration is simply: acc = w^2 * r, where ?w? is the angular velocity in Rad/s, w = Pi * n /30, ?n? is rpm and ?r? is the crank radius.

Another way to look at is to split the circle in half and calculate from 3 and 9 o?clock positions. As we all know at 3 and 9 o?clock crank locations the piston is more than half way down. The
vertical length of the connecting rod is only 8.65 inches; hence the piston is .35 inches lower. The upper travel is 2.5 + .35 = 2.85?, the lower travel is 2.5 - .35 = 2.15?. The time is the same
for both, it follows that the acceleration of the piston varies from that of the crank pin. It is higher around the top and it is lower around the bottom. Easy.

Luckily, the increase and decrease is the same, and is also the same as the ratio of R/L, that is crank radius over connecting rod length. So it becomes an easy calculation for TDC and BDC.

The instantaneous acceleration at any crank angle is a bit more complicated and beyond this Forum, I think.

Rolf Pfeiffer

I agree. That is why I explained it in the article in generalities. 

Paul Lamar
 
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