BMW M54 Harmonic Damper

This will be a multi-post series discussing the function of the harmonic balancer and the shortcomings of the factory unit on the BMW M54 engine, specifically the M54B30.

Why do we need a harmonic damper?

We were all told that the inline-six engine is free of unbalanced forces and moments and is "perfectly balanced" blah-blah... The reality is that it's impossible to have a piston engine with a crankshaft and connecting rods which is perfectly balanced. But let's go deeper and look at where these difficulties come from.

The Reciprocating Piston Engine

We all know the internal combustion engine converts the chemical energy from the fuel into mechanical energy we can use; and the connecting rod links the reciprocating motion of the piston to the rotating motion of the crankshaft.So the piston moves up and down while the crankshaft rotates and the connecting rod has to do both -- sounds complicated. And it is! Now, let's look at the big end of the connecting rod in detail (see Fig. 1). Referring to this p.o.s. MS paint sketch as a figure makes me chuckle...

So we have the crankshaft spinning at constant RPM which means that the time it takes to make each of those 90 degree rotations above is the same. The big yellow circle represents the crankshaft main journal spinning in the main bearing while the blue-black circle represents the rod journal traveling with the rod big end. The stroke is twice the distance between the main journal and the rod journal (blue-yellow to blue-black circle called the crank throw) and is constant at 89.6mm for our M54B30 engine. The rod length is also constant and is 135mm. Let's call the height of the piston at the bottom (bottom dead center, BDC) our reference height and say it is 0mm. From my amazing sketch you can see that when the piston is at top dead center (TDC) the height is equal to twice the crank throw, i.e.the stroke of 89.6mm.
But what happens at the two points 90 degrees before and after TDC? From the good ol' Pythagorean theorem we can calculate the height relative to BDC as h = sqrt(rod length^2 - (stroke/2)^2) - (rod_length-stroke/2)  to get our height above the reference height which means... h=37.15mm. Tada! Okay, I agree it's not obvious. So we start at 0mm and 90degrees later we are at 37.15mm above BDC. But wait! The crank throw is 89.6/2=44.8mm!!! So we take half the time, but traveled less than half the distance. And in the next 90degrees we go from 37.15mm to 89.6mm which is 52.45mm. So we travel more than half the stroke! Okay, if you are still following along you get some kind of gold star B) We just discovered that the piston travels slower on the bottom part of the stroke and faster on the top part of the stroke using only high school geometry. This is very significant. In the next post we will see how the rod to stroke ratio is the key to understanding the dynamics of the piston motion.
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