I am kind of confused when it comes to this topic. I think I kind of understand primary balance. Which is the back and forth rotation of the pistons. But I don't fully understand it or even get secondary engine balance at all.
The simplest way of explaining it would be to say primary, first-order, balance is related to things that vibrate the engine at a frequency equal to the engine speed (e.g., 1000 Hz at 1000 RPM). Secondary, second-order, balance is related to things that have a frequency of twice the engine speed and so on.
In common usage, primary usually refers to sinusoidal and secondary non-sinusoidal vibrations. Of course this all begs the question of what causes the different types of vibrations…
Imagine a typical engine with pistons connected to the crankshaft with connecting rods. While in operation the pistons drive the crankshaft, it may be easier to understand what is happening by thinking about what happens as the crankshaft rotates – we'll see that the vertical movement of the piston is not equal as the piston moves the 180º from 90º after top dead center (ATDC) to 90º before top dead center (BTDC) as it is when it moves the 180º from 90ª BTDC to 90ª ATDC. This difference creates unequal speeds of moving masses and therefore unequal vibrations. Secondary balance is related to these vibrations.
One way to prove this to yourself is to go back to high school geometry and the Pythagorean theorem (a2 + b2 = c2, where a and b are the short sides of right triangle and c is the hypotenuse (the long side)). In thinking about what is happening as the crank turns consider the connecting rod to be c, the crank throw to be a and the displacement of the piston to be b. We can use a 3-4-5 right triangle to make this easy to think about in our heads:
At 90º BTDC or ATDC we have a right triangle formed by the horizontal throw of the crank (a which we can assign to 3 in this case), the vertical displacement of the piston from the center of the crank (b which is 4), and the connecting rod itself (c which will be 5). So the piston is at a position of 4 units above the center of the crank.
At bottom dead center (BDC), where the crank throw is straight down, the piston displacement is 2 (length of connecting rod minus the crank throw). The piston has moved down two units from the 90º position of 4.
At top dead center (TDC), the crank throw is now straight up and the piston's displacement is 8 (length of connecting rod plus the crank throw). The piston has moved up four units from the 90º position of four.
The unequal distance in the two halves of the crank's rotation translates into unequal piston speeds and thus unequal inertia and vibrations.
1Fantastic answer. I think the OP may also need a pic of counterbalancing shaft in relation to the crank to understand how a counterbalancer can flatten out the frequencies. Simply a suggestion.....but this is a great answer that get's to the core. Thank you! Jul 7, 2016 at 18:09
Ok, I'll need to get on that. Finding good illustrations has been harder that I'd have thought…– dluJul 9, 2016 at 2:15
1I dug around a bit. Here's one. It's seems to illustrate what a counter balancer does. dansmc.com/counterbalance3.jpg Jul 9, 2016 at 4:00
1Here's a great image. 66.media.tumblr.com/129ba31bdd7fc5461b159446dd244771/… Jul 9, 2016 at 4:02