Well, part of the issue is due to the conversion of linear
reciprocation versus rotational motion. The crankshaft is run by the
firing of pistons, which move in a linear (one dimensional) path of
motion. No matter how many stroke-cycles (2 or 4) you have, you will
be moving something in a linear path. This piston (and the armature)
have mass. If you have mass, and you change its direction simply back
and forth (and you aren't rotating it), you're going to lose energy.
Let's say that the crankshaft assembly is turning, which has a mass
and rotational frequency. The fact that this assembly is rotating
provides us with a value for "kinetic energy due to rotational
inertia" - if all other factors are left unchanged, this should remain
constant when the engine is up to speed. In other words, if we assume
the engine is up-to-speed when we start observing it, this energy
shouldn't really be changing except due to friction. The rest of this
comment assumes that you have an "ideal" combustion cycle, and you
don't have to worry about heat loss due to entropy.
With each combustion cycle, the piston moves forward and backward.
The basic number-crunching here will assume you have a 4-stroke
If you don't know what a 4-stroke ICE works, check out
http://auto.howstuffworks.com/engine1.htm , which also does a great
job of visually showing you the 4 parts of the engine cycle:
Intake (Piston moves 1 chamber length)
Compression (Piston reverses direction and moves 1 chamber length)
Combustion (Piston reverses direction and moves 1 chamber length)
Exhaust (Piston reverses direction and moves 1 chamber length)
Let's solve it symbolically, shall we?
You have a piston of mass M, and your crankshaft moving with a
rotational speed, (Omega). We know that for every revolution, the
piston has to reverse direction twice, and travel a length L. If the
piston reaches a linear speed, V, between reversals, then it must
carry a kinetic energy of (1/2)*M*V^2. Because the piston stops and
reverses direction, this energy has to go to Zero twice per crankshaft
rotation. Let's also say that your rod-bearing assembly has a radius
of R. Therefore, the maximum kinetic energy of your piston head will
be (1/2)*M*(Omega*R)^2 (this is an approximation; the linear maximum
speed of your piston head will vary depending on the individual specs
of your engine).
This kinetic energy is created and dissipated twice per rotation. So
now you have Ke=[M*Omega*R^2] each time the crankshaft rotates. You
also have a frequency, which is determined by 2*pi*omega assuming your
Omega is in rad/sec. So now, you will have a frequency in
revolutions-per-second, so I would multiply that kinetic energy
loss-per-cycle by your cycles-per-second.
This final result should yield an answer in energy-per-second, more
commonly described as "power". This power efficiency will factor in
to your losses due to internal inertia and kinetic mechanical losses,
which occur due to the breakdown between linear and non-linear axes of
Please let me know if this comment/answer helped you.