Piston motion equations

${\displaystyle l}$ rod length (distance between piston pin and crank pin)

${\displaystyle r}$ crank radius (distance between crank pin and crank center, i.e. half stroke)

${\displaystyle A}$ crank angle (from cylinder bore centerline at TDC)

${\displaystyle x}$ piston pin position (upward from crank center along cylinder bore centerline)

${\displaystyle v}$ piston pin velocity (upward from crank center along cylinder bore centerline)

${\displaystyle a}$ piston pin acceleration (upward from crank center along cylinder bore centerline)

${\displaystyle \omega }$ crank angular velocity

The crankshaft angular velocity is related to the engine revolutions per minute (RPM):

${\displaystyle \omega ={\frac {2\pi \cdot \mathrm {RPM} }{60}}}$

As shown in the diagram, the crank pin, crank center and piston pin form triangle NOP.
By the cosine law it is seen that:

${\displaystyle l^{2}=r^{2}+x^{2}-2\cdot r\cdot x\cdot \cos A}$

Position with respect to crank angle (by rearranging the triangle relation):

${\displaystyle l^{2}-r^{2}=x^{2}-2\cdot r\cdot x\cdot \cos A}$

${\displaystyle l^{2}-r^{2}=x^{2}-2\cdot r\cdot x\cdot \cos A+r^{2}[(\cos ^{2}A+\sin ^{2}A)-1]}$

${\displaystyle l^{2}-r^{2}+r^{2}-r^{2}\sin ^{2}A=x^{2}-2\cdot r\cdot x\cdot \cos A+r^{2}\cos ^{2}A}$

${\displaystyle l^{2}-r^{2}\sin ^{2}A=(x-r\cdot \cos A)^{2}}$

${\displaystyle x-r\cdot \cos A={\sqrt {l^{2}-r^{2}\sin ^{2}A}}}$

${\displaystyle x=r\cdot \cos A+{\sqrt {l^{2}-r^{2}\sin ^{2}A}}}$

Velocity with respect to crank angle (take first derivative, using the chain rule):

${\displaystyle {\begin{array}{lcl}x'&=&{\frac {dx}{dA}}\\&=&-r\sin A+{\frac {({\frac {1}{2}}).(-2).r^{2}\sin A\cos A}{\sqrt {l^{2}-r^{2}\sin ^{2}A}}}\\&=&-r\sin A-{\frac {r^{2}\sin A\cos A}{\sqrt {l^{2}-r^{2}\sin ^{2}A}}}\end{array}}}$

Velocity is negative when x decreases with time(A < 180 degrees) and is positive when x increases with time(A > 180 degrees). The magnitude of the second term is always less than rsinA and hence does not affect the sign (as l>= 2*r).

Acceleration with respect to crank angle (take second derivative, using the chain rule and the quotient rule):

${\displaystyle {\begin{array}{lcl}x''&=&{\frac {d^{2}x}{dA^{2}}}\\&=&-r\cos A-{\frac {r^{2}\cos ^{2}A}{\sqrt {l^{2}-r^{2}\sin ^{2}A}}}-{\frac {-r^{2}\sin ^{2}A}{\sqrt {l^{2}-r^{2}\sin ^{2}A}}}-{\frac {r^{2}\sin A\cos A.(-{\frac {1}{2}})\cdot (-2).r^{2}\sin A\cos A}{\left({\sqrt {l^{2}-r^{2}\sin ^{2}A}}\right)^{3}}}\\&=&-r\cos A-{\frac {r^{2}(\cos ^{2}A-\sin ^{2}A)}{\sqrt {l^{2}-r^{2}\sin ^{2}A}}}-{\frac {r^{4}\sin ^{2}A\cos ^{2}A}{\left({\sqrt {l^{2}-r^{2}\sin ^{2}A}}\right)^{3}}}\end{array}}}$

If angular velocity is constant, then

${\displaystyle A=\omega t\,}$

and the following relations apply:

${\displaystyle {\frac {dA}{dt}}=\omega }$

${\displaystyle {\frac {d^{2}A}{dt^{2}}}=0}$

The equations that follow describe the reciprocating motion of the piston with respect to time. If time domain is required instead of angle domain, first replace A with ωt in the equations, and then scale for angular velocity as follows:

Position with respect to time is simply:

${\displaystyle x\,}$

Velocity with respect to time (using the chain rule):

${\displaystyle {\begin{array}{lcl}v&=&{\frac {dx}{dt}}\\&=&{\frac {dx}{dA}}\cdot {\frac {dA}{dt}}\\&=&{\frac {dx}{dA}}\cdot \ \omega \\&=&x'\cdot \omega \\\end{array}}}$

Acceleration with respect to time (using the chain rule and product rule, and the angular velocity derivatives):

${\displaystyle {\begin{array}{lcl}a&=&{\frac {d^{2}x}{dt^{2}}}\\&=&{\frac {d}{dt}}{\frac {dx}{dt}}\\&=&{\frac {d}{dt}}({\frac {dx}{dA}}\cdot {\frac {dA}{dt}})\\&=&{\frac {d}{dt}}({\frac {dx}{dA}})\cdot {\frac {dA}{dt}}+{\frac {dx}{dA}}\cdot {\frac {d}{dt}}({\frac {dA}{dt}})\\&=&{\frac {d}{dA}}({\frac {dx}{dA}})\cdot ({\frac {dA}{dt}})^{2}+{\frac {dx}{dA}}\cdot {\frac {d^{2}A}{dt^{2}}}\\&=&{\frac {d^{2}x}{dA^{2}}}\cdot ({\frac {dA}{dt}})^{2}+{\frac {dx}{dA}}\cdot {\frac {d^{2}A}{dt^{2}}}\\&=&{\frac {d^{2}x}{dA^{2}}}\cdot \omega ^{2}+{\frac {dx}{dA}}\cdot 0\\&=&x''\cdot \omega ^{2}\\\end{array}}}$

You can see that x is unscaled, x' is scaled by ω, and x" is scaled by ω². To convert x' from velocity vs angle [inch/rad] to velocity vs time [inch/s] multiply x' by ω [rad/s]. To convert x" from acceleration vs angle [inch/rad²] to acceleration vs time [inch/s²] multiply x" by ω² [rad²/s²]. Note that dimensional analysis shows that the units are consistent.

The velocity maxima and minima do not occur at crank angles (A) of plus or minus 90°. The velocity maxima and minima occur at crank angles that depend on rod length (l) and half stroke (r), and correspond to the crank angles where the acceleration is zero (crossing the horizontal axis).

The velocity maxima and minima do not necessarily occur when the crank makes a right angle with the rod. Counter-examples exist to disprove the idea that velocity maxima/minima occur when crank-rod angle is right angled.

For rod length 6" and crank radius 2", numerically solving the acceleration zero-crossings finds the velocity maxima/minima to be at crank angles of ±73.17615°. Then, using the triangle sine law, it is found that the crank-rod angle is 88.21738° and the rod-vertical angle is 18.60647°. Clearly, in this example, the angle between the crank and the rod is not a right angle. Summing the angles of the triangle 88.21738° + 18.60647° + 73.17615° gives 180.00000°. A single counter-example is sufficient to disprove the statement "velocity maxima/minima occur when crank makes a right angle with rod".