Solving a torque equation with two force angles

[JLD Co]

I'm comparing the torque developed by two different engine concepts. One is a new fandangled concept being built that lends itself to a simple torque equation F x r x sine theta (Force, radial vector of the crank, angle of Force) because it has only one angle of Force being applied to the rotating mass. The other is a conventional reciprocating engine with a "square" configuration (equal bore and stroke dimensions). In the conventional engine the above equation works for the crank pin angle (piston rod to crank) but nothing is making sense when I try to combine it with the piston pin angle (piston rod to piston). I know all the angles for any given piston/crank position. I just need the trick to solve for anyone given position... say 90 degrees from TDC. Any mathy engine types available to help me out with this probably elementary problem?

 

[jack action]

Torque is related to rotation and the piston doesn't rotate, so there is not suppose to be any torque there, only a pushing force.

The average power of the piston will be the average force times its average velocity.

The average power of the crankshaft will be the average torque times its average angular velocity (rpm).

Calculated with proper SI units, both values of power will be equal. On the average, the force will relate to the torque by a factor r (radius) and the velocity will relate to the angular velocity by the same factor r.

 

[JLD Co]

A piston, piston rod and crankshaft is a mechanism for converting linear motion into rotational motion...torque. Computing for torque when the angle of force on the lever arm (radial vector) is 90 degrees is simple: F X r (Force times the length of the lever). Computing for torque when the angle of force is anything other than 90 degrees requires the addition of the multiplier sine theta (the sine of the angle of force). But when the force is from two different angles the computation is trickier.
I'm simply trying to determine the torque produced by this mechanism with two angles involved in the "push"... the angle between the vector of the piston and the piston rod (at the piston pin) and then the angle between the radial vector of the crankshaft and the piston rod (at the crank pin).

 

[Nidum]

Have a look here : Engine design pdf

If that does not give you the information you need you then please post a clear diagram of your engine configuration so that we provide more specific help .

 

[jack action]

If I understand you correctly, the following should help you (source):

Where $x_p = l\cos\phi + r\cos\theta = l(\cos\phi + \frac{r}{l}\cos\theta)$ and so the torque is $T = F_g l(\cos\phi + \frac{r}{l}\cos\theta) \tan\phi$ and $\tan\phi = \frac{\frac{r}{l}\sin\theta}{\sqrt{1-\left(\frac{r}{l}\right)^2 \sin^2\theta}}$​

 

[JLD Co]

Now we're talking. There's my two angles. Thank you very much.

 

[JLD Co]

Have a look here : Engine design pdf

If that does give you the information you need you then please post a clear diagram of your engine configuration so that we provide more specific help .

Jack Action gave me what I specifically was looking for but there's is a whole lot of valuable information in your Engine Design pdf. Our team thanks you.

 

[Dr.D]

What Jack has provided will enable you to compute the torque due to gas pressure, but there is more.

The piston mass is being accelerated, and this requires an internal force. The connecting rod CM is being accelerated (both axially and transversely) and there is an angular acceleration of the connecting rod. All of these mass acceleration terms act like forces (they an not true forces, but M*a terms), so they two contribute to the torque acting on the crank at any particular instant.

To deal with the full problem, it is necessary to have a good handle on the kinematics of the entire system. Then formulating the full equation of motion for a single cylinder, including all accelerating masses, will show what effective torque terms are present.

 

[JLD Co]

What Jack has provided will enable you to compute the torque due to gas pressure, but there is more.

The piston mass is being accelerated, and this requires an internal force. The connecting rod CM is being accelerated (both axially and transversely) and there is an angular acceleration of the connecting rod. All of these mass acceleration terms act like forces (they an not true forces, but M*a terms), so they two contribute to the torque acting on the crank at any particular instant.

To deal with the full problem, it is necessary to have a good handle on the kinematics of the entire system. Then formulating the full equation of motion for a single cylinder, including all accelerating masses, will show what effective torque terms are present.

Thank you very, very much. You confirmed my intuitive fear. I'm over my head. I'll consult with a PhD friend who owes me and let him read these responses.
Much obliged.