# Solving a torque equation with two force angles

1. Mar 25, 2017

### 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 any one given position... say 90 degrees from TDC. Any mathy engine types available to help me out with this probably elementary problem?

2. Mar 25, 2017

### 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.

3. Mar 26, 2017

### 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).

4. Mar 26, 2017

### 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 .

Last edited: Mar 27, 2017
5. Mar 26, 2017

### jack action

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 $\tan\phi = \frac{\frac{r}{l}\sin\theta}{\sqrt{1-\left(\frac{r}{l}\right)^2 \sin^2\theta}}$​

6. Mar 27, 2017

### JLD Co

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

7. Mar 27, 2017

### JLD Co

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.

8. Mar 28, 2017

### 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.

9. Mar 28, 2017

### JLD Co

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.