1. Feb 7, 2012

### Ole SeaBee

Hello,

This is for sure not a 'Homework' problem. I have no clue! I understand that a crankshaft looses power from a piston because the force is pressing the Crank outward verses around it's shaft. I have no clue as to how to figure this out. I have looked all over the web and could find nothing that I could understand well enough to come up with the figure I need. When you start talking sin cosine I'm totally lost.

I can give the positions of the Crankshaft and the Rod Arm, I know the abgles and lenghts I'm working with. The Crankshaft has a R of 5.25" It is at 22.5 degrees above horizontal. The Rod lenght is 8.6875" and is at 13.3 degrees above the horizonal. If my 'piston is pushing on the Crank with 1000 lbs of force, How much of this force is going into turning the crank or not into tearing the crankshaft apart????

Thanks,

ed aka Ole SeaBee.

2. Feb 7, 2012

### Q_Goest

Hi ole seabee, welcome to the board. Note that pushing against a wall requires no expenditure of energy. Similarly, if you had a cylinder pushing against an immovable object, it won't expend any energy.

Force is a vector so we can divide it into two forces perpendicular to one another. If we break up the force the piston in an engine is exerting against the crankshaft, it can be broken up into a force in the direction of motion and a force perpendicular to the direction of motion. The force perpendicular to the direction of motion is just like the cylinder pushing against an immovable object and this force isn't expending any energy. Only the force pushing in the direction of motion is expending any energy, so there is no loss of power due to the sideways force of the piston on the crankshaft. The losses are only due to friction, heating and losses through your tail pipe.

3. Feb 7, 2012

### Ole SeaBee

Really, I thought I had read (and understood) somewhere in my searches, that there is a loss of force/power that is not going into turning the crankshaft, at 90 degrees from TDC is the only place where the 1000 lbs is going into turning the crank and the rest of the 180 degree turn has a loss.

Thank You!

4. Feb 7, 2012

### Q_Goest

I design reciprocating pumps and compressors, so the same thing applies to them. I can assure you the side loads do not result in any power losses other than some minor frictional loads.

5. Feb 7, 2012

### Ole SeaBee

Q_Goest, You have just made my day!!!! I am working with 2 different forms of leverages in my project and this was the one thing that I didn't totally understand. If I'll have little or no loss in the crank shaft turn there, I should be able to turn the main lever with less than 16% of the power available even with higher losses I thought it could be done, but this really makes it!

Thank You VERY MUCH!

ed

SeaBees, "We do the improbable right away, the impossible, takes abit longer!!!"

6. Feb 9, 2012

### OldEngr63

@ O_Goest: I wonder if Ole SeaBee is not referring to the variation of torque (I realize that is not what he said) that occurs throughout the IC engine cycle? This is due, of course, to (1) falling cylinder pressure as the cylinder volume increases, and (2) changing mechanism geometry that changes the mechanical advantage between the piston and the crankshaft as the cycle progresses. Without friction, there is, as you said, no loss of energy, but there is definitely a loss of instantaneous torque at the crankshaft.

7. Feb 9, 2012

### Q_Goest

Hmmm..... I guess we'll have to see if ole seabee wants to respond. But yes, I agree, there's a very large variation in the torque produced at the crankshaft as a piston in an engine goes from TDC to BDC.

8. Feb 9, 2012

### Ole SeaBee

Thanks, That might be what I am thinking of, or read about. I am not even sure the differences there!

The crankshaft will be turning a flywheel like in a car transmission, a cam is also mounted on the crankshaft. The cam turns a lever a few degrees. The Crankshaft arm is R - 5.25" out from centerline and the edge of the cam is R - 2.613" from the same centerline. I am trying to see if there will be anything left of the 1000" of pressure/torque from the piston to make the lever turn. The cam makes contact with the lever when the Crankshaft arm is at 22.5 degrees above the horizonal. (The pistons are running horizontally too.)

Thanks,

ed

SeaBees, "We do the improbable right away, the impossible, takes abit longer!!!"

9. Feb 9, 2012

### OldEngr63

Well, it is severe enough, that typically the local torque on the crankshaft actually reverses direction during a portion of the crank cycle. This is one of the reasons why a flywheel is absolutely essential. If you take a very small engine and remove the flywheel, it often will not run because there is not enough energy stored in the system to carry it back up to TDC from BDC. During the compression stroke, crankshaft torque goes negative and torque must be drawn from the flywheel to carry the system through TDC.

10. Feb 9, 2012

### Ole SeaBee

Thank You OleEngr63,

This is being driven abit differently than an IC engine. The cam strikes at both 22.5degrees before TDC then again @ 22.5 before Bottom Dead Center. (In this application it's actually 22.5before Horizonal on both left and right strokes.) My calcs say the lever is going to have 400# resistance to it's turn. (The lever only turns less than 12 degrees.) The cam is on the crankshaft and with it's smaller radius verses the crank arm it looks like we gain 2 X the avalable torque there. So we should need to be getting somewhere near 200# from the piston into the crank. We know the flywheel will carry on the process. I just want to know how much the loss might be at those times.

Thanks,

ed

SeaBees, "We do the improbable right away, the impossible, takes abit longer!!!"

Last edited: Feb 9, 2012
11. Feb 19, 2012

### Ole SeaBee

So can anyone help me with the math to figure these losses??

Thanks,

ed

12. Feb 20, 2012

### Q_Goest

There are no losses in power. Think of the changing geometry of the crankshaft and connecting rod as a continual change in the mechanical advantage of the force being produced by the gas above the piston. So although the torque on the crankshaft will vary as a function of the geometry as the crankshaft rotates, it's only because the mechanical advantage is changing - kinda like what would happen if you put a small baby on a teeter totter and you stood on the fulcrum and ran out to the opposite end. At first you would have small mechanical advantage but your small downward movement would result in a relatively large movement of the baby. The farther out you moved, the more mechanical advantage you would have, so as you got to the end of the board, you'd have lots of leverage but the baby wouldn't be moving as fast.

13. Feb 20, 2012

### FeX32

I agree with the repliers. No loss in power ([tex] force\times distance [/itex] or [tex] torque\times angular velocity [/itex], except for the frictional losses.
Torque on the other hand is a nonlinear function and varies inversely to the velocity relationship between the piston and the crank.

The velocity relationship can be derived to be:

\dot{d}=-a\omega_2\sin(\theta_2)+b\sin(\theta_3)

where the subscript 2 refers to the crank link and the subscript 3 refers to the coupler link. $\dot{d}$ is the velocity of the piston. $a$ is the length of the crank link and $b$ is the length of the coupler link.
The ratio of $\omega_2$ to $\dot{d}$ can be obtained by deriving the velocity relationship between $\omega_2$ and $\omega_3$

This is given by:

\omega_3=\frac{a\cos(\theta_2)}{b\cos(\theta_3)}\omega_2

Last edited: Feb 20, 2012
14. Feb 20, 2012

### FeX32

Anyone have an idea why my TeX is not showing up as compiled on Mac OSX but does in Win 7? Both using Chrome.

15. Feb 20, 2012

### Ole SeaBee

Well, Thanks for your help.. U have put me in my place! I didn't ask the right questions, so I get greek for an answer.

Torque at the above listed position is what I was looking for!

16. Feb 20, 2012

### FeX32

I read that as "geek" first time around lol. But yea greek is right :).
I don't mean offence, only to help out. Those equations are the exact ones you need to determine torque fluctuations at the crank.

17. Feb 20, 2012

### FeX32

But yea if anyone has an idea why the TeX on PF has an issue on different operating systems I would appreciate it. :) (I know you guys likely see it as it should be, but I see it as TeX code using OSX..)

18. Feb 20, 2012

### Ole SeaBee

FeX32,

ok.. As I said earlier. Sin and cos are not something I understand in any way. I am not certain of the amount of torque I will have at the point/position I listed in my first post here. I know there are ways to calculate it, just do not have that kind of education.

Thanks,

ed

19. Feb 20, 2012

### OldEngr63

The Tex is not working with Win XP either. This morning I was unable to open this page at all. Now I can see the page, but the Tex does not work. I think we havee a problem.

20. Feb 20, 2012

### OldEngr63

I think we need to slow down about on the "no loss in power" bit. With no friction, I agree that there is no loss in power, but where in the real world do we find situations without friction? Reciprocating compressors/engines have significant friction to begin, and when we add this strange lever drive mechanism, I see no reason at all to think friction will be reduced.

What is required is to do an internal force analysis, first without friction. Use the forces calculated at the contact points to estimate the friction forces, and then repeat the force analysis, the second time including friction forces. By iterating this analysis, a reasonable estimate of the friction should be obtained, and with it the power losses in operation. We are also talking about a substantial amount of engineering analysis, computation, etc. It is not a trivial effort; you have to really want the result.

21. Feb 20, 2012

### Ole SeaBee

Alrighty then.. Lets work with the slow ole man here abit ok!! First, I have a disc that is R4.5". It has 1000# of resistance to turning on it's axis (at R4.5") and I need it to turn 9 - 15 degrees. I attach a lever to the center of that disc and it is at R11.25" I think I will need 400 lbs of torque to move turn it at that point and that the disc will turn with the 400 lbs. (Forgetting friction for now.)

Now I have a crankshaft that is being rotated by a piston. The piston has an equal 1000 lbs of pressure on it (at this point on, that does not change). The piston linkage is 8.6875" long the end connected to the crankshaft is at 13.3 degrees above horizontal. That linkage is connected to the crankshaft at R5.25" from the cranks CenterLine. At the time the piston linkage is at 13.3 degrees above it's horizontal center line, the crank linkage is at 22.5 degrees above it's horizonal centerline. There is a CAM on the crankshaft that is also at 22.5 degrees, it is R2.613". The cam will strike the lever on the above disc at the R11.25 mark. (The piston figures to be .375" from BDC at that point.)

Do we get anywhere close to 400 lbs of torque from the cam?

Thanks,

ed

PS. I don't know what U are referring to as Tex?

Last edited: Feb 20, 2012
22. Feb 20, 2012

### FeX32

To answer the original post then since you don't like trig.

Solving the equations i wrote in the other post. If you are interested I rearranged them for $\frac{\dot{d}}{\omega_2}$ and then used the equivalence of this equal to Torque/piston force. Pluggin in your angles gives 862.5 ft-lb of torque at the crank.

Not that this is with 1000lbs at the piston. Also note that this is independent of the length of the coupler link (rod), only dependent on the angles. I also assumed that your angles were all positive and starting in the first quadrant.

Cheers,

Last edited: Feb 20, 2012
23. Feb 20, 2012

### FeX32

Thx OldEngr63 about the TeX, seems to work now.
Cheers,

24. Feb 20, 2012

### FeX32

Actually, it is still not working lol. I tried to fix the other post with no luck.

25. Feb 20, 2012

### OldEngr63

@ FeX32: I recently wrote some TeX on another thread and it worked find. What you have written on the previous page of this thread still does not work for me. I think things are still not quite right.