Register to reply

Please Help, I don't have a clue how to calculate this.

by Ole SeaBee
Tags: clue
Share this thread:
Ole SeaBee
#1
Feb7-12, 06:24 PM
P: 14
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.
Phys.Org News Partner Science news on Phys.org
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage
Q_Goest
#2
Feb7-12, 06:41 PM
Sci Advisor
HW Helper
PF Gold
Q_Goest's Avatar
P: 2,903
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.
Ole SeaBee
#3
Feb7-12, 06:54 PM
P: 14
Quote Quote by Q_Goest View Post
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.
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!

Q_Goest
#4
Feb7-12, 07:49 PM
Sci Advisor
HW Helper
PF Gold
Q_Goest's Avatar
P: 2,903
Please Help, I don't have a clue how to calculate this.

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.
Ole SeaBee
#5
Feb7-12, 08:34 PM
P: 14
Quote Quote by Q_Goest View Post
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.
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!!!"
OldEngr63
#6
Feb9-12, 07:01 PM
P: 343
@ 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.
Q_Goest
#7
Feb9-12, 07:37 PM
Sci Advisor
HW Helper
PF Gold
Q_Goest's Avatar
P: 2,903
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.
Ole SeaBee
#8
Feb9-12, 10:13 PM
P: 14
Quote Quote by OldEngr63 View Post
@ 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.
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!!!"
OldEngr63
#9
Feb9-12, 10:31 PM
P: 343
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.
Ole SeaBee
#10
Feb9-12, 11:19 PM
P: 14
Quote Quote by OldEngr63 View Post
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.
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!!!"
Ole SeaBee
#11
Feb19-12, 10:17 PM
P: 14
So can anyone help me with the math to figure these losses??

Thanks,

ed
Q_Goest
#12
Feb20-12, 05:03 PM
Sci Advisor
HW Helper
PF Gold
Q_Goest's Avatar
P: 2,903
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.
FeX32
#13
Feb20-12, 05:43 PM
P: 67
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:
\begin{equation}
\dot{d}=-a\omega_2\sin(\theta_2)+b\sin(\theta_3)
\end{equation}

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

This is given by:
\begin{equation}
\omega_3=\frac{a\cos(\theta_2)}{b\cos(\theta_3)}\omega_2
\end{equation}
FeX32
#14
Feb20-12, 05:53 PM
P: 67
Anyone have an idea why my TeX is not showing up as compiled on Mac OSX but does in Win 7? Both using Chrome.
Ole SeaBee
#15
Feb20-12, 05:57 PM
P: 14
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!
FeX32
#16
Feb20-12, 05:59 PM
P: 67
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.
FeX32
#17
Feb20-12, 06:02 PM
P: 67
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..)
Ole SeaBee
#18
Feb20-12, 06:05 PM
P: 14
Quote Quote by FeX32 View Post
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.
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


Register to reply

Related Discussions
Brain teaser question General Discussion 10
Help. not a clue Introductory Physics Homework 1
No Clue! General Math 8
No clue... Introductory Physics Homework 1
I seriously have no clue Introductory Physics Homework 1