# Thread: 2 stroke piston inertia formulas

1. ## 2 stroke piston inertia formulas

hi all
I need help understanding this formula.
Here's my rewrite of it:

piston assembly inertia = Reciprocating Mass * CRANK RADIUS * W^2*(COS(Wt)+(crank radius/rod length)*COS(2*Wt))

W=angular frequency (crank rotations in 1 second times 6.28
Wt=unknown
CRANK RADIUS=piston stroke/2

Wt on the screenshot is Omega t
Can any one explain it to me?
inertiaformula.gif  Reply With Quote

2. The piston inertia force is independent of the type of engine, two stroke, four stroke or steam....

Omega is the angular velocity in radians per second and Omega (t) or your (Wt) is an instantaneous crank shaft position from T.D.C. To figure the force at T.D.C. substitute zero radians or degrees for (Wt) and for other positions insert your crank angle. The formula is often written with the Greek letter Theta for the angle.

The formula you show is derived by first describing the piston position in terms of the crank angle and the successive derivatives for velocity and acceleration are obtained using the binomial theorem. Your parenthesis show the first two cosine terms of the rapidly converging series.  Reply With Quote

3. OK then please tell me if the result is in Newtons, newtons/meter, or Newtons/meter^2

I'm confused as to the variations in different measures- N/meters, N/meters^2, pound foot, pound foot^2
What is their difference?

What is the most common measure for vibrational force on the crankshaft?  Reply With Quote

4. I'm not sure what your equation defines and without a reference one cannot investigate.

I do recommend you see this online book:

Aircraft Engine Design and Engineering Handbook

Everything you need for engine design and analysis is there.  Reply With Quote

5. The world is mostly doing analysis in SI units.. Google it. Originally Posted by 19jaguar75 OK then please tell me if the result is in Newtons, newtons/meter, or Newtons/meter^2

I'm confused as to the variations in different measures- N/meters, N/meters^2, pound foot, pound foot^2
What is their difference?

What is the most common measure for vibrational force on the crankshaft?  Reply With Quote

6. OK so if I use kilograms for mass, and meters for length then what measure does the equation give? (N/m^2, Newton meter, Joules, etc)  Reply With Quote

7. Google "convert kg to N"

What is your highest level of math education? Your questions are very basic.  Reply With Quote

8. The formula works in any measurement system. force is in Kg*M/sec^2 or Newtons if you work with your units.

I prefer to use Engineering Units and RPM for speed. If you enter mass in pounds, length in inches and engine speed in RPM, multiply by .0000284 to get pounds force. The constant takes care of the pound to slug conversion and the radians per second to RPM conversion.

In metric units you deal with the rotational speed conversion (RPM to radians per second) and changing from millimeters to meters.

Many books have the formula including Mark's Handbook.  Reply With Quote

9. Cragyon, you may be smarter at math than I am but at least I know how to directly answer questions (when I know the answer).
You didn't answer my question. Nor did you give me the clue to do so. Thanks for nothing.

Hudson I think you are telling me that using metric units in that formula results in Newtons.
Trying to understand these units I found this statement on another post on this forum:
“Newtons are units of linear force. Newton-meters measure torque, better understood as rotary force.”
OK but most Americans like to think of force in terms of pounds or psi.
I see from another site that .225 is the conversion factor from Newtons to pounds-force so I guess that is what I should use.  Reply With Quote

10. OK but here's something I perceive as a problem:
For my small 55cc engine with 116.6 grams piston assembly weight (piston, wrist pin, bearing, and the part of the con rod that wraps around the bearing), 9000 RPM, 19mm crank radius, 85mm con rod
the result of the formula at TDC is 2404 Newtons which multiplied by .225 is 541 pounds-force.
I have a hard time believing that such a small engine produces that much linear force that needs to be counter balanced.  Reply With Quote

11. When you start squaring the speed, The numbers begin to grow quickly.

Yes, the inertia force of the reciprocating components you provide produce 2400 N of tension in the rod 150 times per second.

Mass = .1166
R = .019
Engine speed = 150 *2 *pi (rad/sec) = 942.77 rad/sec
Engine speed squared = 888264 (rad/sec)^2
R/L = 19/85 = .2235
Cosine is maximum at angle zero = 1
The parenthesis term at maximum simplifies to (1 + .2235) or 1.2235

Peak reciprocating force = .1166 x .019 x 888264 x 1.2235 = 2408 KgM/sec^2 or Newtons

Now the good news:
1) Except when misfiring, the gas force on top of the piston opposes the inertia force reducing the stress on the rod in normal operation. Note: that this gas force is largely absent on the exhaust stroke of a four stroke engine.

2) The same squared rotational speed acts on the counterweight mass to generate an opposing force.

The usual approach is to balance all of the crankshaft rotational force and half of the reciprocating inertia force in a single cylinder engine.

And yes, it is true that some prefer psi to pascals  Reply With Quote

12. This engine is fairly well balanced and I just calculated the counter balance force which is also way up there. So without counter balance these engines would probably shred themselves right away.
Would the ideal counter balance be that which is somewhere between the correct balance at full throttle and half throttle?
I am reading thru the suggested book right now.
I have yet to see a way to calculate for the con rod that I like so I made up my own way. I divided the rod into 4 sections but now am confused on how to calculate their linear force in line with cylinder. With the piston they are mostly accelerating or decelerating so I need a formula for that. I'm thinking just plain ole F=mass x acceleration. I'm also calculating the 4 sections centrifugal force. Of course I need to use sine or cosine on both to get the right percentage of it that will be in line with the cylinder.
If I can get this all to work out then I hope to have the best computation for this to date. Thanks for your help guys. It's been a few decades since I took college math.  Reply With Quote

13. When I multiply the 1/4 rod section kg by its velocity change x .0209 I get a very miniscule amount.
What am I doing wrong?  Reply With Quote

14. If you don't know basic math and the concept of units and conversions you're way out of element. I nor anybody else owe's you an answer, a solution or education.

You've been helped - you're not getting it. Originally Posted by 19jaguar75 Cragyon, you may be smarter at math than I am but at least I know how to directly answer questions (when I know the answer).
You didn't answer my question. Nor did you give me the clue to do so. Thanks for nothing.

Hudson I think you are telling me that using metric units in that formula results in Newtons.
Trying to understand these units I found this statement on another post on this forum:
“Newtons are units of linear force. Newton-meters measure torque, better understood as rotary force.”
OK but most Americans like to think of force in terms of pounds or psi.
I see from another site that .225 is the conversion factor from Newtons to pounds-force so I guess that is what I should use.  Reply With Quote

15. My piston moves .8mm the last 15 degrees to TDC.
It weighs .1166 kg
Time of 15 degrees is .00028 sec
.0008 meters / .00028 sec = 2.88 m/sec velocity

Using the normal mass x acceleration formula I only get .007 which is too small.
lbf = kg x m/sec velocity change x .0209

When I use the normal formula for piston assy inertia I get over 500 pounds force.
What is wrong here?  Reply With Quote

16. There is over 500 lbf. inertia force according to the values that you provided. See the calculation that I provided.

You originally brought a formula to the forum and asked for help in understanding it. I tried to help explain it and worked the problem for you.

If you leave the mass out of the inertia equation you will obtain the instantaneous acceleration which is over 20000 M/sec^2 at TDC.

I trust you will pardon me if I pass on the four part rod thing.  Reply With Quote

17. Please do share my idea. I think the idea of just adding a percentage of rod weight to the piston assembly and a percentage of the rod weight to the crank is terrible.
OK please bear with me as I try to get this right.
mass x acceleration = force
so
.1166kg x 2.88 m/sec = .336 kg m/sec
and converting to Imperial:
.257 lb x 9.45 ft/sec = 2.4 lb ft/sec
How does 2.4 lb ft/sec relate to the 541 pounds force that the piston inertia formula gives me? (2.4 x 225 - 541)  Reply With Quote

18. As an engineering "outsider" I question all previous assumptions. In this case the assumption that the compression and combustion pressure counter the upward inertia force of the piston. To illustrate why I include this drawing. Please give me your critic (as long as it isn't in favor of conventional thought for the sake of convention).
compr-comb-force.gif  Reply With Quote

19. Gas force up = gas force down. That part is correct. Because the two forces cancel out,through the engine frame, gas pressure has no (Zero) effect on engine balance.

Which is why engine balance is all about cancelling rotary and reciprocating forces.  Reply With Quote

20. At smartconversion there is no option for using meter/second in its force calculator (which I thought was correct), only meter/second^2. Using that I calculate 10368 as the velocity change, not 2.88 which then gives me a result of 272 pound-force which is OK because the velocity I am using is the average during 15 degrees but knowing the piston motion you know that it slows down at an increasing rapid rate as it approaches TDC for an increasing inertia force. I think the piston inertia formula gives the correct result right at the end of its upward movement because its twice that of 272.  Reply With Quote

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