To make it more interesting the highest cylinder wear rate is at TDC where the piston is moving the slowest, not in the middle of the wall during the fastest velocity.
aehaas
aehaas
HTHS
- Thread starter MolaKule
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Is there any simply usable HT/HS conversion table for someone dumb (really dumb I mean) in maths ?
I would like to know the HT/HS of an oil with these specs:
79 cSt viscosity @ 40°C
14 cSt @ 100°C
Flashpoint 250°C
VI claimed: 197
Thx.
I would like to know the HT/HS of an oil with these specs:
79 cSt viscosity @ 40°C
14 cSt @ 100°C
Flashpoint 250°C
VI claimed: 197
Thx.
TDC has the least oil exposure, high pressure, high temperature. There is no motion so a fluid wave cannot build up (same reason that direct acting cam lobes wear less in the mid RPM range) et cetera.
aehaas
[ June 17, 2005, 01:42 PM: Message edited by: AEHaas ]
aehaas
[ June 17, 2005, 01:42 PM: Message edited by: AEHaas ]
To add to this:
The highest cylinder wear rate occurs (quite) near TDC on the down going power stroke when cylinder pressure is still rising.
The wear occurs because the pressure on the piston and the angulation of the con rod presses the piston towards the side of the cylinder opposite that of connecting rod. Since no power is being transfered to the crankshaft, all this pressure pushes the piston towards one side (on compression) then the other (on the power stroke). The wear on the power stoke side is about 4 times the wear on the cylinder on the compression side.
There is no actual wear at exactly TDC since all the forces line up perpendicular to the clinder bore (assuming the cylinders are actually perpendiculat to the crankshaft centerline, and the piston pin is centered in the piston).
Note also, that maximum energy transfer (per unit rotation) from piston to crankshaft occurs around 17-19 degrees ATDC while the combustion pressures are still high (but dropping fast). As the con-rod becomes more and more angulated and energy transfer more efficient, but cylinder pressure is dropping and less energy remains to be transfered.
Here's some data to work with:
Actually it varies between 70 and 90 degrees ATDC in common engines designs depending on rod length.
I believe it would take an infinite length con rod to have it occur at 90 degrees, that wouldn't fit under most hoods
I'm too lazy to derive the equations or look them up today, but I also believe what Doug Hillary was saying was that the piston velocity is max when the con rod is perpendicular to the crank throw, except he said it in a more universally tecnically correct way, bless his heart.
Yep.quote:
Originally posted by XS650:
I believe it would take an infinite length con rod to have it occur at 90 degrees, that wouldn't fit under most hoods
quote:
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To make it more interesting the highest cylinder wear rate is at TDC where the piston is moving the slowest, not in the middle of the wall during the fastest velocity.
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The slower the sliding movement, the less hydrodynamic pressure between parts. Combine this with the pressure from the combustion stroke,and you have metal to metal contact.
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To make it more interesting the highest cylinder wear rate is at TDC where the piston is moving the slowest, not in the middle of the wall during the fastest velocity.
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The slower the sliding movement, the less hydrodynamic pressure between parts. Combine this with the pressure from the combustion stroke,and you have metal to metal contact.
You calculated tangential velocity, or instantaneous linear velocity of the crank throw. That is not piston velocity.
Given a 2 inch radius crank throw, thus a 4 inch stroke, the piston travels 8 inches per crank revolution, or 0.666 foot/rev.
At 78 rad/s = 745 rpm
0.666ft/rev x 745 rev/min = 496 ft/min piston speed, or 8.3 ft/s average piston speed, not 13 ft/s.
Piston speed is not sinusoidal. Maximum is not reached at the 1/2 way point. Maximum piston speed and the position at which it occurs is funtion of stroke and rod length, typically occuring reasonably close to 75 degress from tdc.
The reason for this puzzler is that as the crank rotates, so does the rod. The rod pivots around the crank pin, so the angular rotation of the rod also imparts a velocity to the piston that may be additive to or subtractive from the velocity imparted by the crank pin, depending on where in the rotaional cycle the rod-crank combination is. There are multple ways to solve for this relationship, graphically or numerically.
Some examples:
MolaKule
Staff member
This was a ballpark figure and as I said above, the exact piston speed depends on the crank/piston/wristpin geometrical relations.
I calculated Average piston velocity. Instantaneous piston velocity is a function of crank angle (time).
The theta displacement equation I gave as an example took into account the geometry of the crank/piston/wristpin.
The overall piston motion is sinusoidal, since the piston is almost motionless at say TDC (since velocity is at or near zero and no acceleration) and then acccelerates to maximum velocity at about its halfway point, then decelerates to zero velocity at or near BDC; assuming a perfect geometry case. This reciprocating motion describes a sinusoidal curve.
In Lester Lichty's text, "Internal combustion Engines," see his Chapter 17 on 'Mechanics of Principle Moving Parts' where he shows a graph (Fig. 395) of the piston displacement as a function of time(swept volume per crank angle degree of rotation), Inertia force, and piston velocity as a function of crank angle.
With a crank radius r of 2 in, a rod length L of 8" (for a L/r = 4), and a complete 360 degree rotation, the piston displacement is 4 inches, and the piston velocity curve is sinusoidal. At 4,000 rpm, maximum piston velocity on the sinusoidal velocity curve is 72 fps at 80 degrees and at 280 degrees. The displaement curve looks similar to a Bell curve
"The piston velocity is zero at the beginning of a stroke, reaches maximum near the middle of the stroke, and decreases to zero again at the end of the stroke. Because of the connecting-rod-crank ratio, the piston attains a maximum velocity at about 80 degrees from top dead center," which agrees with your statement.
MolaKule
Staff member
And do you know why?
No, you calculated average crank pin velocity and called it piston velocity.
You said: "v (average) = 78 rad/s X 0.051 m = 3.96 m/s. = 3.96 m/s X 3.281 f/m = 13.0 fps. This is the average linear velocity of the piston. "
You might check your work because you ended up with crank pin velocity, not average pistom velocity.
It's not sinusoidal.
Durley, Kinematics of MAchines, 1903 got it right.
On page 100 in the original tesxt. He draws a correct schematic of a slider crank mechanism and does the math correctly.
http://historical.library.cornell.edu/kmoddl/toc_durley1.html
In engines of typical automotive proportions, the recipricationg motion is quite a ways from being sinusoidal as function of crank rotation (or time)
The Cornell Historical Library has some interesting old technical books, when you run out of other things to do take a look.
MolaKule
Staff member
Nope. The velocity curve of the piston is very sinusoidal. Check out the Lichty reference abopve or any of Ian Taylor's works and you'll see that is the correct description of the piston's velocity curve.
BTW, could not get any of the chapters to download the PDF's.
BTW, could not get any of the chapters to download the PDF's.
A curve that peaks at 76 degrees instead of 90 degrees is not sinusoidal unless you are just making gross generalities
If you are a sying it's sinusoidal because it looks sorta sinusoidal on a low enough resolution display then I understand what you are saying.
This all pales compared to the fact that I can't find my copy of Taylor tonight
I just tried downloading the pdfs again in Firefox and in IE. The are both working now. Maybe their system was on a break.
http://historical.library.cornell.edu/kmoddl/
Is a bit cleaner way to enter their system.
MolaKule
Staff member
Seriously, not making gross generalites, and simply because a curve peaks at 76 degrees instead of 90 does not make it non-sinusoidal.
Think of the geometrical modifications as a "phase" factor to the equation.
Lichty's text is in most university libraries but I don't know if it's on the web.
Lichty also derives the differential equations for velocity and acceleration for the geometry given, and has a diagram of the geomtery as well.
If you would like to see the diffy Q derivation, send a note to [email protected] and I'll send an MSWord file to you with the equations. You can program them into MatLab and see the results.
Think of the geometrical modifications as a "phase" factor to the equation.
Lichty's text is in most university libraries but I don't know if it's on the web.
Lichty also derives the differential equations for velocity and acceleration for the geometry given, and has a diagram of the geomtery as well.
If you would like to see the diffy Q derivation, send a note to [email protected] and I'll send an MSWord file to you with the equations. You can program them into MatLab and see the results.
I thought you had to have a linear combination of sine waves of the same period but different phase shifts in order to have a sine wave with the same period, but a different phase shift?
That sounds right.
However, take the specified curve and subtract it from a pure sinusoid and you see 3rd order and 5th order sine waves (and higher) from the articulation of the finite length connecting rod.
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