The obscure technical question thread

[Deech]

Question: When a piston crosses top dead centre and starts to reverse its directon, for an instant its velocity has to equal zero. (Can't change direction without stopping)

So does that mean the entire rotating assembly of an engine stops completely twice per revolution (on a flat-plane crankshaft), or do the connecting rods stretch; letting the crankshaft keep rotating, but for the pistons come to a halt?

It's just something that's got me stumped
Thanks in advance

[kircher]

The crankshaft continues to rotate, the instant that the pistons stop and reverse is just that; an instant. It doesn't pause for any length of time. And in a flat plane crank, like in a 4 or ferrari style v8 the pistons will all pause at the same time, in the 8, one bank and then the other bank 90 degrees later, given the V angle is 90 degrees. This is all theoretical though and doesn't take bearing clearances or any flexing into account. I think in 4 cylinder engines this pausing of all pistons and reversing is a major reason for the inherent unbalance in an inline 4.

[Deech]

Interesting, thanks for that.

[Airmon]

No because you're comparing linear velocity to angular velocity. At the instant the piston reaches a velocity of 0, the crank for instance, will also have a 0 velocity in the same direction as the piston, but will still have a velocity perpendicular to it.

[kircher]

If you're mathematically minded, the "vertical" component of the cranks rod journal can be described by the cosine function. When you are at TDC, you are at the 'instant' when it is stationary, then it gradually accelerates faster and faster to maximum speed at 90 degrees, then slows again until 180 degrees (BDC) where it instantly pauses and continues upward again and the cycle continues. It's a case of Simple Harmonic Motion, just like a pendulum. The con rod however doesn't directly convert this vertical motion to the piston, so it will vary depending on the rod:stroke ratio. In the hypothetical situation where you have an infinitely long connecting rod, the vertical motion of the crank journal will be directly transmitted to the piston. A shorter rod will cause higher piston acceleration, but the average speed will remain the same. It's been a while since I've done this sort of math, but if you want to know piston position relative to TDC and BDC the 'easy' way you can draw a triangle with the rod length as one side, half the stroke as one, and the pistons position relative to the crank centreline as the unknown and use the sine rule to work it out. The only known angle in the calculation is degrees from TDC.

[madmelon]

kircher is spot on. Top and bottom dead centre are both points of stoppage and also the points of maximum acceleration towards the opposite end.

Regarding the imbalance of 4 cylinder engines, that's caused by the difference in acceleration away from each dead centre due to the rod length. ie. 10 degrees of rotation of the crank away from TDC will not produce the same piston movement as 10 degrees of rotation away from BDC. The only engines that get close to being balanced out are 6 or 12 cylinder engines due to the three "wave forms" generated but even they can't be truly perfectly balanced as the induction, compression, power and exhaust strokes all add to or subtract from the inertial forces present.

[gtfpv]

top thread . this has been dicussed for decades and is very very common in discussion .
the previuos posts have explained thisalready . but there is a point where the crank is rotating and the piston is stationary . easier to understand if you are turning the crank by hand .
another amuseing thought is when the engine is running . the forces on the piston vary during gas firing . there is a point where the crankshaft acually reverses everytime the gas charge is fired on every cylinder . ( mind boggling ) hence the need for HARMONIC balancers . sorry to go off topic . but that is the 2nd most talked about topic in relation to the 1st .

[kircher]

Are there cam grinders that take your rod:stroke ratio into account when doing custom grinds? It's my belief that with a higher rod ratio, you wouldn't need as much overlap to get the same effect, due to the longer dwell near TDC.

[madmelon]

Originally Posted by kircher

Are there cam grinders that take your rod:stroke ratio into account when doing custom grinds? It's my belief that with a higher rod ratio, you wouldn't need as much overlap to get the same effect, due to the longer dwell near TDC.

No idea but I remember reading an article on the AFR Heads website about the 350 Chev "should have build"- a 0.125" larger bore and a slightly smaller stroke with the same or longer rod. It made very impressive power and it was able to use rather a high compression ratio.

Harmonics are a big problem in crankshafts but it's not the primary harmonic that does the most damage- it's usually the 3rd and 4th harmonics that cause fatigue failures due to their larger number of cycles (factors of 3 and 4 higher respectively). There area huge number of ways of attempting to balance and dampen this out, depending on the type of engine. The old radial aircraft engines are particularly interesting. It's a real challenge to engineers to try and eliminate radial and torsional harmonics without making the components stupidly complex or overy heavy (ie poor responsiveness), even with todays computer simulations available.

[Wally]

Originally Posted by kircher

Are there cam grinders that take your rod:stroke ratio into account when doing custom grinds? It's my belief that with a higher rod ratio, you wouldn't need as much overlap to get the same effect, due to the longer dwell near TDC.

They should take the geometry into account. As the rod/stroke ratio decreases, the maximum instantaneous velocity closes in on TDC either side. The piston's velocity coming away from TDC pulls in mixture faster, so the intake cam event especially, should match it.

Question: When a piston crosses top dead centre and starts to reverse its directon, for an instant its velocity has to equal zero. (Can't change direction without stopping)

So does that mean the entire rotating assembly of an engine stops completely twice per revolution (on a flat-plane crankshaft), or do the connecting rods stretch; letting the crankshaft keep rotating, but for the pistons come to a halt?

It's just something that's got me stumped
Thanks in advance

Reminds me of the old fixed wheel bike. Would have been interesting if it stopped every time the pedal was at TDC.

Apart from some minor dwell from bearing lash, crank deflection, rod flex, piston side slap, bigend and gudgeon moments, etc the time interval where the piston transitions from relative +ve to -ve velocity is extremely short lived. I can see where you're coming from though; if the crank pin/throw is still moving, but the piston is stationary, something must be taking up the slack of the tensile and compressive forces and rod stretch is a candidate. I'd go with the combination slack of all the clearance fit bits 'n bobs.

[kircher]

Another thing I read when I was doing my own research on rod:stroke ratios was that with a longer rod (higher rod ratio) the piston dwells longer at TDC, but shorter at BDC, and a shorter rod (lower rod ratio) will dwell longer at BDC, yet shorter at TDC, relative to the longer rod combo. This doesn't make sense mathematically to me. The piston travels in one dimension, so is always travelling on the one plane, and the crank journal rotates in circular motion, with the rod joining and converting the circular motion into the linear motion of the piston. With the piston always on the one plane, and the rod length and stroke being constant, shouldn't the dwell at BDC and TDC be identical?

[Wally]

Originally Posted by madmelon

No idea but I remember reading an article on the AFR Heads website about the 350 Chev "should have build"- a 0.125" larger bore and a slightly smaller stroke with the same or longer rod. It made very impressive power and it was able to use rather a high compression ratio.

Harmonics are a big problem in crankshafts but it's not the primary harmonic that does the most damage- it's usually the 3rd and 4th harmonics that cause fatigue failures due to their larger number of cycles (factors of 3 and 4 higher respectively). There area huge number of ways of attempting to balance and dampen this out, depending on the type of engine. The old radial aircraft engines are particularly interesting. It's a real challenge to engineers to try and eliminate radial and torsional harmonics without making the components stupidly complex or overy heavy (ie poor responsiveness), even with todays computer simulations available.

I kind of agree with you, but don't underestimate first and second order harmonics/forces, especially when applied to an engine other than a six cylinder.

Simple harmonic motion doesn't generally occur in a combustion engine cylinder, because the piston travel isn't equal per crank radian travelled. The second harmonic is twice the cosine frequency of the first harmonic at an acceleration value equal to the rod length/half the stroke length (R/L). Because the second harmonic is exacly half the frequency of the primary, TDC forces are cummulative while BDC forces are subtractive. On a six there are sufficient counter forces to equal things out. On a four there is sufficient imbalance to cause some pretty nasty damage, third, fourth and nth order harmonics aside.

[kircher]

Originally Posted by Wally

I kind of agree with you, but don't underestimate first and second order harmonics/forces, especially when applied to an engine other than a six cylinder.

Simple harmonic motion doesn't generally occur in a combustion engine cylinder, because the piston travel isn't equal per crank radian travelled. The second harmonic is twice the cosine frequency of the first harmonic at an acceleration value equal to the rod length/half the stroke length (R/L). Because the second harmonic is exacly half the frequency of the primary, TDC forces are cummulative while BDC forces are subtractive. On a six there are sufficient counter forces to equal things out. On a four there is sufficient imbalance to cause some pretty nasty damage, third, fourth and nth order harmonics aside.

Ok, I gotta get off my *** and do a Mechanical Engineering degree. This all sounds interesting to me.

[Wally]

Originally Posted by kircher

Another thing I read when I was doing my own research on rod:stroke ratios was that with a longer rod (higher rod ratio) the piston dwells longer at TDC, but shorter at BDC, and a shorter rod (lower rod ratio) will dwell longer at BDC, yet shorter at TDC, relative to the longer rod combo. This doesn't make sense mathematically to me. The piston travels in one dimension, so is always travelling on the one plane, and the crank journal rotates in circular motion, with the rod joining and converting the circular motion into the linear motion of the piston. With the piston always on the one plane, and the rod length and stroke being constant, shouldn't the dwell at BDC and TDC be identical?

You probably need to plot it on some graph paper to appreciate what happens. On short rod ratio's the acceleration curve in particular is gnarly around TDC, while long rod ratio's it is smooth. Remember that although the crank may be travelling at a fix speed, the piston isn't because the rod is not operating at a fixed angle.

[kircher]

Originally Posted by Wally

You probably need to plot it on some graph paper to appreciate what happens. On short rod ratio's the acceleration curve in particular is gnarly around TDC, while long rod ratio's it is smooth. Remember that although the crank may be travelling at a fix speed, the piston isn't because the rod is not operating at a fixed angle.

Yes, I have graphed it on paper, I drew a triangle with one side being half the stroke, the other being the length of the rod, and the unknown side being the distance of the piston pin centre to the crank centreline. I put the angle from TDC between the unknown side and the half stroke side if that makes any sense. I also made a spreadsheet using the sine rule to calculate and graph piston motion, and my results tend to contradict what I've read elsewhere. Particularly what I read here http://www.stahlheaders.com/Lit_Rod%20Length.htm. It says

Long Rod is faster at BDC range and slower at TDC range.

and

Short Rod is slower at BDC range and faster at TDC range.

But my own calculations say that the motion will be identical at BDC and TDC, just in opposite directions, or negative, if that makes sense. I think that web site might be wrong, or saying something that is different to what I'm saying.

[Wally]

I'll have a read when I get time. but just a word of warning "short rod" is not necessarily the same as small rod/stroke ratio.

[kircher]

Originally Posted by Wally

I'll have a read when I get time. but just a word of warning "short rod" is not necessarily the same as small rod/stroke ratio.

yes, I know - it's a ratio.

[Wally]

OK, time for roughie maths. I'm working with the radius (half stroke), rod length, and radians.

Let's take two different setups, both fantasy, but distinct enough show the difference. Say one at 3.0 (100mm stroke (r=50), 300mm rod (L)) and another at 1.2 (100 stroke, 120mm rod).

3.0 (r/L = 0.17)
45° ABDC distance to travel to TDC = 87.4 mm
90° = 54.2mm
135° = 16.7mm
175° = 0.22mm

1.2 (r/L = 0.42)
45° = 90.7mm
90° = 61 mm
135° = 20 mm
175° = 0.27mm



As you can see the long rod covers more distance around BDC and has less distance to travel to TDC. So they travel faster around BDC and slower around TDC.

[kircher]

Thanks for that

[Deech]

Question answered and then some!

Can anyone recommend any decent books for starting out on the physics of engines (camshaft specs, piston kinetics etc?)
More study is needed.

[kircher]

I have read "automotive mechanics" by ed may and les simpson, volume 1 and 2 and I believe they were very good books. They're older, but engineering principles don't really change over time. It educated me a lot on the basic principles of an internal combustion engine. Others may have better recommendations, but that's a good start I think. There should be some pretty good info at your local library, even if it is in the reference section.

[madmelon]

"Design of Machinery" by Robert L. Norton.

not to be confused with "Machine Design", also by Norton. Both very good books but the former specifically deals with cams, cranks and all sorts of motion devices. These are the prescription texts at Wollongong Uni for two Mech Eng 2nd year subjects.

Cams are vastly more complicated than they first seem. If starting from displacement, you must differentiate 3 times, obtaining velocity, acceleration and jerk. Jerk is the one that most people don't know about and it's actually quite important for camshaft design to keep the followers, valves, etc happy for a long time.

[Wally]

My recommendation is to initially consult with some old hands who own machine shops; I'm sure there are some members here who have some good solid knowledge too. Stroke them enough and they'll reveal the rules of thumb. Once you have these locked in your head you know your calcs will be in the right ballpark.


Study up on kinematics.

[Wally]

Originally Posted by madmelon

....... If starting from displacement, you must differentiate 3 times, obtaining velocity, acceleration and jerk. Jerk is the one that most people don't know about and it's actually quite important for camshaft design to keep the followers, valves, etc happy for a long time.

Ah yes Jolt/Jerk. how did the third order derivatives go:

. Momentum equals mass times velocity
. Force equals mass times acceleration
. Yank equals mass times jerk
. Tug equals mass times snap
. Snatch equals mass times crackle
. Shake equals mass times pop

[whynot]

Have a look at this animation; it may help put your question into perspective. It visualises what airmon said. Notice how the speed of the crank (measured in radians) stays the same. However the liner velocity of the shaft is always changing. The rate of change is at its highest as the shaft crosses the zero axis, only to slow down as it approaches the peaks, then stops for an instant, before reversing direction.

Run a few numbers through the calculator embedded in the page. It is an interesting exercise.




http://www.rkm.com.au/ANIMATIONS/ani...sine-wave.html

[kircher]

Originally Posted by whynot

Have a look at this animation; it may help put your question into perspective. It visualises what airmon said. Notice how the speed of the crank (measured in radians) stays the same. However the liner velocity of the shaft is always changing. The rate of change is at its highest as the shaft crosses the zero axis, only to slow down as it approaches the peaks, then stops for an instant, before reversing direction.

Run a few numbers through the calculator embedded in the page. It is an interesting exercise.




http://www.rkm.com.au/ANIMATIONS/ani...sine-wave.html

Great animation. It really makes it easier to visualise the SHM aspect of circular motion