Great video! One thing to add for those who need a more explanation. The reason the piston travels farther during the 1st and 4th 90 degrees of rotation than the 2nd and 3rd ninety degrees is because, even though the distance from TDC to horizontal center line and horizontal center line to BDC is the same (in the example it is 0.5), the rod base has to move out from vertical center which brings the piston head down more than when it moves back right. Think about it this way. If the piston line is a fixed length, the bottom of the piston line is fixed on the x axis (horizontal center line) and the top of the piston line is fixed on the y axis (vertical center line). As you move the bottom of the piston line away from the y axis (right or left) the top of the piston line moves down the y axis. As you move the bottom of the piston line back towards the y axis the top of the piston line moves up the y axis. Back to the engine. This means that the first 90 degrees of rotation not only moves the rod base down the y axis, but also the rod base moves to the left away from the y axis. Both movements cause the piston head to move down. The 2nd 90 degrees of rotation moves the rod base down, but the movement of the rod base back to the right towards the y axis, tries to raise the piston head. The net effect is less piston head movement downward than the 1st 90 degrees of movement. The 3rd 90 degrees of roation sees the rod base moving out (right) from the y axis which causes the piston head to go down, but the rod base is also moving upwards, the net effect being the same amount of piston head movement as the 2nd 90 degrees of rotation, just in the opposite direction. Lastly, the 4th 90 degrees of rotation sees the rod base coming back in towards the y axis and the rod base moving upwards, thus the piston head moves upward having the same piston head movement as the 1st 90 degrees of rotation, just in the opposite direction.
Thanks for taking the time to explain it your way. I really wasn't able to wrap my mind around this, but in the middle of reading your comment it finally clicked for me.
I never loved the mere "math explanation" of a phenomenon, so I always looked for the physical reason causing secondary forces in my studies but nobody seemed to understand it for real to clear my doubts...so, we're basically talking about a "fourier-ization" of the resulting force, but now i know where it comes from so big up and thank you!
I just recently got into cars and how they work, your videos have helped me understand. You explain everything to a T and u are also really clear and thorough with everything you explain. Thank you so much for helping me understand and making it interesting.
Very interesting and well presented. I'm wanting to balance some cylinder two stroke cranks so I'll be using your very helpful videos and I'll be asking questions when I run into difficulties.Thanks for help.
I freaking love the power of the car community, I had a small doubt about the secondary imbalance that was bugging me for a few days and the comment section cleared it out.
Great! You explained it so good that school teachers (not only mechanics, techs & engineers) should use it for Pythagorean theorem, sinusoid curve, and a little of trigonometry in 7th or 8th grade... I have seen so many graduated students that "can't apply anything to nothing"... (obviously not the case of engineering, math, physics...) Thank you one more time, my teenager is also your fan...
Math is so important becouse once you find the equivalent expression of some dinamical process, you apply the simple tool of Math to it and can conclude a lot of characteristics and consecuences of that process which will be hard difficult (if not impossible) just applyng plain reasoning.
It’s important for scientists and engineers. People who want to become said professionals know this or eventually realize it when studying math. Most people don’t need to know geometry or algebra, let alone advanced mathematics. It is useful however to learn math as it will help develop analytical and critical thinking skills.
The reason of the piston moving faster going up then down would be because of the angle of the rod. If the rod was straight the whole time while piston going up and down then the numbers would be as expected. Correct?
my reply is old relative to your comment but whatever if you've already known the answer, other readers might benefit from my answer which is 'the reason the piston goes up faster than down is because of the length of the rod relative to the radius of the crankshaft'. That's my explanation I hope it helps and if there's anything wrong with it please let me know.
@@theadel8591 ur right that other readers might benefit from ur answer, could you expand on that statement though. Would it mean that the speed of the piston going up and down could be equalised if the crankshaft radius was increased?
A bit of hand waving to get the green curve, but generally correct. Another way to explain it is to say that the primary imbalance is _defined_ to be the momentum that the piston would have with a very long connecting rod, moving with a velocity that is the cosine of the rotation (the purple curve.) As you point out, the velocity is not a cosine due to the shortness of the rod. The secondary imbalance (green) is defined as the difference between the actual momentum (red curve) and the cosine curve. How do you get a term that oscillates at twice engine speed? It's in the little triangle on the whiteboard. The sum of the squares of the sides of a triangle equals the square of the hypotenuse. One of the sides is a cosine, and a cosine of an angle, squared, is 1 + the cosine of twice that angle. Twice the angle means it "rotates" twice as fast.
I don't understand one thing. The whole piston velocity imbalance you're talking about occurs only one time per revolution, doesn't it? We have fast-slow-slow-fast (using your circle) and starting and TDC. It's a sine that happens once per revolution. That would happen twice per revolution if we had fast-slow-fast-slow. Can you please explain the reasoning behind this? Thanks for the videos, keep up the good work.
I got confused by this too. I think it's made easier by thinking about it in terms of where the vertical force is going: - In primary imbalance, we have a force of N pushing DOWN on the shaft until BDC (intake/combustion), then we have a force of -N on the shaft pulling UP on the shaft until TDC (compression/exhaust). Ultimately, we have one change in force every rotation of the shaft (N changes to -N) - secondary imbalance, we break it down further: - TDC: Force xN pushing down on the shaft until 90° - 90°: Force N pushing down on the shaft until BDC - BDC: Force -xN pushing up on the shaft until 270° - 270°: Force -N pushing up on the shaft until TDC Note that: - Our variable 'N' isn't the same one as in primary imbalance. I just used the same symbol to make it easier to read. - Our variable 'x' is just some fraction. It'll vary depending on the diameter of the crank shaft and the length of the piston rod. You see that the force changes twice as often (xN to N to -xN to -N) in one rotation as compared to our example primary imbalance. An easier way to see it is to take diagram No. 3 and split it vertically in 2. On the right we have downward force, and on the left we have upward force, so it's DOWN fast-slow UP slow-fast At least, this is my interpretation of it. I welcome feedback.
I never realized this, thanks! I have, however, theorized about power production from an elliptical crank shaft motion. I got the idea from seeing pro cyclists using elliptical chain rings (the front gear attached to the pedals). The oval gears allow the pedaling force needed at the top and bottom of the stroke to be less (because you're less efficient at those positions) and normal in the forward and backward positions. Have you ever heard of this being used in a car? If not, what are your thoughts on it (just the idea, not necessarily how it would be implemented)? Thanks!
+Dani2wheels I had the same idea after watching this video, googled for it, and found my way back to your comment :) The first problem is that the piston doesn't drive the crank via a chain (like your legs do on a bike) so it will inevitably drive it in circles unless you connect it eccentrically like in a Wankel engine. [searches google for "eccentric crank engine balance" instead] I found a rather back-of-the-envelope patent for a design like that: www.ipo.gov.uk/p-find-publication-getPDF.pdf?patentNo=GB2449506&DocType=A&JournalNumber=6236
As always great way explain... and great topic... but this causes more questions...i always belive it was only to make the engine more steady and smoth idle...
As is usually the case, this guy knows enough about what he's talking about to pull the wool over the eyes of the typical youtube viewer. Certainly it is true that in order to understand secondary imbalance you have to understand asymmetry in piston motion. But just because you understand one of the elementary pieces of a bigger puzzle does not mean that you understand the whole puzzle. The explanation he gives is partly correct, but at about 3:48, he jumps in over over his head. This is what he does in most of his videos. He understands a little bit of a bigger puzzle, and he explains the part that he understands, but makes a mess of the bigger thing he's ostensibly explaining, because he doesn't really understand it. Here, as soon as he starts talking about the primary and secondary forces, he is over his head and does not really know what he's talking about. The mathematical approach to vibrational motion is harmonic analysis, where the vibrational motion is represented by an infinite series of pure sinusoid terms starting (the fundamental) at frequency the same as the crankshaft rotational frequency. Then 2x that frequency, then 3x that frequency, and so on. "Primary balance" refers to the case where there is no vibrational motion component at the 1st order, i.e., no vibration at frequency the same as the crankshaft's rotational frequency. The primary example of an engine that has perfect primary balance but that lacks perfect secondary balance is the in-line four using the flat 180-degree crankshaft. With this engine, there is no vibrational motion at the crankshaft's rotational frequency. There is no rectilinear motion (back and forth in a straight line) at this fundamental frequency, and no rotational motion (end-over-end rocking motion) at this frequency. The two pairs of pistons (the inner pair and the outer pair) do not generally move at the same speed (because one pair is in the lower part of the stroke and the other pair is in the upper part of the stroke), which means that the momentum in one direction does not perfectly cancel with the momentum in the other direction. The aggregate piston mass thus exhibits rectilinear vibrational motion at frequency starting at 2x the crankshaft rotational frequency. As such, the inline four has perfect primary balance (because there is no vibrational motion of any sort at the crankshaft's rotational frequency), but it does not have perfect secondary balance (because there is vibrational motion at frequencies that are integer multiples of the crankshaft's rotational frequency). The reason the lowest frequency of vibrational motion in this engine is 2x the crankshaft rotational frequency is that when the two pairs of pistons pass each other near (but not exactly at) the midpoint of the stroke, the direction of motion (and direction of momentum) reverses. This occurs because the pair that was moving faster becomes the slower pair and vice versa. This is a simplified description of what actually happens, but this is close enough to explain that the direction of the rectilinear, aggregate piston momentum reverses once with each 180-degree rotation of the crank and twice with each 360-degree rotation. Therefore the lowest frequency of this rectilinear (aggregate) motion is 2x the crankshaft rotational frequency, and it is for this reason that it is considered a secondary imbalance and not primary imbalance. To make the picture more complete, consider the example of the inline triple, which is more like a boxer twin. Rectilinear vibrational motion cancels among the three pistons, however there is end-over-end rocking motion at the same frequency as crankshaft rotation, and as such, the inline triple does not achieve perfect primary balance. The same is true for the boxer twin. But if you put two inline triples end-to-end, the end-over-end rotational (rocking) motion cancels between the two halves. You end up with no rectilinear motion (at any multiple of the crankshaft rotational speed) and no end-over-end rotational motion (at any multiple of the crankshaft rotational speed). Thus, the inline six doesn't merely achieve perfect primary balance (which would make it no better than the inline four); it achieves perfect balance, primary and secondary.
d = Position of piston in cylinder relative to crankshaft centreline r = Crank throw radius = Stroke / 2 L = Conrod centre-centre distance theta = Crank angle (theta = 0 at TDC) *d = r cos(theta) + sqrt(L^2 - r^2 sin^2(theta))* [Primary Component] + [Secondary Component] And if you know your double-angle formulas, sin^2(theta) = (1 - cos(2 theta)) / 2. The primary represents the up/down motion of the piston in the cylinder. The secondary component represents the side-to-side component inherent in the crankpin's circular motion, and since sin^2(theta) has double the frequency of cos(theta), the side-to-side component induces an additional vibration occuring at twice the crankshaft speed. The side-to-side component is where the secondary vibration comes from.
Great example graphing the points of the travel, it really anchored the 2ndry balance concept. Is it accurate to say the secondary force graph isn't a perfect sine wave, but greater in magnitude above the x-axis vs below it?
A major benefit of horizontally opposed engines is that the secondary forces are mostly balanced essentially for free. Non-sinusoidal acceleration due to the crankshaft geometry (translation of rotary to linear motion via Pythagoras) requires harmonic forces.
It is stated that a 60 Degree V6 will have a rocking motion since it is basically (2) 3 cylinder engines. However I think that assumes the piston pairs are sharing connecting rod journals. What is the balance characteristics of a 60 Degree V6 that used separate connecting rod journals for each rod? These separate journals are then allowed to be spaced for optimum balance. I think the early 1960s GMC 305 V6 had that kind of layout.
I wish Jason would revisit this topic in a bit more detail and include "rod to stroke ratio" and "piston dwell time" at btc and tdc as a result of crank stroke.
Omg! I allways tought inline-4 engines and flat plane V8s were perfectly balanced. Now I know that there is up and down vibrations from the secondary inbalance. Thancc
Can you make a video on why straight line engines are good for acceleration? Your videos make me sound like I know what i'm talking about when I talk cars to friends :D
Good explanation on the secondary imbalance. Just my opinion, instead of saying the force as the contributing factor of the imbalance, perhaps you could say it is due to the momentum created by the piston because it involves both the mass and the velocity
I'm having some trouble understanding the concept of the secondary balance. Can you recommend me some literature or something that can help me? Thank you and keep the good work!
Now correct me if I'm wrong, Mr.fenske, and I realise that the graph on the board is hand drawn and simplified, but am I right by saying that since the acceleration from 90 degrees to 270 from TDC is LESS than 270, back to 90, then sholdn't the Force, (F=ma) also be less on the graph from 90 to 270? It doesn't change the fact that the net force, will yield a similar graph as shown and the point is made about secondary balance, but as I don't know any better, I'm quite confused by your acceleration graph where the peaks and valleys are of the same height. Thanks for the wonderful videos.
Thanks for great video for simple explanation what causes secodaru inbalance but why reveils after sixties I mean 1960 before that nothing was done to balance a S.I. In Ford A, T, A, engines don´t have any counter weights. Is this something that reveills it self after 5000 RPM.
I was wondering if you could do a video on how to have a high flow exhaust/intake that is also quiet. The reason: Some states have laws stating citizens cannot have an exhaust that is significantly louder than the stock exhaust system without risk of a ticket.
I totally get that the piston moves further from 90 btdc to 90 atdc than it does in the other "bottom" 180 deg of crank rotation, but I still don't understand why the frequency of secondary imbalance is twice the primary. 1.) Why doesn't this difference in piston speed merely result in a larger primary imbalance (with regard to single-cyl engine for now) and 2.) Can anyone explain why a parallel twin engine with a 270 degree crank arrangement is said to have better secondary balance?
So the secondary forces are all the forces that are not caused by the piston head itself, or all the forces that are not moving in the same direction as the piston?
What does the torsional balancer cancel out? The 60 and 120 degree V12 have balanced first and secondary forces, but isn’t there another force that needs to be balanced that isn’t?
Thank you very much for this educational video. Do you happen to have a video about engines redline? How to get engine to Rev higher? That would be a very interesting topic, because you can increase your horsepower. If not maybe this would be a good future video idea, a video about how to get engines to rev higher. Especially engines with only two valves per cylinder.
If you are talking about a stock engine, the only way to do is by lightening the cranktrain. Crankshaft, conrods, pistons, valves etc. Porting the head may help breathing issues slightly, but the main things will be the manifolds, intake and exhaust. Once they breath freely, mild forced induction - that of a centrifugal supercharger or a turbo with large housing - will also be an advantage.
Question, can the difference in the imbalance be corrected, to equal to zero, by manipulating the rod-ratio? In other words, finding the sweet spot that would make the upward force equal to the downward force.
The secondary imbalance is created due to the pistons moving at different velocities, if I'm correct at what I've grasped from the video. But what if I have a 4 cylinder inline?. Two of it's pistons and gonna move down while the other two go up. So when two pistons go up with higher velocity the other two will go down with the same velocity, balancing the secondary forces. The same would happen when the two pistons will come down with lesser velocity and the other two will go up with the same velocity. This way, everything's balanced. [I have no knowledge about how this all works, but just imagined this stuff. Correct me if I'm wrong plz....:)]
the problem is more correctly defined as different accelerations by the piston. The act of the piston going over top dead center is different (faster) than at bottom dead center, due to the connecting rod geometry described in the video. Thus the top dead center (TDC) part of the cycle creates a higher "g-force" than the bottom dead center part (BDC). typical inline 4 cylinder engines balance the primary forces (2 pistons up and 2 pistons down), but the difference in G forces between TDC and BDC creates a smaller secondary imbalance. As each pair of pistons goes through TDC (which happens twice per revolution) a small inertia remains. Thus a typical 4 cylinder engine has a slight secondary imbalance. This is usually ignored for engines that operate at lower engine RPM but in high revving engine applications its possible to balance it with a Secondary balance shaft that is driven at twice engine RPM.
Are balancing shafts only/more important with low conrod length to stroke ratio. A 4 cylinder with a ratio of 2 seems to be quite balanced and does not have balancing shafts. I am looking at de-stroking a BMW B48. It currently has a ratio of 1.56 but if I reduce the stroke to 70mm from 94.6 it becomes 2.29. Would you think it still requires balancing shafts. B48 Con-rod centre to centre is 148mm, de-stroked it would be 160.3mm.
Anther great video. I don't quite agree with your description of piston velocity though. I think piston velocity is zero at both TDC and BDC. At the half way up or down rotation point the piston speed is at maximum. So the piston is not slower in the bottom half than the too half. The secondary imbalance is because the piston speed is out of phase with the primary balance.
Yes, it is stopped at TDC and BDC, but the piston moves faster (a further distance in a shorter amount of time) during the top 180 degrees of rotation than the bottom 180 degrees of rotation.
Engineering Explained i'm fascinated by your videos so don't take this as a criticism. I can't see how the distance travelled by the piston is any different in the top 180deg compared to the bottom. If it was, then the crank rotation wouldnt be circular! In a single cylinder engine you could argue the piston is travelling faster in the 180deg of the power stroke but for anything with more than 4 cylinders i think the radial velocity of the crank is the same throughout the 4 strokes.
Lol, i know you get a lot of requests, but some day could you talk about suspension geometry, like CG, roll center, roll couple, things like that. That would be awesome :D
hey if you have a boosted engine,each piston gets same boost pressure? normal intake manifold normal pistons.Same with single fogger nitro system.Sorry for the bad English!
It took me 20 minutes just sitting here after the video ended to figure out what on earth you were talking about with piston speed being different in the top 180 degrees of the arc than the bottom 180 degrees. It seems like it's all the same distance and the same shape going the same direction, so it should be the same speed in each half. But, circles aren't squares. Am I right about this notion? If you draw a diamond inside the circle (or a square standing on a corner, if you prefer), you get straight paths from 0-90-180-240 degrees on the circle. If the connecting rod followed those straight lines, speed would be constant. But because it's following a circular path around those four points, the speed (acceleration?) changes with the circle in relation to the flat plane described by the diamond shape. So the highest piston speed would actually be at 45 degrees and 285 degrees, presuming 0 degrees is TDC. ...Is that right...?
I think you're very confused on the matter. The piston velocity cannot ever be constant because the piston is reciprocating. The velocity of the piston is constantly changing through the cycle and consequently the piston is always accelerating.
Shouldn't the slow areas be at 0 and 180 and the fast at 90 and 270? Seems more intuitive than looking at it from the "bottom up". After all the same thing is happening if you take your measurements from the "top down". It's really just the difference in the relative mass of a constant speed counter weight and a variable speed piston that causes this right?
great explanation!! just one question though. where did you get the proportions to your model of 'diameter 1, con-rod 2' ? are these dimensions proportional to most engines? i mean, do all engines' con-rods travel faster from 0(deg) to 180(deg) BECAUSE of these dimensions? or do the con-rods travel this distance faster due to the power stroke
I made up the dimensions so the math would be simple and easy to understand, the proportions are not representative of the real world (at least, they have no basis from the real world). The speeds could be faster or slower depending on the setup.
Engineering Explained you are awesome! Thanks.i definitely want to know more about real applications where it can be broken down to the measurements. Thanks
interestingly, Smokey Yunick did an article about con rod length/stroke ratios. It was about 45 yrs ago, but I think it was about keeping the length as short as possible but that increased the cylinder wall loading, increasing the chances of ventilating the block.
The longer the connecting rod, the less the secondary force, right ? I assume : If the connecting rod is infinite long, x will be almost equal to 2, and then no secondary force. Is that right ? How does the engineer determine the ratio of (radius of crankshaft/connecting rod) ?
The difference in acceleration is explained based on the different "heights" from the bottom of the of crank circle to the top of the piston. I just can't understand why do we have to take into account this virtual height (virtual because it doesn't correspond to any actual physical component or assembly). Since all the parts involved have a fixed length and are not deformable, the path travelled by the rod has to be constant and the sum "rod+crank radius" is always the same value!
U say that u have a stronger force upwards and a weaker force downwards but is it actually not fully true because it is like that if you look one pair of the opposite 90 degree travelings but if you look at the other 2 travelings it's actually vice versa (i mean the downward force is stronger and the upper is weaker)? Otherwise nice vid once again :) ! Keep it up.
This all makes sense, but I think the way you explain it is a little off. The "forces" themselves occur once per crank revolution. The forces can be calculated by converting the equation shown here (en.wikipedia.org/wiki/Engine_balance#Secondary_.28non-sinusoidal.29_balance) into the time domain and taking the second derivative of x with respect to time. If you perform a fourier sine series expansion on the resulting equation, you would find that terms 3, 4, 5... are negligible. In such an expansion, each term is a higher order than the last (oscillates twice as fast as the fundamental, then three times as fast, four times, etc.). Therefore, you can say that the piston inertial forces can be APPROXIMATED by two sine waves added together. Because the second term in the series, the second sine wave, is second order one could say it "occurs" twice per crank revolution. The thing to take away is that the second order wave is JUST A MATHEMATICAL CONSTRUCT. The actual forces on the piston cyclically occur ONCE per crank revolution.
could you please explain how the upward and downward forces happen twice per revolution? it looks like it is the same as the piston (one downward force and one upward force) per revolution. i understand that the upward force is greater than the downward force, but i dont see how that is occurring 2x more frequently than the primary force of the piston. thanks! EDIT: for instance, why is there a positive secondary force at 180?
Bear in mind that there is a deceleration just prior to TDC and an equal acceleration just after - the point is that the deceleration and acceleration prior to and after BDC are weaker. What this means is that, per crankshaft rotation, you have strong deceleration prior to TDC, strong acceleration afterwards, then weaker deceleration towards BDC and weaker deceleration after, hence four secondary moments per crankshaft rotation. His diagram to the upper right corner is confusing as it makes it seem as though there are two halves of the rotation during which the velocity of the piston is constant, where, in fact, what it shows is that the pairs of decelerating and accelerating moments are not identical, two being rapid and two slower.
the secondary forces occur twice per revolution because of the irregular rotation of the crank, when the piston is on the top dead center to half way to bottom dead center the crank will rotates faster and then decelerate halfway down and half way upward. Due to the changes of rotational speeds it creates vibrations. So the movement of the crank from 0 to 90 degrees is fast, 90 to 180 degrees is fast, then 180 to 270 degrees is slow and 270 to 360 degrees is slow. So, it occurs twice per revolution, half is fast and half is slow and due to this it produce vibrations.
1:54 It´s not two because the rod gets drawn *out* (of vertical) on the first half, which makes cylinder go a longer way. If you would just pull the rod to the side the cylinder would also move down, that´s where the additional way is coming from. On the second half it´s the other way around, here the rod gets drawn in (to vertical). If you would just pull the rod inwards, the cylinder would go up, that´s the effect that makes the second way shorter.
So the additional travel of the piston from the horizontal movement during TDC to 90 is downward and adds to the downward vertical travel of the piston from the top of the crankshaft height to the center axis height of the crankshaft, and from 90 to BDC this same additional travel of the piston from the horizontal movement during 90 to BDC is upward and subtracts from the downward vertical travel of the piston from the center axis height of the crankshaft to the bottom of the crankshaft height. Consider the movement of the piston by moving the base of the connecting rod to the side and then back again to the original position to see how the piston moves and why this first adds to the travel during the TDC to 90 and then subtracts from the travel during the 90 to BDC.
I have a question, well actually I have two! 1) Doesn't the second upper quarter, which also travels fast, counter the first upper quarter? If the first upper quarter is travelling up and over, the second upper quarter is coming over and down with equal speed. If the two fast quarters were on one side (ie the upstroke) that would make sense, but the diagram to me makes it appear like there would be unequal sideways force, rather than vertical. Can you explain this please? Thank you. 2) I'm also a little confused about the issue re: Pythagorean. What equation do you get for BDC>270 triangle?
Your Math is off X=2-.4365. or X=1.5635. so half of the dots are in the wrong place in the 2nd diagram. I think you are giving a cascading error. This is a better explanation of the problem than Philip H. Smith in "The Design and Tuning of Competition Engines" but it is not all there. What are the books you read, that gave you the ideas you present here on U-Tube?
The Con rod only has a one inch throw. You may have done the Pythagorean theorem correctly as a math operation. But there is no room for the 2.4365 inches of travel with the one inch throw crank that you put in the drawing. Pride goes before the fall.
James Thornton He said 2.4365" is the height of the piston pin at 90° and 270°, not the travel. The piston travels 1.1306" up+down from 270° over TDC to 90°, but travels only 0.8730" down+up from 90° under BDC to 270°. That means the piston travels faster through TDC and slower through BDC if the crank is rotating at a constant speed. Engineering Explained What you described in the video is not a secondary force, it is simply a non-sinusoidal primary force due to the geometry. A secondary (also second order) force is caused from the reciprocating components, meaning the crank pin and a function of the connecting rod. Not only are you lifting it up and down in a sinusoidal wave, but laying a rod at an angle then straightening it up is also adding a force from the center of mass rising and falling. At TDC and BDC the inertia from the rod is pulling up on the crank but at 90 and 270 the rod is pushing down on the crank. To help illustrate this, forget about the crank causing an up/down motion and think about your pen on the table and make this motion with your pen: _ / | \ _ and let the cap rise up in a line and keep the tail end slide on the table. When you straighten it this causes and upwards momentum. Do it fast enough and the pen will want to keep moving up because the center of mass has moved upwards, and you'd have to pull it back down. This is what is happening to the connecting rod. This is what a secondary force is, and it happens twice per revolution. Also, you should talk more about the vibrations due to the side-to-side motion of the connecting rods as this has significant effects on all engine configurations.
Blaylock1988 It is a second order being described *very well. The piston motion of a crank slider mechanism is made up of an infinite harmonic. It is usually enough to consider only the 1st and 2nd component of this in terms of engine balancing.
Certain engine designs have an advantage in balancing. For example my favorite engine, the straight six with a star configuration crank, is inherently balanced up until the 6th order. Whereas all I-4 engines are not inherently balanced at 2nd, 4th, 6th, etc. So those need to be balanced using other methods.
I can’t quite get my head around where the double frequency comes from. Each revolution, the piston is fast once and slow once. Wouldn’t this create a single peak and a single valley? Where does the double frequency come from? I think it’s the fact that from 0-90 the piston is accelerated by the crank, from 90-180 the piston is slowed the crank, from 180-270 the piston is accelerated by the crank and from 270-360 the piston is slowed by the crank. But this has nothing to do with the different distance travelled by the piston between 0-180 and 180-360. I’ve googled around but can’t find it.
Or is the green line simply the result of subtracting the blue line (theroretical forces with infinite connecting rod) from the red line (actual forces)? I think this is true regardless, but I wonder if there is a different explanation around as to where the double frequency comes from.
Does air resistance due to intake & exhaust cycles have any influence on this balancing dynamic? Does alloy of this crankshaft bearings have any influence on engine life, like aluminium vs. copper based alloy, & different alloys, & so on.
The electric starter motor turns in one direction beginning the engine's rotation. The issue is fuel, as fuel is usually mixed with air before it enters the combustion chamber. With direct injection, it's theoretically possible, but would be difficult for four strokes. Much easier to do with two strokes because there is less work to do in a cycle. Detroit diesels were known to sometimes reverse direction.
the piston traveled from 0 to 90 degree (from tdc) is larger when compared to 90 to 180 degree (bdc), now at what angle will the piston travel be equal in length?
+Avinash T You have to know the ratio of the rod length to stroke for the engine in question (the example given of 2:1 will probably be "close", it's 1.85:1 in my engine). Draw the triangle (as above) using trig, also substitute the correct number for x (rod length ratio + 0.5) and solve for the angle. FYI: for oblique triangles the formula c^2 = a^2 + b^2 (right triangle) becomes the more general c^2 = a^2 + b^2 - 2(ab)cosC, where C is the angle you seek.
I think the concept is still true but using your same geometry I think it would be better to measure from the top of the piston and say that the piston moves 1.1 in the top half of rotation and only 1 in the bottom half.
Great video! One thing to add for those who need a more explanation. The reason the piston travels farther during the 1st and 4th 90 degrees of rotation than the 2nd and 3rd ninety degrees is because, even though the distance from TDC to horizontal center line and horizontal center line to BDC is the same (in the example it is 0.5), the rod base has to move out from vertical center which brings the piston head down more than when it moves back right.
Think about it this way. If the piston line is a fixed length, the bottom of the piston line is fixed on the x axis (horizontal center line) and the top of the piston line is fixed on the y axis (vertical center line). As you move the bottom of the piston line away from the y axis (right or left) the top of the piston line moves down the y axis. As you move the bottom of the piston line back towards the y axis the top of the piston line moves up the y axis.
Back to the engine. This means that the first 90 degrees of rotation not only moves the rod base down the y axis, but also the rod base moves to the left away from the y axis. Both movements cause the piston head to move down. The 2nd 90 degrees of rotation moves the rod base down, but the movement of the rod base back to the right towards the y axis, tries to raise the piston head. The net effect is less piston head movement downward than the 1st 90 degrees of movement. The 3rd 90 degrees of roation sees the rod base moving out (right) from the y axis which causes the piston head to go down, but the rod base is also moving upwards, the net effect being the same amount of piston head movement as the 2nd 90 degrees of rotation, just in the opposite direction. Lastly, the 4th 90 degrees of rotation sees the rod base coming back in towards the y axis and the rod base moving upwards, thus the piston head moves upward having the same piston head movement as the 1st 90 degrees of rotation, just in the opposite direction.
Thanks for taking the time to explain it your way. I really wasn't able to wrap my mind around this, but in the middle of reading your comment it finally clicked for me.
Jordan Fischer beautiful explanation, thank you
My friend thanks a lot . I wasn t able to fìgure out why this was happening but due your explanation everything became crystall clear
i cant understand. :< or maybe im just dumb
Okay got it.
I love this channel, it is impressive how clear your explanations can be, so thank you, great job Jason.
Greatly explained. Shortly, yet so clearly. You are one of a kind!
I never loved the mere "math explanation" of a phenomenon, so I always looked for the physical reason causing secondary forces in my studies but nobody seemed to understand it for real to clear my doubts...so, we're basically talking about a "fourier-ization" of the resulting force, but now i know where it comes from so big up and thank you!
I just recently got into cars and how they work, your videos have helped me understand. You explain everything to a T and u are also really clear and thorough with everything you explain. Thank you so much for helping me understand and making it interesting.
i just want to thank you for all the videos you have made. You have taught me a lot. your work is definitely appreciated. :)
Very interesting and well presented. I'm wanting to balance some cylinder two stroke cranks so I'll be using your very helpful videos and I'll be asking questions when I run into difficulties.Thanks for help.
So THAT'S why there's balance shafts in my LD9! And it answers the question of why they turn at twice engine rpm. Great vid!
Yes indeed!
I freaking love the power of the car community, I had a small doubt about the secondary imbalance that was bugging me for a few days and the comment section cleared it out.
Wow, great video! I had never thought about this before.
Always wondered what the secondary balance was - thanks for posting!
Great! You explained it so good that school teachers (not only mechanics, techs & engineers) should use it for Pythagorean theorem, sinusoid curve, and a little of trigonometry in 7th or 8th grade... I have seen so many graduated students that "can't apply anything to nothing"... (obviously not the case of engineering, math, physics...) Thank you one more time, my teenager is also your fan...
Great to hear, thank you both for watching!
You were right Mrs. 10th Grade teacher, math is important.
Math is so important becouse once you find the equivalent expression of some dinamical process, you apply the simple tool of Math to it and can conclude a lot of characteristics and consecuences of that process which will be hard difficult (if not impossible) just applyng plain reasoning.
It’s important for scientists and engineers. People who want to become said professionals know this or eventually realize it when studying math. Most people don’t need to know geometry or algebra, let alone advanced mathematics. It is useful however to learn math as it will help develop analytical and critical thinking skills.
this isnt 10th grade math...
1:30 Something I do understand!
I'm back!
Can you do a video on "rocking couple" of a in-line4 crossplane engine?
Rocking moments in engines are caused by the net torque about the centre of mass of the pistons being non-zero.
u r so good at explaining.thnx for all of ur videos.
The reason of the piston moving faster going up then down would be because of the angle of the rod. If the rod was straight the whole time while piston going up and down then the numbers would be as expected. Correct?
my reply is old relative to your comment but whatever if you've already known the answer, other readers might benefit from my answer which is 'the reason the piston goes up faster than down is because of the length of the rod relative to the radius of the crankshaft'. That's my explanation I hope it helps and if there's anything wrong with it please let me know.
@@theadel8591 ur right that other readers might benefit from ur answer, could you expand on that statement though. Would it mean that the speed of the piston going up and down could be equalised if the crankshaft radius was increased?
so what is the solution for this effect or vibration have adopted manufacturers and engineers .... Hello from Venezuela, I love your channel
A Balance shaft module. You will find a BSM in most 4 cylinder auto engines.
A bit of hand waving to get the green curve, but generally correct. Another way to explain it is to say that the primary imbalance is _defined_ to be the momentum that the piston would have with a very long connecting rod, moving with a velocity that is the cosine of the rotation (the purple curve.) As you point out, the velocity is not a cosine due to the shortness of the rod. The secondary imbalance (green) is defined as the difference between the actual momentum (red curve) and the cosine curve.
How do you get a term that oscillates at twice engine speed? It's in the little triangle on the whiteboard. The sum of the squares of the sides of a triangle equals the square of the hypotenuse. One of the sides is a cosine, and a cosine of an angle, squared, is 1 + the cosine of twice that angle. Twice the angle means it "rotates" twice as fast.
I don't understand one thing. The whole piston velocity imbalance you're talking about occurs only one time per revolution, doesn't it? We have fast-slow-slow-fast (using your circle) and starting and TDC. It's a sine that happens once per revolution. That would happen twice per revolution if we had fast-slow-fast-slow.
Can you please explain the reasoning behind this? Thanks for the videos, keep up the good work.
I got confused by this too.
I think it's made easier by thinking about it in terms of where the vertical force is going:
- In primary imbalance, we have a force of N pushing DOWN on the shaft until BDC (intake/combustion), then we have a force of -N on the shaft pulling UP on the shaft until TDC (compression/exhaust).
Ultimately, we have one change in force every rotation of the shaft (N changes to -N)
- secondary imbalance, we break it down further:
- TDC: Force xN pushing down on the shaft until 90°
- 90°: Force N pushing down on the shaft until BDC
- BDC: Force -xN pushing up on the shaft until 270°
- 270°: Force -N pushing up on the shaft until TDC
Note that:
- Our variable 'N' isn't the same one as in primary imbalance. I just used the same symbol to make it easier to read.
- Our variable 'x' is just some fraction. It'll vary depending on the diameter of the crank shaft and the length of the piston rod.
You see that the force changes twice as often (xN to N to -xN to -N) in one rotation as compared to our example primary imbalance.
An easier way to see it is to take diagram No. 3 and split it vertically in 2. On the right we have downward force, and on the left we have upward force, so it's DOWN fast-slow UP slow-fast
At least, this is my interpretation of it. I welcome feedback.
@@harryham1 Although I didn't pose the question, I was able to benefit from your very comprehensive explanation, thank you :)
@@arnedorr6904 Thanks for letting me know that!
Going through some serious stuff at the moment, and that really brightened me up.
Very well explained. Solid job.
I never realized this, thanks! I have, however, theorized about power production from an elliptical crank shaft motion. I got the idea from seeing pro cyclists using elliptical chain rings (the front gear attached to the pedals). The oval gears allow the pedaling force needed at the top and bottom of the stroke to be less (because you're less efficient at those positions) and normal in the forward and backward positions. Have you ever heard of this being used in a car? If not, what are your thoughts on it (just the idea, not necessarily how it would be implemented)? Thanks!
+Dani2wheels I had the same idea after watching this video, googled for it, and found my way back to your comment :)
The first problem is that the piston doesn't drive the crank via a chain (like your legs do on a bike) so it will inevitably drive it in circles unless you connect it eccentrically like in a Wankel engine.
[searches google for "eccentric crank engine balance" instead]
I found a rather back-of-the-envelope patent for a design like that:
www.ipo.gov.uk/p-find-publication-getPDF.pdf?patentNo=GB2449506&DocType=A&JournalNumber=6236
crossplane crank
Wow, I always wondered about the elliptical front sprocket, but not enough to actually question why.
excellent job Jason!
very nice lesson
As always great way explain... and great topic... but this causes more questions...i always belive it was only to make the engine more steady and smoth idle...
As is usually the case, this guy knows enough about what he's talking about to pull the wool over the eyes of the typical youtube viewer. Certainly it is true that in order to understand secondary imbalance you have to understand asymmetry in piston motion. But just because you understand one of the elementary pieces of a bigger puzzle does not mean that you understand the whole puzzle. The explanation he gives is partly correct, but at about 3:48, he jumps in over over his head. This is what he does in most of his videos. He understands a little bit of a bigger puzzle, and he explains the part that he understands, but makes a mess of the bigger thing he's ostensibly explaining, because he doesn't really understand it. Here, as soon as he starts talking about the primary and secondary forces, he is over his head and does not really know what he's talking about.
The mathematical approach to vibrational motion is harmonic analysis, where the vibrational motion is represented by an infinite series of pure sinusoid terms starting (the fundamental) at frequency the same as the crankshaft rotational frequency. Then 2x that frequency, then 3x that frequency, and so on. "Primary balance" refers to the case where there is no vibrational motion component at the 1st order, i.e., no vibration at frequency the same as the crankshaft's rotational frequency. The primary example of an engine that has perfect primary balance but that lacks perfect secondary balance is the in-line four using the flat 180-degree crankshaft. With this engine, there is no vibrational motion at the crankshaft's rotational frequency. There is no rectilinear motion (back and forth in a straight line) at this fundamental frequency, and no rotational motion (end-over-end rocking motion) at this frequency. The two pairs of pistons (the inner pair and the outer pair) do not generally move at the same speed (because one pair is in the lower part of the stroke and the other pair is in the upper part of the stroke), which means that the momentum in one direction does not perfectly cancel with the momentum in the other direction. The aggregate piston mass thus exhibits rectilinear vibrational motion at frequency starting at 2x the crankshaft rotational frequency. As such, the inline four has perfect primary balance (because there is no vibrational motion of any sort at the crankshaft's rotational frequency), but it does not have perfect secondary balance (because there is vibrational motion at frequencies that are integer multiples of the crankshaft's rotational frequency). The reason the lowest frequency of vibrational motion in this engine is 2x the crankshaft rotational frequency is that when the two pairs of pistons pass each other near (but not exactly at) the midpoint of the stroke, the direction of motion (and direction of momentum) reverses. This occurs because the pair that was moving faster becomes the slower pair and vice versa. This is a simplified description of what actually happens, but this is close enough to explain that the direction of the rectilinear, aggregate piston momentum reverses once with each 180-degree rotation of the crank and twice with each 360-degree rotation. Therefore the lowest frequency of this rectilinear (aggregate) motion is 2x the crankshaft rotational frequency, and it is for this reason that it is considered a secondary imbalance and not primary imbalance. To make the picture more complete, consider the example of the inline triple, which is more like a boxer twin. Rectilinear vibrational motion cancels among the three pistons, however there is end-over-end rocking motion at the same frequency as crankshaft rotation, and as such, the inline triple does not achieve perfect primary balance. The same is true for the boxer twin. But if you put two inline triples end-to-end, the end-over-end rotational (rocking) motion cancels between the two halves. You end up with no rectilinear motion (at any multiple of the crankshaft rotational speed) and no end-over-end rotational motion (at any multiple of the crankshaft rotational speed). Thus, the inline six doesn't merely achieve perfect primary balance (which would make it no better than the inline four); it achieves perfect balance, primary and secondary.
This old videos are imo the og stuff
Keep up the good work man.
Congrats!! I like the hatch more than the sedan. Turbo lag noticeable?
Great explanation! finally I got it!
d = Position of piston in cylinder relative to crankshaft centreline
r = Crank throw radius = Stroke / 2
L = Conrod centre-centre distance
theta = Crank angle (theta = 0 at TDC)
*d = r cos(theta) + sqrt(L^2 - r^2 sin^2(theta))*
[Primary Component] + [Secondary Component]
And if you know your double-angle formulas, sin^2(theta) = (1 - cos(2 theta)) / 2.
The primary represents the up/down motion of the piston in the cylinder. The secondary component represents the side-to-side component inherent in the crankpin's circular motion, and since sin^2(theta) has double the frequency of cos(theta), the side-to-side component induces an additional vibration occuring at twice the crankshaft speed. The side-to-side component is where the secondary vibration comes from.
Great example graphing the points of the travel, it really anchored the 2ndry balance concept. Is it accurate to say the secondary force graph isn't a perfect sine wave, but greater in magnitude above the x-axis vs below it?
could you do a video on the inline 6 engine and how the primary and secondary forces are perfectly balanced in that engine layout?
Yes indeed, fairly soon!
A major benefit of horizontally opposed engines is that the secondary forces are mostly balanced essentially for free.
Non-sinusoidal acceleration due to the crankshaft geometry (translation of rotary to linear motion via Pythagoras) requires harmonic forces.
❤️ thanks such a great content for free
Thank you for posting this
It is stated that a 60 Degree V6 will have a rocking motion since it is basically (2) 3 cylinder engines. However I think that assumes the piston pairs are sharing connecting rod journals.
What is the balance characteristics of a 60 Degree V6 that used separate connecting rod journals for each rod? These separate journals are then allowed to be spaced for optimum balance. I think the early 1960s GMC 305 V6 had that kind of layout.
I wish Jason would revisit this topic in a bit more detail and include "rod to stroke ratio" and "piston dwell time" at btc and tdc as a result of crank stroke.
Omg! I allways tought inline-4 engines and flat plane V8s were perfectly balanced. Now I know that there is up and down vibrations from the secondary inbalance. Thancc
My comment is probably the only one marked in a video from three years. 10x
Can you make a video on why straight line engines are good for acceleration? Your videos make me sound like I know what i'm talking about when I talk cars to friends :D
Good explanation on the secondary imbalance. Just my opinion, instead of saying the force as the contributing factor of the imbalance, perhaps you could say it is due to the momentum created by the piston because it involves both the mass and the velocity
To put it into perspective, isn't it due to the velocity that it's called secondary forces ?
Or am I totally wrong with that ?
I'm having some trouble understanding the concept of the secondary balance. Can you recommend me some literature or something that can help me?
Thank you and keep the good work!
Great information, thanks for sharing.
Now correct me if I'm wrong, Mr.fenske, and I realise that the graph on the board is hand drawn and simplified, but am I right by saying that since the acceleration from 90 degrees to 270 from TDC is LESS than 270, back to 90, then sholdn't the Force, (F=ma) also be less on the graph from 90 to 270? It doesn't change the fact that the net force, will yield a similar graph as shown and the point is made about secondary balance, but as I don't know any better, I'm quite confused by your acceleration graph where the peaks and valleys are of the same height. Thanks for the wonderful videos.
Nice videos, have you covered harmonic balancers and flywheels ?how about two stroke outboard engines
Thanks for great video for simple explanation what causes secodaru inbalance but why reveils after sixties I mean 1960 before that nothing was done to balance a S.I. In Ford A, T, A, engines don´t have any counter weights. Is this something that reveills it self after 5000 RPM.
so is there also a ratio of piston travel and stroke that could be the perfect sweet spot so that those secondary forces can be engineered out?
Rare information, thank you for explanation!
wow me like down here in uganda for u are real a blessing to the world
NAKROPA FAROUK Happy to hear it, thanks for watching!
I was wondering if you could do a video on how to have a high flow exhaust/intake that is also quiet. The reason: Some states have laws stating citizens cannot have an exhaust that is significantly louder than the stock exhaust system without risk of a ticket.
I totally get that the piston moves further from 90 btdc to 90 atdc than it does in the other "bottom" 180 deg of crank rotation, but I still don't understand why the frequency of secondary imbalance is twice the primary. 1.) Why doesn't this difference in piston speed merely result in a larger primary imbalance (with regard to single-cyl engine for now) and 2.) Can anyone explain why a parallel twin engine with a 270 degree crank arrangement is said to have better secondary balance?
So the secondary forces are all the forces that are not caused by the piston head itself, or all the forces that are not moving in the same direction as the piston?
This is ingenious.
I think this has some connection with how the speed at the driven shaft of a universal joint is not uniform due to the angle at the joint
What does the torsional balancer cancel out? The 60 and 120 degree V12 have balanced first and secondary forces, but isn’t there another force that needs to be balanced that isn’t?
Thank you very much for this educational video. Do you happen to have a video about engines redline? How to get engine to Rev higher? That would be a very interesting topic, because you can increase your horsepower. If not maybe this would be a good future video idea, a video about how to get engines to rev higher. Especially engines with only two valves per cylinder.
If you are talking about a stock engine, the only way to do is by lightening the cranktrain. Crankshaft, conrods, pistons, valves etc. Porting the head may help breathing issues slightly, but the main things will be the manifolds, intake and exhaust. Once they breath freely, mild forced induction - that of a centrifugal supercharger or a turbo with large housing - will also be an advantage.
Question, can the difference in the imbalance be corrected, to equal to zero, by manipulating the rod-ratio? In other words, finding the sweet spot that would make the upward force equal to the downward force.
Thanks for the video! Going to watch it now :)
dont forget that when piston move down it pushes by burning gases and then it moves up piston compress gases ,so its also have big effect on balance
The secondary imbalance is created due to the pistons moving at different velocities, if I'm correct at what I've grasped from the video. But what if I have a 4 cylinder inline?. Two of it's pistons and gonna move down while the other two go up. So when two pistons go up with higher velocity the other two will go down with the same velocity, balancing the secondary forces. The same would happen when the two pistons will come down with lesser velocity and the other two will go up with the same velocity. This way, everything's balanced. [I have no knowledge about how this all works, but just imagined this stuff. Correct me if I'm wrong plz....:)]
the problem is more correctly defined as different accelerations by the piston. The act of the piston going over top dead center is different (faster) than at bottom dead center, due to the connecting rod geometry described in the video. Thus the top dead center (TDC) part of the cycle creates a higher "g-force" than the bottom dead center part (BDC). typical inline 4 cylinder engines balance the primary forces (2 pistons up and 2 pistons down), but the difference in G forces between TDC and BDC creates a smaller secondary imbalance. As each pair of pistons goes through TDC (which happens twice per revolution) a small inertia remains. Thus a typical 4 cylinder engine has a slight secondary imbalance. This is usually ignored for engines that operate at lower engine RPM but in high revving engine applications its possible to balance it with a Secondary balance shaft that is driven at twice engine RPM.
Philip Ball Thank you :)
Are balancing shafts only/more important with low conrod length to stroke ratio. A 4 cylinder with a ratio of 2 seems to be quite balanced and does not have balancing shafts. I am looking at de-stroking a BMW B48. It currently has a ratio of 1.56 but if I reduce the stroke to 70mm from 94.6 it becomes 2.29. Would you think it still requires balancing shafts. B48 Con-rod centre to centre is 148mm, de-stroked it would be 160.3mm.
Anther great video. I don't quite agree with your description of piston velocity though. I think piston velocity is zero at both TDC and BDC. At the half way up or down rotation point the piston speed is at maximum. So the piston is not slower in the bottom half than the too half. The secondary imbalance is because the piston speed is out of phase with the primary balance.
Yes, it is stopped at TDC and BDC, but the piston moves faster (a further distance in a shorter amount of time) during the top 180 degrees of rotation than the bottom 180 degrees of rotation.
Engineering Explained i'm fascinated by your videos so don't take this as a criticism. I can't see how the distance travelled by the piston is any different in the top 180deg compared to the bottom. If it was, then the crank rotation wouldnt be circular! In a single cylinder engine you could argue the piston is travelling faster in the 180deg of the power stroke but for anything with more than 4 cylinders i think the radial velocity of the crank is the same throughout the 4 strokes.
The crank motion is circular, and the distance the piston moves is greater on the upper 180. This video proves it with simple geometry.
Mind doing a video on harmonic dampers like the fluidampr
Please, make a video for cylinder offset engines balance.
Lol, i know you get a lot of requests, but some day could you talk about suspension geometry, like CG, roll center, roll couple, things like that. That would be awesome :D
Ahhh, these are indeed on the to-do list!
hey if you have a boosted engine,each piston gets same boost pressure? normal intake manifold normal pistons.Same with single fogger nitro system.Sorry for the bad English!
It took me 20 minutes just sitting here after the video ended to figure out what on earth you were talking about with piston speed being different in the top 180 degrees of the arc than the bottom 180 degrees. It seems like it's all the same distance and the same shape going the same direction, so it should be the same speed in each half.
But, circles aren't squares. Am I right about this notion? If you draw a diamond inside the circle (or a square standing on a corner, if you prefer), you get straight paths from 0-90-180-240 degrees on the circle. If the connecting rod followed those straight lines, speed would be constant. But because it's following a circular path around those four points, the speed (acceleration?) changes with the circle in relation to the flat plane described by the diamond shape. So the highest piston speed would actually be at 45 degrees and 285 degrees, presuming 0 degrees is TDC. ...Is that right...?
I think you're very confused on the matter. The piston velocity cannot ever be constant because the piston is reciprocating. The velocity of the piston is constantly changing through the cycle and consequently the piston is always accelerating.
thank you! cant wait to see what it takes to ballance out my i4
secondary force for every 45 degree or there about window of change of forces between crank path and piston path which explains the green line forces
Shouldn't the slow areas be at 0 and 180 and the fast at 90 and 270? Seems more intuitive than looking at it from the "bottom up". After all the same thing is happening if you take your measurements from the "top down". It's really just the difference in the relative mass of a constant speed counter weight and a variable speed piston that causes this right?
great explanation!! just one question though. where did you get the proportions to your model of 'diameter 1, con-rod 2' ? are these dimensions proportional to most engines? i mean, do all engines' con-rods travel faster from 0(deg) to 180(deg) BECAUSE of these dimensions? or do the con-rods travel this distance faster due to the power stroke
I made up the dimensions so the math would be simple and easy to understand, the proportions are not representative of the real world (at least, they have no basis from the real world). The speeds could be faster or slower depending on the setup.
Engineering Explained you are awesome! Thanks.i definitely want to know more about real applications where it can be broken down to the measurements. Thanks
interestingly, Smokey Yunick did an article about con rod length/stroke ratios.
It was about 45 yrs ago, but I think it was about keeping the length as short as possible but that increased the cylinder wall loading, increasing the chances of ventilating the block.
The longer the connecting rod, the less the secondary force, right ?
I assume :
If the connecting rod is infinite long, x will be almost equal to 2,
and then no secondary force. Is that right ?
How does the engineer determine the ratio of (radius of crankshaft/connecting rod) ?
You're a genius!
Please explained crankshaft torsional vibration in three engine types: flat 6 , V6 and Inline 6
Hopefully my future videos will have what you're looking for!
Engineering Explained And also for Boxer engine, please!!!
The difference in acceleration is explained based on the different "heights" from the bottom of the of crank circle to the top of the piston. I just can't understand why do we have to take into account this virtual height (virtual because it doesn't correspond to any actual physical component or assembly).
Since all the parts involved have a fixed length and are not deformable, the path travelled by the rod has to be constant and the sum "rod+crank radius" is always the same value!
Yup, just want to know the real impact on removing the balance shafts!
BTW how's ur teggy? Engineering Explained
It's a champ!
U say that u have a stronger force upwards and a weaker force downwards but is it actually not fully true because it is like that if you look one pair of the opposite 90 degree travelings but if you look at the other 2 travelings it's actually vice versa (i mean the downward force is stronger and the upper is weaker)? Otherwise nice vid once again :) ! Keep it up.
Does gravity play any role in this? We have seen some inverted airplane piston engines in the past, could this be one reason for those applications?
This all makes sense, but I think the way you explain it is a little off. The "forces" themselves occur once per crank revolution. The forces can be calculated by converting the equation shown here (en.wikipedia.org/wiki/Engine_balance#Secondary_.28non-sinusoidal.29_balance) into the time domain and taking the second derivative of x with respect to time. If you perform a fourier sine series expansion on the resulting equation, you would find that terms 3, 4, 5... are negligible. In such an expansion, each term is a higher order than the last (oscillates twice as fast as the fundamental, then three times as fast, four times, etc.). Therefore, you can say that the piston inertial forces can be APPROXIMATED by two sine waves added together. Because the second term in the series, the second sine wave, is second order one could say it "occurs" twice per crank revolution. The thing to take away is that the second order wave is JUST A MATHEMATICAL CONSTRUCT. The actual forces on the piston cyclically occur ONCE per crank revolution.
Can you make a video regarding floating valve spring?
Why are balance shafts not used on 2 stroke gasoline engines?
Most 2 strokes are relatively small displacement so the piston mass and stroke length are small producing lower imbalance forces.
could you please explain how the upward and downward forces happen twice per revolution? it looks like it is the same as the piston (one downward force and one upward force) per revolution. i understand that the upward force is greater than the downward force, but i dont see how that is occurring 2x more frequently than the primary force of the piston. thanks!
EDIT: for instance, why is there a positive secondary force at 180?
***** could you please explain more. what exactly occurs at 90 and 270? it looks like a downward force changes to an upward force at 180 to me.
Bear in mind that there is a deceleration just prior to TDC and an equal acceleration just after - the point is that the deceleration and acceleration prior to and after BDC are weaker. What this means is that, per crankshaft rotation, you have strong deceleration prior to TDC, strong acceleration afterwards, then weaker deceleration towards BDC and weaker deceleration after, hence four secondary moments per crankshaft rotation. His diagram to the upper right corner is confusing as it makes it seem as though there are two halves of the rotation during which the velocity of the piston is constant, where, in fact, what it shows is that the pairs of decelerating and accelerating moments are not identical, two being rapid and two slower.
the secondary forces occur twice per revolution because of the irregular rotation of the crank, when the piston is on the top dead center to half way to bottom dead center the crank will rotates faster and then decelerate halfway down and half way upward. Due to the changes of rotational speeds it creates vibrations. So the movement of the crank from 0 to 90 degrees is fast, 90 to 180 degrees is fast, then 180 to 270 degrees is slow and 270 to 360 degrees is slow. So, it occurs twice per revolution, half is fast and half is slow and due to this it produce vibrations.
why secondary forces are 2x per revolution? why from 0deg to 90deg secondary force curve goes down, and from 90deg to 180deg curve goes up?
Purely mathematically rod length =1/2 stroke for complete secondary balance. Too bad it is practically unrealistic
1:54 It´s not two because the rod gets drawn *out* (of vertical) on the first half, which makes cylinder go a longer way. If you would just pull the rod to the side the cylinder would also move down, that´s where the additional way is coming from.
On the second half it´s the other way around, here the rod gets drawn in (to vertical). If you would just pull the rod inwards, the cylinder would go up, that´s the effect that makes the second way shorter.
So the additional travel of the piston from the horizontal movement during TDC to 90 is downward and adds to the downward vertical travel of the piston from the top of the crankshaft height to the center axis height of the crankshaft, and from 90 to BDC this same additional travel of the piston from the horizontal movement during 90 to BDC is upward and subtracts from the downward vertical travel of the piston from the center axis height of the crankshaft to the bottom of the crankshaft height. Consider the movement of the piston by moving the base of the connecting rod to the side and then back again to the original position to see how the piston moves and why this first adds to the travel during the TDC to 90 and then subtracts from the travel during the 90 to BDC.
I have a question, well actually I have two!
1) Doesn't the second upper quarter, which also travels fast, counter the first upper quarter? If the first upper quarter is travelling up and over, the second upper quarter is coming over and down with equal speed. If the two fast quarters were on one side (ie the upstroke) that would make sense, but the diagram to me makes it appear like there would be unequal sideways force, rather than vertical. Can you explain this please? Thank you.
2) I'm also a little confused about the issue re: Pythagorean. What equation do you get for BDC>270 triangle?
If You use a displaced knob in crankshaft? Yo counterarrest the exceess of lateral displacement?
Your Math is off X=2-.4365. or X=1.5635. so half of the dots are in the wrong place in the 2nd diagram. I think you are giving a cascading error. This is a better explanation of the problem than Philip H. Smith in "The Design and Tuning of Competition Engines" but it is not all there. What are the books you read, that gave you the ideas you present here on U-Tube?
My math is correct. The height is 2.4365 (as you've mentioned, and as I've written on the white board). It's the pythagorean theorem.
The Con rod only has a one inch throw. You may have done the Pythagorean theorem correctly as a math operation. But there is no room for the 2.4365 inches of travel with the one inch throw crank that you put in the drawing. Pride goes before the fall.
James Thornton He said 2.4365" is the height of the piston pin at 90° and 270°, not the travel. The piston travels 1.1306" up+down from 270° over TDC to 90°, but travels only 0.8730" down+up from 90° under BDC to 270°. That means the piston travels faster through TDC and slower through BDC if the crank is rotating at a constant speed.
Engineering Explained What you described in the video is not a secondary force, it is simply a non-sinusoidal primary force due to the geometry.
A secondary (also second order) force is caused from the reciprocating components, meaning the crank pin and a function of the connecting rod. Not only are you lifting it up and down in a sinusoidal wave, but laying a rod at an angle then straightening it up is also adding a force from the center of mass rising and falling. At TDC and BDC the inertia from the rod is pulling up on the crank but at 90 and 270 the rod is pushing down on the crank.
To help illustrate this, forget about the crank causing an up/down motion and think about your pen on the table and make this motion with your pen: _ / | \ _
and let the cap rise up in a line and keep the tail end slide on the table.
When you straighten it this causes and upwards momentum. Do it fast enough and the pen will want to keep moving up because the center of mass has moved upwards, and you'd have to pull it back down. This is what is happening to the connecting rod. This is what a secondary force is, and it happens twice per revolution.
Also, you should talk more about the vibrations due to the side-to-side motion of the connecting rods as this has significant effects on all engine configurations.
Blaylock1988 It is a second order being described *very well. The piston motion of a crank slider mechanism is made up of an infinite harmonic. It is usually enough to consider only the 1st and 2nd component of this in terms of engine balancing.
Certain engine designs have an advantage in balancing. For example my favorite engine, the straight six with a star configuration crank, is inherently balanced up until the 6th order. Whereas all I-4 engines are not inherently balanced at 2nd, 4th, 6th, etc. So those need to be balanced using other methods.
I can’t quite get my head around where the double frequency comes from. Each revolution, the piston is fast once and slow once. Wouldn’t this create a single peak and a single valley? Where does the double frequency come from? I think it’s the fact that from 0-90 the piston is accelerated by the crank, from 90-180 the piston is slowed the crank, from 180-270 the piston is accelerated by the crank and from 270-360 the piston is slowed by the crank. But this has nothing to do with the different distance travelled by the piston between 0-180 and 180-360. I’ve googled around but can’t find it.
Or is the green line simply the result of subtracting the blue line (theroretical forces with infinite connecting rod) from the red line (actual forces)? I think this is true regardless, but I wonder if there is a different explanation around as to where the double frequency comes from.
something going wrong with the equation at 2:00, X should equal 1.9365 and add it to 0.5 (half the length of the circle) to get 2.4365, am i right ??
Abdelrahman Yakout yes it is
Does air resistance due to intake & exhaust cycles have any influence on this balancing dynamic?
Does alloy of this crankshaft bearings have any influence on engine life, like aluminium vs. copper based alloy, & different alloys, & so on.
Air - perhaps, but ultimately negligible. Material selection is always important, yes.
Great video .but i've a question : how to make the engine rotate in the same one direction and not in the opposite direction
The electric starter motor turns in one direction beginning the engine's rotation. The issue is fuel, as fuel is usually mixed with air before it enters the combustion chamber. With direct injection, it's theoretically possible, but would be difficult for four strokes. Much easier to do with two strokes because there is less work to do in a cycle. Detroit diesels were known to sometimes reverse direction.
Authur Jackson thanks
inertia forces depend of cosx and cos2x for first and secondary forces. that we have this graphic presentation. ;)
whats the effect of secondary and primary forces in a firing order of a engine?
help plz
the piston traveled from 0 to 90 degree (from tdc) is larger when compared to 90 to 180 degree (bdc), now at what angle will the piston travel be equal in length?
+Avinash T You have to know the ratio of the rod length to stroke for the engine in question (the example given of 2:1 will probably be "close", it's 1.85:1 in my engine). Draw the triangle (as above) using trig, also substitute the correct number for x (rod length ratio + 0.5) and solve for the angle. FYI: for oblique triangles the formula c^2 = a^2 + b^2 (right triangle) becomes the more general c^2 = a^2 + b^2 - 2(ab)cosC, where C is the angle you seek.
How would you get a scholarship in engineering?
please answer this !!
apply for one.
Punishing people with an innate skill because of their family income. You've a long way to come to civilized society yet.
wat now
We can think of secundary forces as thouse caused by increasing and decreasing the instantaneous rotational speed during each revolution..
I think the concept is still true but using your same geometry I think it would be better to measure from the top of the piston and say that the piston moves 1.1 in the top half of rotation and only 1 in the bottom half.
can you please tell me the top 3 most balanced engines ?Im guessing it goes inline6 then crossplane v8 from some of your other videos.
+trebombs4life I6, V12, H6. V12 will have the smoothest power delivery of those listed.
+trebombs4life Although it is seldom used, the V16 is perfectly balanced regardless of the V angle. It is probably the smoothest engine ever produced.