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Why would you raise roll center?What happens if you put longer upper ball joints in and leave the lowers alone?What happens if you put longer lowers in and leave the tops alone?Does both of these raise the roll center?Also what if you put longer ones in top and bottom?Just trying to wrap my head around front end geometry stuff! Thanks guys
The topic of roll centers is a godsend to automotive writers. It creates endless controversy, and we'll probably be writing articles about it forever. I've written quite a few, and even produced a video that's mostly about the subject, which is still available. Here, I will assume that the reader is at least a little familiar with the concepts of front view projected control arms, front view instant centers, and front view force lines (the lines from contact patch centers to instant centers, whose intersection is conventionally taken as the "kinematic roll center"). Quick answer to the question of whether roll centers matter is: yes, but most people don't define them properly, and many don't have a very clear understanding of what we mean by a roll center, and what we use it for when analyzing car behavior. Slightly longer answer: actually, it's the anti-roll or pro-roll properties of the suspension linkage that matter, and also its anti-pitch or pro-pitch properties. It is potentially useful to describe these properties in terms of roll and pitch centers, but only if those accurately reflect or predict the actual moments generated by the wheel pairs at issue. Taking the force line intersection, or the height of that point, as the roll center, does not do that, except when the force line slopes are equal and opposite – i.e. when they cross in the middle of the car. The force line slopes matter, and the ground plane forces matter, and the moments generated as a result matter. But you can't analyze the effects by looking at where the force lines cross each other. It is entirely possible to dispense with the whole concept of roll centers, and some think that's the best approach. However, most people still use the concept. Use it for what? Properly, as a shorthand, easy-to-visualize way of expressing the relationship between sprung mass lateral inertia force at one end of the car, and the anti-roll moment induced in the suspension linkage in response to that lateral inertia force. It is a notional coupling point between the notional two-wheel suspension system and the sprung mass, for lateral forces: an imaginary roller in a vertical slot. Since the roller doesn't transmit vertical forces, its lateral location doesn't matter, and can just as well be considered undefined. But its height matters. Its height, Hrc, is correctly assigned in terms of predicting geometric anti-roll moment MxG when the sprung mass lateral inertia force FySM, times Hrc, equals a moment equal and opposite to MxG: Hrc (FySM) = – MxG (1a) This can also be written: Hrc (FySM) + MxG = 0 (1b) The second form is sort of an algebraic expression of the way a roll center is described in SAE J670e: a point at which we can exert a lateral force on the car and no roll will result. The SAE definition is somewhat problematic, but at least its current version includes a caveat that it is not to be construed as a true instantaneous center of rotation. With independent suspension, geometric anti-roll comes from positive or negative support forces induced in the linkage upon application of transverse (y axis) force at the tire contact patches. These support forces, FzGR and FzGL, depend on the magnitude of the ground-plane force and the slope of the force line, and nothing else. The moment created by these forces is the difference between them, times half the distance between them – the distance between them being the track width, Ly. MxG = (FzGR – FzGL) Ly/2 (2) Hrc then has to be the value that makes the moments cancel. Substituting the right side of Equation (2) into Equation (1a): Hrc (FySM) = – (FzGR – FzGL) Ly/2 (3a)Or: Hrc = – (FzGR – FzGL) Ly / (2FySM) (3b)Or: Hrc = – (FzGR/FySM – FzGL/FySM) Ly / 2 (3c) Note that this implies that the roll center height depends only on the relationship of the right and left induced support forces to the total lateral force, and the track width. The sprung mass inertia force FySM is equal and opposite to the sum of the opposing transverse forces at the two contact patches, FySMR and FySML. FySM = FySMR + FySML (4) Finally, the induced support forces and the lateral ground-plane forces at each wheel are in the same ratio as the rise and run of the force line. If the angle of elevation of the force line is θ, that ratio is tanθ. So for right and left wheels, we have tanθR and tanθL. FzGR = FySMR tanθR (5a) FzGL = FySML tanθL (5b) Substituting into Equation (3c): Hrc = – (FySMR tanθR /FySM – FySML tanθL /FySM) Ly / 2 (6a)Or: Hrc = – ((FySMR/FySM) tanθR – (FySML/FySM) tanθL) Ly / 2 (6b)Or: Hrc = ((FySML/FySM) tanθL – (FySMR/FySM) tanθR) Ly / 2 (6c) FySMR/FySM and FySML/FySM are the portions of total transverse force for the right and left wheels, respectively. Graphically, we can find Hrc in a front view by:1. constructing a vertical line at a distance from the right wheel equal to that wheel's portion of the total lateral force times the track – I call this the resolution line;2. finding the intercepts of the right and left wheel force lines with that resolution line;3. averaging the heights of those intercepts. We can also find the total jacking force for the wheel pair by adding the support forces: FzG = FzGR + FzGL (7) In roughly symmetrical independent suspensions, it does hold that more geometric roll resistance implies more upward jacking. However, it is actually possible to have an independent suspension with a roll center above ground (net geometric anti-roll) and downward jacking! This occurs when the majority of the anti-roll force comes from the inside wheel, despite that wheel having lesser ground-plane force. That is possible when the inside wheel has a lot of anti-roll geometry and the outside one has little, or slight negative, anti-roll geometry. This can occur in a strut suspension in a rolled condition. It is also possible with beam axle suspension to add upward jacking while subtracting geometric roll resistance, and vice versa. This happens in a NASCAR-style stock car when we lower just the right end of the Panhard bar. The rear jacks up in a left turn, yet geometric load transfer is less than if we raised the right end of the bar to make it level. Conversely, if we raise the right end of the bar above level, leaving the left end unchanged, we increase roll center height and geometric load transfer, yet the rear jacks down. It is quite possible to have a force line intersection outside the car, and toward either the inside or the outside of the turn, and have net pro-roll, or a roll center below ground. It is also possible to have a force line intersection below ground and outside the car toward either the inside or outside of the turn, and have net anti-roll, or a roll center above ground. This does not imply that the geometry doesn't matter. It implies that the force line intersection cannot be taken as the roll center. Some thinkers note that the jacking force can itself induce roll moments, in the presence of off-center sprung mass c.g.'s or spring splits, and try to come up with ways of assigning roll center height that include spring split and/or c.g. location. I do agree that jacking-induced roll and pitch moments are real, and in some cases big enough to matter. However, they act on the entire mostly-rigid sprung mass, as does elastic roll resistance. They cannot be analyzed in isolation for the front and rear, or right and left in the case of pitch. Therefore it is not analytically productive to try to incorporate them in calculation of geometric properties of the front or rear, or right or left wheel pair suspensions. Is it possible to spring the car so stiffly that roll centers don't matter, or don't matter much? Yes. But it has to be really stiff – stiff enough so that it essentially has no suspension except the tires. Short of the point where we produce a big go-kart, we do not reduce the geometric component of the load transfer by reducing roll. We reduce roll displacement, but not elastic or geometric roll moment. It is also possible to spring just one end of the car stiffly enough so that geometric anti-roll doesn't matter much at that end. In that case, roll center height at the other end becomes very important. An instance of that would be a stock car with a coil-bind or stiff bump rubber setup at the front. Once the front is bottomed, there isn't going to be much further elastic load transfer at the rear, because the car can't roll much, but the rear will still have increasing geometric load transfer if lateral load increases, and that will be determined by the Panhard bar height.Used with permission from the author Mark Ortiz
After looking at the drawing that let r eat posted and thinking in my head longer ball joints top or bottom will produce more angle in the upper a arm and that will raise roll center. Is this correct?
Yes, because it makes it intersect quicker and at a higher point you can tune your roll centers by using one side or the other. Lower Roll center should help the front cut better. Around 2" is a good start, I move my lower pickup points on metric cars, because they are longer and dont effect camber as much depends on rules.
It needs to be measured in the dynamic state, not steady state, as there is a huge difference between the two.
Also, even though you can physically lay it out on the ground, it is much easier using computer software.
If you have a plumbob (sp) and some chalk, it will make things a whole lot easier.
Generally if you get your static right and your camber curves where you want them you will be close but you need to know dynamically where your going BUT that being said its hard to do dirt dynamically because the cars dive and roll and that migrates your roll center differently. You would have to know your roll amount and shock travels in dive and roll and at what points through your corner to know exactly where you are going to its harder then asphalt where you don't really setup for roll anymore just dive. Really your upper a arms get your camber curves and those also control how quick and far and when your roll centers migrate. Learn to tune your Roll Center based on your needs not what it says on paper get your baseline and tune to what the car wants and know how your moving it. sometimes putting yourself in a box does more harm then good
Ok so does putting taller ball joints in the bottom raise roll center or shorter? I would think taller
Any longer ball joint, top or bottom, raises roll center.
Ok so if I shorten let's say the top is it the same effect as shortening the bottom. Yes it lowers roll center but is it the same handling adjustment or will the top effect the car differently then the bottom? I ask because mine currently has the tall lowers on a dirt works dw7. It doesn't seem to get over as far as a lot of people on the rf
To get the car to roll more, you could shorten the right rights and/or lengthen the lefts, which basically softens the rf by lengthening the moment arm.
I would first consult the chassis builder to make sure you have the correct front end parts and setup first.
Lengthening the bottom will require you to lower ride height to have the same angle in your lower control arm. Lengthening the top ball joint does not. And if you are lowering the ride height, the car won't want to roll over as much.Quote:
Originally Posted by Duckhnter83
Are all threaded ball joints the same size thread? Could someone recommend a shorter threaded ball joint maybe 1/4 to 1/2 in shorter then the k772
https://www.lefthanderchassis.com/v2...dgroup=2512660Quote:
Originally Posted by Duckhnter83
I believe you have that backwards correct me if I'm wrong. If you go to taller bottom ball joints you will have to raise your ride heights to get the same original ride height. Because the spindle mounts above the lower a arm.Quote:
Originally Posted by speedbuggy
We're probably saying the same thing, but you're referencing ride height and I'm referencing Lower Control Arm angle.Quote:
Originally Posted by Duckhnter83
If all you did was put in longer lower ball joints and nothing else, ride height, measured from the frame to the ground, will be lower. If you were to raise the car back to its original ride height, the LCA angle will be different, which might be good or bad, depending on what your original angle was.
Most recommendations I see suggest that the inner LCA pivot point and the ball joint pivot point are level with each other to create a smooth camber gain curve.
Ok yes we are speaking the same. My durtworks currently has the 1" raised from stock lower ball joints and at ride heights the lower control arm is level. My uppers though are at around 15 and 17 degrees. I was thinking of bringing them down around 10 degrees. What is a good angle for upper control arms?