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Question about double-wishbone suspensions
What's the general configuration of the upper and lower wishbones in a double-wishbone suspension? When the suspension is not compressed are the upper and lower wishbones generally parallel to one another like in the first picture or are they non-parallel like in the second picture?
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They are largely parallel, but the trick is the upper arm is shorter, or the mounting points are arranged sot he upper and lower wishbones travel in a different arc, causing the wheel to gain negative camber as the suspension compresses. As I understand it, it is more how the movement affects it, rather than how it starts out.
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I've been trying to do the calculations to graph the changes in the wheel plane for a double wishbone suspension like in the first picture. Here's what I get for an upper wishbone length of 8, lower wishbone length of 13 and hub length of 7 http://www.freewebs.com/likeoldsportscars/Wheel_Plane.mht . The red plane corresponds to when the suspension is not compressed and the green plane corresponds to when the lower wishbone has rotated upwards 30 degrees. The calcuations for the wheel plane is pretty straight forward in this case but when you consider a double wishbone suspension like in the picture below with the wheel hub connected to the arms with a couple of ball joints the math gets more complicated.
![]() Last edited by nine_one_4; 03-15-2006 at 06:06 PM.. |
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Typically it isn't good to start out parallel. It is even worse to orient angled as shown.
A good starting point would be lower arm parallel to ground, upper arm is shorter and slopes down from the upper ball joint toward the center of the car. Somtimes it can be better to have the lower arm slope upward a bit (though slight). You want to set up your roll centers based on the vehicle cg height. Then watch the motion of the roll centers through articulation to be sure they don't move too far away and create unwanted roll. There are also a lot of other concerns, such as jacking, rear RC height v. front RC height and how that relates to oversteer/understeer, etc. Get the suspension design book by Milliken and Milliken and that would be a good place to start understanding the importance of instance centers and how they interact with creating roll centers. Good luck! I am working on my rear suspension design as we speak (though staying with front struts to minimize the costs at this point - they have worse evils with RC migration). Of course, I rely heavily on my chassis guy since he has the experience - I am just trying to be an "educated consumer" and it does help to know!
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Thanks for your input. I looked at the front suspensions on the F1 cars as they were coming out of the garages for qualifying at Malaysia and the wishbones seem to be as you describe with the lower one being parallel and the upper one angling down a bit. I have a few suspension books bookmarked at Amazon and I'll add the one you mentioned if it's not already bookmarked. I might also pick up Solidworks because somebody said they have been modeling a double wishbone suspension on it and I wouldn't have to do all the math.
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Before going with solidworks, there are a number of different suspension-specific tools out there for reasonable money
here is a good link to a suspension site http://www.eng-tips.com/threadminder.cfm?pid=800 read through some of the threads - there is a particular software that comes up often that is supposed to be good and like $250 IIRC - don't have time to search it out for you right now, but it would be good to go through and read some of these threads too. Here is how archaic I am - I still use graph paper, that thing with the pencil to make circles (I think called a compass) and a protractor. Then I just draw out the extremes to determine where the RC went to and get camber gain/loss. ![]()
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Thanks for the link - I've bookmarked the site. I think a lot of people use this software . I've also found this one but I have no idea how good it is. This is the software I'd really like, though.
Btw, I also used a compass, protractor and paper but mostly to double check if my calculations are valid. I can't graph the results, though, for the case where the wheel hub is connected to the wishbones with balljoints like in this photo. ![]() Last edited by nine_one_4; 03-20-2006 at 07:32 PM.. |
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nine_one_4, Why not? The later pic is the exact same as your original line drawings. The hub just has an offset (and possibly some static camber built in.) Neither of those affect the camber change. Don't draw your red lines out the the hole for the hub, draw them stright from balljoint center to balljoint center. The purple hub could twist up, down, and around 5' but as long as the balljoint c to c stays the same it doesn't matter.
BTW, your upper and lower lines should project tangent to the axis on which the A-arm moves. Not along one leg as you've shown. Your exagerating the length of the arm... SMD
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Ok, so then the hub can not rotate vertically relative to the wishbones if they are held fixed, correct? So the hub must be connected to the wishbones with something other than balljoints. Are those heim joints?
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They are usually ball joints on production cars. But a ball joint is nothing but a heim w/ a tapered bolt though it in single shear. They both have full angular freedom. The point is that the SHAPE of the upright (OR the arms) has nothing to do w/ the free body diagrams. Only where its pivot points/axes are located.
Excuse my poor pic, I don't have any editing programs @ work: ![]() The red upright is equivalent to the green upright. SMD
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I’m not following what you’re saying. Assume that my diagram below represents the right front suspension of a car and that you are looking at it from behind the front wheel. If the car is turning left on a flat surface, then the body of the car will rotate clockwise with respect to the road and because the balljoints allow the hub to have full angular freedom where they connect to the wishbones wouldn’t the resulting forces allow the hub to rotate somewhat about the z axis relative to the wishbones? If that’s the case, then wouldn’t the plane of the wheel would be determined differently than if the hub was mounted to the wishbones with heim joints that only allowed the hub to rotate about the y axis and not about the z axis?
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The hub is supposed to rotate around the Z axis--that's how steering happens. The movement around that axis is controlled by the tie-rods.
--DD
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Dave, you missed this part of my previous post.
"Assume that my diagram below represents the right front suspension of a car and that you are looking at it from behind the front wheel." |
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First, Z is USUALLY the vertical axiz but its just a notation.
Ok, now your involving caster (if the upper and lower wishbone pivots are either not vertical or the arms cause that to happen turning up/down travel). Most people seperate caster and camber curves for simplicity. Instead of the two graphs, you COULD create a full 3d surface describing the changes but I've never seen it done (though top teams like F1 surely do it). Its not as usefull as it appears, becasue bumpsteer would cause you to travel on an curved path across the surface and not purely vertically. Plus, they are largely decoupled to begin with. I think your sort of getting ahead of yourself. The best thing is to go pick up a book that covers this (Phuns "how to make a car handle" is a good primer, and the miliken(sp?) book is the do-all-be-all source). BTW, a heim allows rotation about the bolts axis (though you have to keep the bolt fom loosening). Its nothing more than a spherical bearing "on a stick". SMD
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Sorry, Alfred, I got my axes confused.
The hub should not be able to rotate around your Z axis, because it is located by two fixed points--the ball joints. I'm not sure where extra rotation would come in, unless you are accounting for looseness or 'slop" in the ball joints. Now, if you mean does the hub give you a camber change as the suspension articulates, then the answer is yes. In fact, the unequal-length wishbones are specifically intended to induce a camber change which (in part or in whole) compensates for the "lean" of the mounting points, which attach to the body of the car which does lean in turns. If you had equal-length parallel wishbones, the wheel would always be perfectly parallel to the body of the car. Which is fine until the car leans--toward the outside of the turn! In which case, the outer wheels (the heavily loaded ones) go into positive camber, which is exactly what you don't want! With unequal-length wishbones that are not parallel, you can reduce that amount of camber gain, sometimes getting it exactly to zero or even having it go slightly into negative camber! (Which would be what you wanted for maximum grip, I believe, but is usually impractical on most cars.) Cut out four pieces of cardboard and attach them with thumbtacks. One will represent the body of the car, one the hub carrier, and the other two the upper and lower suspension links. Play with the lengths of the links and the position of the mounts on the body. See what happens when the body leans, or when the wheel goes up and down relative to the body. That can give you a clearer understanding of what is going on. --DD
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I probably didn't explain it well but my last diagram is just a simplified version of the one below (and viewed from the other side of the tire). Also the non-pivot ends of the wishbones in my diagram have the same x coordinate when the arms are parallel. The z axis comes out of the screen and only the lower wishbone rotates about the z axis itself - the upper wishbone rotates about the line that is parallel to the z axis and that passes through the upper wishbone's pivot point.
So, refining what I said earlier, if the car is turning left on a flat surface, then the body of the car will rotate clockwise with respect to the road and because the balljoints allow the hub to have full angular freedom where they connect to the wishbones wouldn’t the resulting forces allow the hub to rotate somewhat about a line parallel to the z axis? If that’s the case, then wouldn’t the plane of the wheel would be determined differently than if the hub was mounted to the wishbones with heim joints that only allowed the hub to rotate about a line parallel to the y axis? ![]() ![]() Last edited by nine_one_4; 03-24-2006 at 08:38 AM.. |
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Quote:
With equal-length wishbones, you will get no rotation relative to the body, which is usually not what you want in a turn. --DD
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Dave, I understand what you're saying and that's what the results of my calculations as graphed here show http://www.freewebs.com/likeoldsportscars/Wheel_Plane.mht .
Actually, after trying to visualize it with your idea of cardboard and thumb tacks I think it may be as simple as what smdubovsky said in his first post above - that the hub would just add some static camber and the wheel plane is otherwise fully determined by the ends of the wishbones. |
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nine_1_4:
W/ regards to your last picture, Yes the plane the wheel IS offset. But, all of the ANGULAR motions it makes are the same no matter what the offset is (making the analysis of camber and toe easier). The offset WILL result in the centerpoint of the wheel moving in the XYZ axes. This will affect things like scrub and weight transfer as the suspension moves. To further complicate things these happen anyway even w/ 0 offset since the wheel has width and usually has caster & static camber - which will casue the contact patch to move around. SMD
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Quote:
Edit: I just found some information here that says that when a double wishbone suspension moves the spindles "...also have a slight side-to-side motion caused by the arc that the wishbones describe around their pivot points. This side-to-side motion is known as scrub. Unless the links are infinitely long the scrub motion is always present." So does that mean that both wishbones would not rotate about axes parallel to the z-axis? |
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