gt1guy
Apparently a racist
That makes sense on both counts.
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Linked Suspensions Bible
That makes sense on both counts.
93blackxj
Well-known member
That's not all engine torque. Right front raised as well. If it was all engine torque right side would compress and left would raise. Since both front shocks extend there is obviously a pitching torque acting on the chassis transmitted through the links. But I do agree a good bit is from the driveline.
93blackxj
Well-known member
What surprised you about the results?
Can you explain the "link torque about chassis"? If 0 torque is no squatting, wouldn't suspension 1 be creating less clockwise torque about CG than 2? Link vertical force at chassis makes sense. Seems like to me torque would be less on suspension 1 since it has less vertical force and squatting more under accel since it is a much lower AS.
Edit: Is the torque value the "reaction" torque from the ground pushing on the tire? I guess this is what you meant in this statement.
And since suspension 1 is squatting more, since some of that load transfer is going through the springs? Is this basically saying that lower AS transfers more force to the ground?
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Treefrog
Book Wheeler
That S2 with AS over 100, based on the torque, still wants to squat.
"Link torque about chassis" is the amount of torque about the sprung CG caused by the forces from the links.
From anti squat values, S1 will squat more. So I would expect more torque on the sprung mass. Which is what the numbers show.
Vertical force is only part of it. There is also the horizontal forces. The math I'm using is perpendicular distance from the link's extended line.
93blackxj
Well-known member
I understand this but I think I am misunderstanding your sign convention. To me, the upper tension force is the negative moment (Clockwise) and lower compression force is positive (CCW). Making the total moment on the CG counter clockwise. I did the numbers myself, and either way you come up with the same value. It just confused me for a minute.
What about the IC location? Suspension 1 IC is located way out front of the CG with 305LB vertical force. Suspension 2 IC is behind the CG with vertical force of 725lbs. That is about the only explanation that I see for suspension 2 to lift
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Treefrog
Book Wheeler
I uploaded the spreadsheet so people can experiment for themselves. I can upload so pictures of high anti with an IC out front later.
93blackxj
Well-known member
Ive played around with it. Something definitely isn't adding up to what I would expect
Treefrog
Book Wheeler
Edit now that I'm not outside hoping to see a tornado: I was not expecting for the torque to support squating at higher of AS values.
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Treefrog
Book Wheeler
TLDR: We can't find the 3d IC the same way we found the 2d one. So we have to look backtrack on the path to getting the 2d IC for a method of finding the 3d IC. Jump to after the line for some discussion on why it is relevant and what it means.
Hopefully, anyone who has made it this far in the bible understands what the IC is. To expand the IC, there is two ways to view it, geometrically and on a body in motion. The first one looks at how the system moves while the second looks at how it is moving. And per the physics definition of IC, it only exists as a point if the motion is planar. If the motion is not planar, it is an instantaneous rotational axis. A body in motion is rotating and sliding along this axis at the point in time. The equations for the instantaneous rotational axis are based around a body in motion. A 2d IC is fairly easy to find. The roll axis not so much. And we want a geometric point about which the axle (or chassis) is orbiting not a line of rotation.
So how do we find this point? Do to so, I'll describe how we find find the 2d IC, a value that is defined by motion, but is often solved statically. To do so, lets take a trip back to Calculus 1. In particular, how we the find the solve the slope of a curve. The slope of a curve can be defined by the rise over run of two points on this curve. As the X distance between these points decreases, the slope gets more accurate. It should be most accurate when the difference in X is 0. But that means we have to divide the rise by a run of zero. So, what we do is use an infinitely small, but not zero, x difference.
Now, lets go back to the static 2d IC. We can apply this same concept to finding an IC. Using the side view of our suspension as an example, we move it up slightly and down slightly and take note of a the location of a point on the axle after movement. These two points and the location of the point at the static location can be used to draw a circle. The center of the circle, is the IC. As the movements get smaller, the IC becomes more accurate. To do this without moving the suspension (run = 0 in the curve example), we take two points that move in a known way. In our case, a circle. With an infinitely small movement on a circle we move in a straight line, the tangent. We also have a line, from our point to the circle center. In an instantaneous image, the tangent applies to any circle with it's center on that line. With 2 lines, the intersection is a point that we can draw 2 circles that meet the known tangents. This intersection is the IC.
Now lets apply this to our 3d system. We move the axle around a little bit and get the center of orbit. In the previous example, we were able to draw a circle. But in 3d space we move about a sphere. Much like a 2d circle can be defined by 3 points, a sphere can be defined by 4. So we can move the axle a bit. One wheel up, then down, and the other wheel up and down. This works with any arbitrary point. The smaller our movements, the better the result.
But what if, like 2d, we take it at points of known movement. We should be able to find it without moving anything. And it just so happens that a linked suspension needs 4 links to be considered constrained. And those 4 links each have a point that moves with the axle. But this does not work. Because the 4 lines never intersect, we cannot find common point of movement for the known points.
So we go back to the slight movements and finding a sphere. This is not too hard. From our 4 points, we make 2 different groups of 3 points and make 2 circles with them. We then take make a line normal to the circles through their centers. Where the two lines intersect is the center of our sphere.
Just like we can see how the suspension moves around the chassis, we can see how the chassis moves around the axle. I wonder if we can treat this like the side view IC and draw lines from the contact patches to figure out a 3d equivalent to anti values. I think that looking at the how the chassis orbits around the axle is likely more important than looking at the inverse since our axles are fixed to the ground. Much like the 2d IC, it is probably a good idea to keep it from switching which end of the vehicle it is on. It seems like keeping the chassis' center of movement as small as possible will help keep the vehicle predictable.
Hopefully, anyone who has made it this far in the bible understands what the IC is. To expand the IC, there is two ways to view it, geometrically and on a body in motion. The first one looks at how the system moves while the second looks at how it is moving. And per the physics definition of IC, it only exists as a point if the motion is planar. If the motion is not planar, it is an instantaneous rotational axis. A body in motion is rotating and sliding along this axis at the point in time. The equations for the instantaneous rotational axis are based around a body in motion. A 2d IC is fairly easy to find. The roll axis not so much. And we want a geometric point about which the axle (or chassis) is orbiting not a line of rotation.
So how do we find this point? Do to so, I'll describe how we find find the 2d IC, a value that is defined by motion, but is often solved statically. To do so, lets take a trip back to Calculus 1. In particular, how we the find the solve the slope of a curve. The slope of a curve can be defined by the rise over run of two points on this curve. As the X distance between these points decreases, the slope gets more accurate. It should be most accurate when the difference in X is 0. But that means we have to divide the rise by a run of zero. So, what we do is use an infinitely small, but not zero, x difference.
Now, lets go back to the static 2d IC. We can apply this same concept to finding an IC. Using the side view of our suspension as an example, we move it up slightly and down slightly and take note of a the location of a point on the axle after movement. These two points and the location of the point at the static location can be used to draw a circle. The center of the circle, is the IC. As the movements get smaller, the IC becomes more accurate. To do this without moving the suspension (run = 0 in the curve example), we take two points that move in a known way. In our case, a circle. With an infinitely small movement on a circle we move in a straight line, the tangent. We also have a line, from our point to the circle center. In an instantaneous image, the tangent applies to any circle with it's center on that line. With 2 lines, the intersection is a point that we can draw 2 circles that meet the known tangents. This intersection is the IC.
Now lets apply this to our 3d system. We move the axle around a little bit and get the center of orbit. In the previous example, we were able to draw a circle. But in 3d space we move about a sphere. Much like a 2d circle can be defined by 3 points, a sphere can be defined by 4. So we can move the axle a bit. One wheel up, then down, and the other wheel up and down. This works with any arbitrary point. The smaller our movements, the better the result.
But what if, like 2d, we take it at points of known movement. We should be able to find it without moving anything. And it just so happens that a linked suspension needs 4 links to be considered constrained. And those 4 links each have a point that moves with the axle. But this does not work. Because the 4 lines never intersect, we cannot find common point of movement for the known points.
So we go back to the slight movements and finding a sphere. This is not too hard. From our 4 points, we make 2 different groups of 3 points and make 2 circles with them. We then take make a line normal to the circles through their centers. Where the two lines intersect is the center of our sphere.
Just like we can see how the suspension moves around the chassis, we can see how the chassis moves around the axle. I wonder if we can treat this like the side view IC and draw lines from the contact patches to figure out a 3d equivalent to anti values. I think that looking at the how the chassis orbits around the axle is likely more important than looking at the inverse since our axles are fixed to the ground. Much like the 2d IC, it is probably a good idea to keep it from switching which end of the vehicle it is on. It seems like keeping the chassis' center of movement as small as possible will help keep the vehicle predictable.
Treefrog
Book Wheeler
As I play around with the code some more, I am finding that which small movements you pick matter a lot. The results for moving the wheels up and down while holding the opposite one steady greatly varies from moving both wheels up and down and moving the wheels in opposite directions. Also, the 2d method of finding the IC should apply to even suspension movement on a symmetric suspension, but it does not seem to line up.
Treefrog
Book Wheeler
So, I should probably share some pictures of what the movement centers look like. So here is a standard front and rear triangulated 4 link. I am pretty sure that the geometry is the default in the 4 link calc. I think the takeaway here is the variety of shapes of surfaces that the axle orbits about. In case the text is too small, yellow is the front.
And here is what the chassis's movement looks like to the axle:
I still do not really understand why the 2D IC and movement centers did not agree. Maybe it has to do the axle not twisting about the 2d IC.
And here is what the chassis's movement looks like to the axle:
I still do not really understand why the 2D IC and movement centers did not agree. Maybe it has to do the axle not twisting about the 2d IC.
Treefrog
Book Wheeler
While rereading this thread wondering if there was anything I had forgotten, learned more about, or had learned enough to figure out, I came across the discussion a few of us had a bit over three years ago regarding portals and antisquat. At the time we had reached the conclusion that portals changed antisquat values, and that the offsetting of the axle tube has an affect. And that was about it.
Even though I barely believe in antisquat anymore, it is still an easy reference value. So I figured this would be a good one to revisit. I wanted to find the antisquat drawing for something with portals. In particular solid axles, though an independent suspension drawing should be easy to find as a offshoot. If we have the drawing, we can get antisquat.
It has occurred to me that the offset does not play any special roll. That is, the links being higher up does not change how we draw things.
Portals change where we start the line that goes to the IC.
It struck me that the key was in looking at why we start the anti line at the wheel hub for independent acceleration and at the contact point for independent braking and solid axle. But we are not going to the wheel hub or contact point, we are going to where the force is applied. But why is the force at the contact patch when reacting to torque? It is because that torque the housing or knuckle reacts to is the acceleration force times a distance from the wheel center. In the case of a ring and pinion gear or brakes, that distance is the tire radius. For an independent suspension the force still exists, but there is no torque. So the distance must be 0.
So we should be able to apply this force and distance approach to portals. The force is fixed in magnitude and direction. Therefore we can only control the distance.
When the gearboxes input and output shafts rotate opposite directions, the housing torque is the sum of the magnitudes in the direction of the output shaft's torque. If they rotate the same direction, the reaction torque is the output - input.
The reaction torque around the ring gear is opposite in direction and equal in magnitude to the axle shaft torque. Combine this with the gearbox torque from above, and deal with some sign stuff and we get equations for equivalent distances.
Solid axles:
Even gear portals: d = 1+2/GR
Odd gear portals: d = 1-2/GR
Positive is above the tire.
Independent:
Even gear portals: d = 1+1/GR
Odd gear portals: d = 1-1/GR
Positive is above the tire.
Even though I barely believe in antisquat anymore, it is still an easy reference value. So I figured this would be a good one to revisit. I wanted to find the antisquat drawing for something with portals. In particular solid axles, though an independent suspension drawing should be easy to find as a offshoot. If we have the drawing, we can get antisquat.
It has occurred to me that the offset does not play any special roll. That is, the links being higher up does not change how we draw things.
Portals change where we start the line that goes to the IC.
It struck me that the key was in looking at why we start the anti line at the wheel hub for independent acceleration and at the contact point for independent braking and solid axle. But we are not going to the wheel hub or contact point, we are going to where the force is applied. But why is the force at the contact patch when reacting to torque? It is because that torque the housing or knuckle reacts to is the acceleration force times a distance from the wheel center. In the case of a ring and pinion gear or brakes, that distance is the tire radius. For an independent suspension the force still exists, but there is no torque. So the distance must be 0.
So we should be able to apply this force and distance approach to portals. The force is fixed in magnitude and direction. Therefore we can only control the distance.
When the gearboxes input and output shafts rotate opposite directions, the housing torque is the sum of the magnitudes in the direction of the output shaft's torque. If they rotate the same direction, the reaction torque is the output - input.
The reaction torque around the ring gear is opposite in direction and equal in magnitude to the axle shaft torque. Combine this with the gearbox torque from above, and deal with some sign stuff and we get equations for equivalent distances.
Solid axles:
Even gear portals: d = 1+2/GR
Odd gear portals: d = 1-2/GR
Positive is above the tire.
Independent:
Even gear portals: d = 1+1/GR
Odd gear portals: d = 1-1/GR
Positive is above the tire.
Treefrog
Book Wheeler
In post 125 I looked at to suspensions that give the same 2d results. And the results were rather inconclusive as to the differences. It was a few percent at most. So I ran the suspensions again to see if the orbit centers are different.
Up first is what the rear axle orbits around. The full up travel points are labeled. Normal is both uppers and lower at full width. The other is changed to half width.
And how the chassis orbits around the rear axle. The full up travel points are labeled. Normal is both uppers and lower at full width. The other is changed to half width.
Up first is what the rear axle orbits around. The full up travel points are labeled. Normal is both uppers and lower at full width. The other is changed to half width.
And how the chassis orbits around the rear axle. The full up travel points are labeled. Normal is both uppers and lower at full width. The other is changed to half width.