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JD'sTT55
10-16-2011, 09:56 AM
Like the title states I'm debating on how much to offset my 3rd link to the passenger side. I've read that offsetting it helps compensate for torque during acceleration but too much may cause it to do the opposite during braking. I realize there are variables for the exact location but what would be a good window to stay between?
Thanks in advance. Jeremy.

exwestracer
10-16-2011, 10:02 AM
In a road racing application you will almost always see the top link centered in the chassis. The further you offset it, the more difference you will see in handling and acceleration between right and left hand corners. To get the link to make a considerable difference in acceleration would probably have a serious effect on handling.

Randy67
10-16-2011, 05:39 PM
I've seen a couple Trans-Am racing rear suspensions with an offset upper link. Brian Hobaughs (very bad arse car) car has it offset as well: http://www.lateral-g.net/forums/showthread.php4?t=23997 You can see it in post #4 pretty good. I don't think Mike Maier would have done this if it didn't work well.

I have the upper link offset as well, it is just to the left of the housing (not going to try and weld it to cast iron). Once the car is going I will know how it works.

Bryce
10-16-2011, 07:07 PM
In a road racing application you will almost always see the top link centered in the chassis. The further you offset it, the more difference you will see in handling and acceleration between right and left hand corners. To get the link to make a considerable difference in acceleration would probably have a serious effect on handling.

Basically these are my thoughts too and the reason I built a centered 3-link.

The jaguar (C-type?) used a offset upper link and was successful. My intermeccanica omega design by mark donahue and holman moody has an offset 3-link.

exwestracer
10-17-2011, 03:00 AM
Basically these are my thoughts too and the reason I built a centered 3-link.

The jaguar (C-type?) used a offset upper link and was successful. My intermeccanica omega design by mark donahue and holman moody has an offset 3-link.

The Jags broke the top link mount often. May have been a structural issue, though. Sure, offsetting the top link will work, still not the best from a design standpoint...

Norm Peterson
10-18-2011, 04:11 AM
Perhaps fortunately, the "penalty" for offsetting the upper link under braking is less severe than the benefit under acceleration. The effect is based on the amount of longitudinal force that can be developed at the rear wheels, which is going to be at least 55% of vehicle weight during acceleration but only 25% or so under braking. As the tires get grippier, the 55% acceleration term potentially goes up (assuming sufficient power is available) and the 25% braking term would need to go down to avoid overbraking the rear.

With respect to compliance steer effects during either acceleration or braking, I think offsetting the upper link would call for relatively more rigid lower link bushings or (preferably) rod ends.

There is no requirement for an offset upper to cancel 100% of the acceleration torque reaction . . .


Norm

David Pozzi
10-18-2011, 09:55 AM
Near the end of the Trans-Am series, I looked over cars racing at Laguna Seca. Most of them used an offset upper link.

exwestracer
10-18-2011, 02:37 PM
Near the end of the Trans-Am series, I looked over cars racing at Laguna Seca. Most of them used an offset upper link.

Well, tough to argue with that! David, did you happen to notice if they were all pretty much the same offset? I'm wondering if that had to do with the chassis mfr...?

David Pozzi
10-18-2011, 08:04 PM
There was one MFR that built a lot of them. I can't remember the name. I took pics of a 3 link in a Corvette & posted them in this forum several years ago. The front end of the link had a sliding clamp on a roll cage tube. I've also driven a car with the top link offset & it worked well with no adverse effects. The offset was 8" to 10" I would guess.
David

exwestracer
10-19-2011, 05:16 AM
There was one MFR that built a lot of them. I can't remember the name. I took pics of a 3 link in a Corvette & posted them in this forum several years ago. The front end of the link had a sliding clamp on a roll cage tube. I've also driven a car with the top link offset & it worked well with no adverse effects. The offset was 8" to 10" I would guess.
David

Possibly Pratt & Miller. They used a unique method of triangulating the windshield area. Kind of a small pic, but I believe I see the vertical tube where the upper link slider mounted. No arguing with their success...
50377

David Pozzi
10-19-2011, 11:27 AM
Here is the 3 link thread with pics I took: https://www.pro-touring.com/showthread.php?1109-Three-link-suspension-used-in-TA-race-cars

MadBill
02-09-2012, 01:55 PM
Hi. I should probably have taken more time to search before making my first post, but since this thread topic exactly matches my question...

How does one determine the required offset for a 3 link in order to reduce/eliminate the unloading of the right tire under power? My gut feel is that it relates to axle ratio, rear track and angle of the (assumed upper) 3rd link, but...

BTW, this is for a road racer, and since virtually all of our courses are right hand, the resultant unloading of the left front under braking will be offset by trail braking-induced weight transfer.

<EDIT>
OK, I've already found more, e.g. Bryce's 65 Falcon thread, and I realize my problem with Billy Shope's calculator (I too got "NaN"s for results) was likely that I was targeting anti-squat values too low for it. [The input I'm getting is that for high powered cars like ours (2150# dry, 600 HP) is that 10-20% is in the ball park. We've already experienced wheel hop under braking with values ~ 35%] I guess I want to know if it can be calculated based only on the slope and location of the upper arm, irrespective of % A-S.

exwestracer
02-09-2012, 05:36 PM
I can't see why the split would be anything other than proportional to the percentage of track width that you offset the link.

If you had a travel indicator on each rear shock, you could compare the load/unload travels on a good launch to figure out the rear roll moment. The weight transfer could be expressed as a percentage of total rear axle weight and used to figure out how much offset would just counteract it. I think.

MadBill
02-09-2012, 06:18 PM
I S.W.A.G.ed that the way it worked was that the vertical component of the torque-induced load in the 3rd link acts to push down on the right side axle, e.g. if the link was at 30° and the axle torque reaction created say a 2,000 lb. tensile load, the vertical component would be 2000 x sin 30° or 1,000 lb., and so the added load at the right contact patch would be 1000 x (distance from link to left contact patch)/track width. In this scenario, less anti-squat would likely mean a lower angle for the third link and thus more offset required to get the same cancellation force..

BrianP
02-09-2012, 07:46 PM
Billy Shope has a website for calculating linkage loads for 3 links with offset linkages.

go to page 18 at http://www.racetec.cc/shope/

MadBill
02-09-2012, 08:49 PM
Per a post by Billy in the "3 Link for 65 Falcon" thread, that calculator doesn't work for less than 51% A-S and, depending on the resolution re my thread on calculating A-S, that's more than I want to use.

MadBill
02-11-2012, 02:57 PM
Something Ray said a couple of posts back re checking shock travel lit a bulb! :idea:

How about this approach for determining percentage:
1. Set the car on stands with the rear coilovers removed and the wheels resting on weigh scales, simulating static ride height.
2. Lock the brakes to prevent wheel rotation and then with a long wrench or a lever bolted to the pinion flange, apply CW torque to the pinion gear.
3. Note the relative change of L & R scale readings from the unloaded case to determine offset effect and adjust link angle and/or lateral location to achieve desired result or to populate a table for future reference.

exwestracer
02-11-2012, 03:19 PM
I don't think locking the chassis down on stands will give you an accurate result. In effect, you have eliminated the rear roll center and any possibility of torque roll, which is what you are trying to offset in the first place. Interesting concept... I wonder if there is a way to lock the drive wheels tightly enough and SAFELY "brake stand" the car on the scales?

Norm Peterson
02-11-2012, 03:31 PM
I think you'd need a T-handle sort of wrench, so that there are no unbalanced forces being applied to the pinion. Otherwise the force that you apply at the end of whatever you're using for leverage will also contribute to the changes in wheel loads.

I also think you'd want to block the front tires instead of setting the rear brakes. Applying the rear brakes will "short-circuit" the axle torque right back through the axle tubes directly to the pumpkin to resist the pinion's attempt to rotate upward/downward. The torque would remain within the axle assembly, no external link or arm loads would occur from it, and the offset of the upper link or TA wouldn't matter.


Norm

exwestracer
02-11-2012, 04:19 PM
I think you'd need a T-handle sort of wrench, so that there are no unbalanced forces being applied to the pinion. Otherwise the force that you apply at the end of whatever you're using for leverage will also contribute to the changes in wheel loads.

I also think you'd want to block the front tires instead of setting the rear brakes. Applying the rear brakes will "short-circuit" the axle torque right back through the axle tubes directly to the pumpkin to resist the pinion's attempt to rotate upward/downward. The torque would remain within the axle assembly, no external link or arm loads would occur from it, and the offset of the upper link or TA wouldn't matter.


Norm

DOH! Good point!

MadBill
02-11-2012, 06:41 PM
I don't think locking the chassis down on stands will give you an accurate result. In effect, you have eliminated the rear roll center and any possibility of torque roll, which is what you are trying to offset in the first place. Interesting concept... I wonder if there is a way to lock the drive wheels tightly enough and SAFELY "brake stand" the car on the scales?

Hmmm... Well, first the roll center is defined by the Watts link location, which would still be in place (horizontal on the bottom of the diff) and second, I'd be twisting the DS exactly the way the engine does; it's just that the torque reaction would show up relative to the sprung weight of the axle assembly instead of the corner weights.

Good thought re brake-torquing the car! however Norm's right, either way the brakes would cancel out the twisting force.

Plan 'B': Fit A frames with bases of say 3 ft. length in place of the wheels and set the car, springs intact, on two scales per side, one each at the front and rear of each frame base. This way, axle torque would be resisted by vehicle weight and would show up as differential loading on the scales.

Theoretically though, I believe the results should be linear from 0 ft-lb. on up, so the 'Armstrong' method of inputting torque should suffice. It couldn't hurt to use a T bar, but as the torque reaction at the pinion is pretty well centered in the track, the resultant load on the scales should be equal.

exwestracer
02-12-2012, 04:48 AM
Bill,
My point in the earlier post was that the roll torque you are compensating for is coming from the engine block trying to twist in the chassis. Unless you model it with that force applied, you will get A/S; but you won't have the differential spring loading you get on launch.

Norm Peterson
02-12-2012, 05:18 AM
. . . I believe the results should be linear from 0 ft-lb. on up
I agree about the linearity of this.

I doubt that Billy Shope's application or any other similar derivation would work if it wasn't linear, and if you're pushing so much torque at it that things are starting to deform under that load far enough to invalidate the assumptions that you're making, you should probably be mounting the axle solidly to the chassis anyway.



It couldn't hurt to use a T bar, but as the torque reaction at the pinion is pretty well centered in the track, the resultant load on the scales should be equal.
The torque part of it will divide properly, or it will at least divide up the same as it does when the torque comes down the driveshaft.

It's the single pulling force that changes things.

If you're pulling downward, it'll add a downward force at the pinion that isn't there when you're driving, and it will also be divided between the rear tires. Depending on the length of your lever, it would be possible that the scale reading on the RR would remain unchanged while the LR would show about double the change that torque alone would cause.

If you're pulling horizontally, you'll be adding a horizontal force at the pinion, which will resolve exactly like the lateral load transfer from unsprung mass does when you're investigating that. If you're pulling horizontally from below the axle, this effect will work against what you're trying to see. Above the axle and it will add.

Think [YourPullingForce] x [PinionHeightAboveGrade] / [RearTrack].


Any other direction would involve some of each of the above.


Norm

MadBill
02-12-2012, 08:26 AM
Bill,
My point in the earlier post was that the roll torque you are compensating for is coming from the engine block trying to twist in the chassis. Unless you model it with that force applied, you will get A/S; but you won't have the differential spring loading you get on launch.

Good point. The torque reaction in the chassis would be reflected in a change in front wheel loadings, so scales would be required there to determine the exact percentage of cancellation. However, if all that was required was to observe the relative effect, the change in rear wheel loads alone would show this. (On further thought, wouldn't the resultant asymmetric loading just be the mirror image of the rear's?)

Norm Peterson
02-12-2012, 08:37 AM
Now we're getting chassis rigidity and the front vs rear distribution of roll stiffness involved. A perfectly rigid chassis with no front roll stiffness - or a wheelstanding car where all of the engine torque reaction has no choice but to go rearward - will oppose the right to left rear tire load shift that comes from driveshaft torque with exactly the same amount of chassis torque in the opposite direction. In such cases, I don't think that offsetting a 3rd link or TA could gain anything at all.

Ladder bar suspensions, the practice of removing or de-activating front sta-bars, and the large by huge drag race rear "antiroll bars" are efforts intended to approximate the above.


Norm

MadBill
02-12-2012, 08:41 AM
I agree about the linearity of this.

I doubt that Billy Shope's application or any other similar derivation would work if it wasn't linear, and if you're pushing so much torque at it that things are starting to deform under that load far enough to invalidate the assumptions that you're making, you should probably be mounting the axle solidly to the chassis anyway.



The torque part of it will divide properly, or it will at least divide up the same as it does when the torque comes down the driveshaft.

It's the single pulling force that changes things.

If you're pulling downward, it'll add a downward force at the pinion that isn't there when you're driving, and it will also be divided between the rear tires. Depending on the length of your lever, it would be possible that the scale reading on the RR would remain unchanged while the LR would show about double the change that torque alone would cause.

If you're pulling horizontally, you'll be adding a horizontal force at the pinion, which will resolve exactly like the lateral load transfer from unsprung mass does when you're investigating that. If you're pulling horizontally from below the axle, this effect will work against what you're trying to see. Above the axle and it will add.

Think [YourPullingForce] x [PinionHeightAboveGrade] / [RearTrack].


Any other direction would involve some of each of the above.


Norm

Since it is the relative change in L/R loading we're looking for, the pinion downward load from say applying 100# downward force on a 24" bar would also be divided evenly or almost so between the contact patches. To play it safe though, you could use two torque wrenches pulled in opposite directions to cancel each other. Plan 'C': Leave the driveshaft in place, slip a 2 x 4 across the frame rails under it near the front end for support and apply the torque to the front yoke. The 'Torquemaster' stands in the car for the test to ensure no further quibbles re accurately reproducing actual conditions.

exwestracer
02-12-2012, 08:52 AM
Now we're getting chassis rigidity and the front vs rear distribution of roll stiffness involved. A perfectly rigid chassis with no front roll stiffness - or a wheelstanding car where all of the engine torque reaction has no choice but to go rearward - will oppose the right to left rear tire load shift that comes from driveshaft torque with exactly the same amount of chassis torque in the opposite direction. In such cases, I don't think that offsetting a 3rd link or TA could gain anything at all.

Ladder bar suspensions, the practice of removing or de-activating front sta-bars, and the large by huge drag race rear "antiroll bars" are efforts intended to approximate the above.


Norm

If we stand at the front of the car and watch our typical Chevy transmit torque to the axle, the driveshaft is trying to extend the LR and compress the RR, while the BLOCK is trying to extend the LF and compress the RF. I believe those forces are compounding, not offsetting?

MadBill
02-12-2012, 10:02 AM
I think they are a matched set, per Newton's Third Law": "For every action there is an equal and opposite reaction." if we crank up the LR spring say 3 turns, the RF load increases also and the LF and RR decrease.

Slightly off topic, but how much and where does the reaction occur, if for example, the engine is rubber mounted and the transmission is in first gear with say a 3:1 ratio? How about if engine and tranny are hard-mounted? What if we had a 1:1 spur gear set on the tranny output shaft, reversing the driveshaft rotation?

exwestracer
02-12-2012, 11:21 AM
The reaction occurs wherever there is a limit to freedom of motion. The rubber will absorb some of the torque as we know, but that is still where the reaction load is fed into the chassis. The reverse rotation thing was tried for awhile in circle track racing (spinning the ENGINE the opposite way, mostly open wheel cars), with some success. In those light cars, the engine (block) torque reaction did quite a bit to counteract chassis roll in the corners. I believe reverse rotation been banned just about everywhere though...

Norm Peterson
02-12-2012, 12:38 PM
If we stand at the front of the car and watch our typical Chevy transmit torque to the axle, the driveshaft is trying to extend the LR and compress the RR,
In terms of forces, that's adding load to the LR by stealing it from the RR.



while the BLOCK is trying to extend the LF and compress the RF.
Absolutely.

But since the actual input is really a powertrain torque reaction being applied to the chassis (either as couple forces at the engine mounts or as a pure torque on a motor plate), it will divide front to rear in proportion to the relative front and rear suspension roll stiffnesses.

Strictly speaking, the chassis torsional stiffnesses between the engine mounts and the front suspension, vs that between the engine mounts and the rear suspension, should also be included, though their effect would be minor unless somewhere along the chassis you have a section that's really soft in torsion.



I believe those forces are compounding, not offsetting?
Continuing with the line of thought regarding this torque as a chassis torsion to be distributed according to where the stiffnesses are, not all of the powertrain torque reaction is taken by the front suspension, not even if the engine mounts were to be located ahead of the front suspension. Driving the RF down relative to the LF eventually becomes driving the RR of the chassis and therefore the RR suspension down relative to the LR and re-planting the RR tire to some extent. This also means that some of the "planting" of the LR from driveshaft torque is being "undone" by the upward movement at the LF dragging the LR of the chassis up with it a bit as well. Overall, this is a good thing since two equally loaded tires will have more total grip (and in this situation will naturally want to drive straight instead of having some yaw moment being developed from unequal forward bite).


Norm

exwestracer
02-12-2012, 01:11 PM
I believe you are correct about the forces, Norm. I was just looking at it in terms of both reactions promoting chassis roll, rather than one resisting the other.

I still think measuring rear travels on a "real" launch would be the most accurate way to predict how far the top link would need to be offset. Now that I think about it some more, I remember an article in Circle Track (maybe) years ago about a stock car that only used a ladder bar offset to the right, and a single link at axle centerline on the left. They were trying it to get some roll steer effect going, but the idea is basically the same.

Norm Peterson
02-13-2012, 05:22 AM
The reaction occurs wherever there is a limit to freedom of motion. The rubber will absorb some of the torque as we know, but that is still where the reaction load is fed into the chassis. The reverse rotation thing was tried for awhile in circle track racing (spinning the ENGINE the opposite way, mostly open wheel cars), with some success. In those light cars, the engine (block) torque reaction did quite a bit to counteract chassis roll in the corners. I believe reverse rotation been banned just about everywhere though...
Got to thinking about this a bit. It appears that in this situation the intent is to equalize front tire loading rather than rear to mitigate a push on corner exit when the front is also going light. Rear tire loading would be made somewhat more unequal and would also be generating a yaw moment that would oppose the push without having to put the axle itself into loose steer. Sound even close?


Norm

exwestracer
02-13-2012, 05:34 AM
Got to thinking about this a bit. It appears that in this situation the intent is to equalize front tire loading rather than rear to mitigate a push on corner exit when the front is also going light. Rear tire loading would be made somewhat more unequal and would also be generating a yaw moment that would oppose the push without having to put the axle itself into loose steer. Sound even close?


Norm

That's it in a nutshell, Norm. If the driver didn't have to rely on "hitting the stagger" to get the car off the corner, he was more in control of his own fate... I believe Mr. Yunick got the ball rolling on reverse rotation LONG before anyone else even heard about it.