Register now to get rid of these ads!

4-link debate

Discussion in 'The Hokey Ass Message Board' started by sideweeler, Apr 4, 2013.

  1. I thought that's exactly what I WAS trying to do.

    At the axle centerline, NOT the contact patch of the tire, the upward rotation force against the pinion shaft is exactly the same as the torque applied to the axle shaft (Isaac Newton...equal and opposite reaction...???). If not, where does the "rest" of the torque disappear to? EDIT: I am not talking about driveshaft torque here. Gear reduction has nothing to do with it. If I am putting 1000 lb/ft into rotating the ring gear down, then I also have 1000 lb/ft trying to lift the pinion shaft straight up.

    As far as the laws of physics go, I agree 100% that they do NOT change. So why would the force (obviously tension) applied to the top link in the video suddenly reverse to "reduced compression" just because the brackets were welded to the axle instead of floating? It makes NO difference whether the suspension is a 3 link, parallel 4 link, or triangulated 4 link; the basic force coming from pinion climb is always in the same direction.

    How that force is applied to the CHASSIS is where I think we've come unstuck. Your original diagram with all 4 links in compression does apply to an IRS system (my words, not yours), as the pinion shaft is disconnected from the suspension linkage because there is no solid axle housing transferring the force to the brackets and links. It also applies to the OUTER links on a dirt track floater system, where (as you said) the pinion force is once again disconnected from the axle housing by the floater bearings.

    As soon as we weld the upper brackets to the solid axle housing (anywhere above axle centerline), the pinion climb is now resisted by the upper links.

    And Al, this:
    "Although experience is valuable, it isn't the same thing as knowledge."

    HAS to be the quote of the month... My "experience" in changing out pulled heim joints on the top links of race cars obviously hasn't translated into any "knowledge" of why that might have happened...

    I'm not being defensive, my point was that while I use Google and the internet for information, my own observations come from things that I have seen and had in my own hands. I know, for a fact, that if the upper links on a solid axle suspension run uphill to the frame, and the lowers are level, the car will squat under hard acceleration.

    Which is where we started, so I'm done.
    Last edited: Apr 25, 2013
  2. blowby
    Joined: Dec 27, 2012
    Posts: 6,592

    from Nicasio Ca

    These axle cam videos are better than a Danika Patrick Go Daddy commercial. Here is an asphalt racer with standard triangulated four link. You can see the rear end rotating under acceleration and braking. Whether the top links are being put into tension or just less compression is hard to say. The car sure rides smooth.
  3. Perfect. THANK YOU.

    Full screen it and watch the upper link bushings closely. You will see that the bushing sleeve on the axle housing pulls BACK from the link under acceleration and moves FORWARD when the driver lets off the throttle. Since the bushings would really LIKE to be centered (as they are at rest), I think that spells it out. Also note that the upper links are running downhill to the chassis, and the suspension extends (anti-squat) under acceleration. If that doesn't convince people, nothing will...
    Last edited: Apr 25, 2013
  4. Morrisman
    Joined: Dec 9, 2003
    Posts: 1,600

    from England

    One way to see if your system binds is to remove the bolt on one bushing, move the suspension up or down, or angled, and see how hard it is to get the bolt back in. I did this on my rubber bushed triangulated rear end, and it was easy enough. Now if the bushings had been hard poly, or two inches wide.....

  5. Kerrynzl
    Joined: Jun 20, 2010
    Posts: 2,284

    One thing about those videos is you can see into them what you want.

    I cannot see the top of the diff head rotating back under torque, I see the underside of the diff rotating forward under torque [ even though they’re both rotating the same direction ]
    If you look at it closely the rear-end is actually moving back when the throttle is off [unless for some unlikely coincidence the sound is out of phase to the movement ]

    Here is a video of the FORWARD driving force from the tyres [ it is only 20 secs of sloooow motion so it won’t bore anybody ]
    You will notice the tyres distorting and moving forward BEFORE any torque reaction causes lift.

    On a drag car with the shocks angling back from behind the rear Axle, it is very common to make adjustments to the extension side of shocks because the lower links are trying to “drive under”
  6. Kerry, I am not trying to "prove I'm right", I'm merely trying to illustrate the fundamental forces applied to the rear axle housing under hard acceleration. Those forces are the same regardless of what type of suspension system is attaching the housing to the chassis. Here is an excellent video (if a tad dark) showing what happens to the axle as power is applied.
    You can clearly see the pinion rotate UP as the driver applies power against the brakes, and the whole time the car is accelerating down the strip. Since the power is applied to the tire along the axis of the axle shaft, how can the top of the housing be rotating forward (pushing)? Yes, I understand this is a leaf spring car; but it does not change the force the ring gear exerts against the pinion and the housing.

    If we attached forward links to the top of the housing of the car in this video, I cannot see how they could be loaded in compression.
  7. Dave Zapatka
    Joined: Oct 14, 2009
    Posts: 74

    Dave Zapatka

    That's NO joke !
  8. hoop98
    Joined: Jan 23, 2013
    Posts: 1,363

    from Texas

    Any suspension will work if you don't let it....
  9. Kerrynzl
    Joined: Jun 20, 2010
    Posts: 2,284



    It is quite simple!

    The torque at the tyre footprint [ and the equal and opposite thrust at the axle centerline ] is greater than the torque reaction of the pinion [ due to mechanical torque multiplication ].
    It is very easy to see this happen with a“Birdcage” with a separate torque arm. [ where the torque reaction is separated from compression loads ]
    With a birdcage the thrust is divided between upper and lower links
    But with the Torque Arm, the torque reaction is also divided between upper and lower mounting/pivot points [ the upper being tension ]

    Now when you combine these loads together in a typical 4-link suspension , you are adding torque reaction forces to already existing acceleration forces [ the thrust or compression loads on the links ]

    So the torque reaction loads get subtracted from the upper compression loads [ but because the compression loads are greater than tension loads, it merely reduces these loads ]
    On the lower link the torque reaction loads get added to the lower compression loads. [ this is what I see in the videos ]

    The difference between the upper compression and lower compression loads still create exactly the same lift at the instant centre [ this is also the centre for the acceleration forces ]

    I lost a few $$$ many many years ago in a bet ,because I had the opposite view to what I had now [ I had difficulty picturing this also ]:D

    I’m not trying to get into a chest thumping exercise , quite the opposite.
    If you pm me, I will send you some calculations [ so save boring people that don’t like math ]
  10. ago
    Joined: Oct 12, 2005
    Posts: 2,200

    from pgh. pa.

    How about 3 link, with 2 lower links and one upper with a pan hard? This I believe would be a freer moving suspension.

  11. CutawayAl
    Joined: Aug 3, 2009
    Posts: 2,144

    from MI

    All the options have at least some pros and cons. A 3-link is a good good design with minimal shortcomings. The biggest down sides are that most cars don't have a natural mounting point for the third/upper link so that has to be fabricated. And, in some cases the upper link mounting point will intrude into the car's interior space(as upper 4-link mountings may do).
  12. Good points. The typical result on any car that needs a back seat is a VERY short upper link that gives a lot of pinion angle change with any significant amount of suspension travel.
  13. If we are talking about rotational torque on the pinion shaft, this is correct. However, if we are looking at the lift force on the pinion shaft, it is simply not possible to "lose" any of the equal reverse torque force that the ring gear applies to the housing through the pinion bearings.

    Let's say for simplicity that we have a 24" tall tire. Then the lever arm from the axle centerline to ground is 12". If we apply 1000 ft lb of torque through the axles, we have that same 1000 ft lbs at the ground. If we were to measure lift on a 12" long bar attached to the side of the pinion case, we would see that same 1000lb trying to lift the bar up into the floor. If we restrain the housing with links (non-birdcage 4 link for example), that same force is applied to the links; it has nowhere else to go...

    Yes, the forces are divided between the upper and lower links, and the length of the brackets from axle centerline determines the division of that torque load between the links.
    Most ALL performance oriented solid axle suspensions are designed to have the lower links level at ride height (as pictured). Therefore, the lower links are providing almost zero lift force, as the thrust load path is parallel to the ground. Yet, we still have the ability to lift the chassis up at the rear (anti-squat) with the upper links running downhill to the frame.

    Where is the lift component coming from if not tension on the upper links? Don't get me wrong, I'm a firm believer in math and engineering. I'm also a firm believer in what is right in front of my eyes. If the top links (as shown above) are under any sort of resultant compression force from forward acceleration, that force would be pushing DOWN on the chassis, given the link angle. This is simply not what happens with this type of suspension.

    You're right, maybe we should take this discussion to PM. I'm sure most of the readers will just weld the brackets on where it's most convenient anyway...
  14. While I feel that you're probably right about most folks, I still am enjoying the education you are providing by leaving the discussion in the open forum instead of PM.
  15. blowby
    Joined: Dec 27, 2012
    Posts: 6,592

    from Nicasio Ca

    The answer to the upper link force is found here. I will not provide the answer, I'm much too, umm, humble for that. Instead I will let someone else do it. :)


    Interactive Four-Bar Linkage Dynamic Force Analysis

    Given position, velocity, acceleration, and inertia properties such as mass and mass moment of inertia for each moving link of a four-bar linkage, we are able to perform force analysis for the linkage. Three static equilibrium equations, in terms of forces in the X and Y directions and moment about the center of gravity of the link, can be written for each link. The free body diagram is also available. For link 2, we get

    F12x + F32x + Fg2x = 0 (1)
    -m2g + F12y + F32y + Fg2y = 0 (2)
    Ts + (-rg2) x F12 + (r2-rg2) x F32 + Tg2 = 0 (3)
    where rg2=rg2*exp(i(theta2+delta2)) is the position vector from joint Ao to the center of gravity of link 2. F12 and F32 are the joint forces acting on link 2. Fg2 and Tg2 are the inertia force and inertia moment of link 2. m2 is the mass of link 2 and Ts is the driving torque. For link 3 we get
    F23x + F43x + Fg3x = 0 (4)
    -m3g + F23y + F43y + Fg3y = 0 (5)
    (-rg3) x F23 + (r3-rg3) x F43 + Tg3 = 0 (6)
    where rg3 = rg3*exp(i(theta3+delta3)) is the position vector from joint A to the center of gravity of link 3. F23 and F43 are the joint forces acting on link 3. Fg3 and Tg3 are the inertia force and inertia moment of link 3. m3 is the mass of link 3. For link 4 we get
    F34x + F14x + Fg4x = 0 (7)
    -m4g + F34y + F14y + Fg4y = 0 (8)
    (-rg4) x F14 + (r4-rg4) x F34 + Tg4 + Tl = 0 (9)
    where rg4=rg4*exp(i(theta4+delta4)) is the position vector from joint Bo to the center of gravity of link 4. F14 and F34 are the joint forces acting on link 4. Fg4 and Tg4 are the inertia force and inertia moment of link 4. m4 is the mass of link 4 and Tl is the torque of external load. Equations (3), (6), and (9) can be expressed as
    Ts - rg2*cos(theta2+delta2)*F12y + rg2*sin(theta2+delta2)*F12x
    + [r2*cos(theta2) - rg2*cos(theta2+delta2)]*F32y - [r2*sin(theta2)
    - rg2*cos(theta2+delta2)]*F32x + Tg2 = 0 (10)

    -rg3*cos(theta3+delta3)*F23y + rg3*sin(theta3+delta3)*F23x
    + [r3*cos(theta3) - rg3*cos(theta3+delta3)]*F43y - [r3*sin(theta3)
    - rg3*cos(theta3+delta3)]*F43x + Tg3 = 0 (11)

    -rg4*cos(theta4+delta4)*F14y + rg4*sin(theta4+delta4)*F14x
    + [r4*cos(theta4) - rg4*cos(theta4+delta4)]*F34y - [r4*sin(theta4)
    - rg4*cos(theta4+delta4)]*F34x + Tg4 + Tl = 0 (12)
    Note that Fijx = -Fjix and Fijy = -Fjiy, equations (1-9) can be written as nine linear equations in terms of nine unknowns. They can be expressed in a symbolic form
    Ax = b (13)
    where x = the transpose of (F12x, F12y, F23x, F23y, F34x, F34y, F14x, F14y, Ts) and is a vector consisting of the unknown forces and input torque, b = the transpose of (Fg2x, Fg2y-m2g, Tg2, Fg3x, Fg3y-m3g, Tg3, Fg4x, Fg4y-m4g, Tg4 + Tl) and is a vector that contains extenal load plus inertia forces and inertia torques.
    The interface below allows the user to find the joint forces and the required input torque given the link lengths, theta1, one additional position, one angular velocity, one angular acceleration, the external load and the inertia properties for the mechanism.

    Unit Type:
    Link lengths (m or ft): r1: r2: r3: r4:
    Link masses (kg or slugs): m2: m3: m4:

    Link moments of inertia (kg-m^2 or lbf-ft-sec^2): ig2: ig3: ig4:
    Position Vector for Center of Gravity

    Magnitudes (m or ft): rg2: rg3: rg4:

    Mode for all angles: theta1:
    Select and input the known angle (theta2, theta3, or theta4):
    Departures from Link Positions: delta2: delta3: delta4:

    External load (N-m or ft-lbf) on joint (1,4): tl:

    Select and input the known angular velocity (omega2, omega3, or omega4):

    Select and input the known angular acceleration (alpha2, alpha3, or alpha4):
  16. hoop98
    Joined: Jan 23, 2013
    Posts: 1,363

    from Texas

    My eggs have lots of omega3 it says!

    The angle of the descendant is equal to the cube root of the entropy of the cibus divided by the dam in the roadside drainage canal....
  17. Motomike43
    Joined: Jan 13, 2013
    Posts: 156


    He is right. Its supposed to be that way. it helps keep the rear Racked side to side. The links will not bind. It will have full travel. They have been building them like that for a long time. Lots of brand new trucks still use it today. Even the front and rear end of a new jeep wrangler is like that. It works well.
    If the links are all set straight at the axle you need a 5th link to rack the rear. (Panhard Bar)
  18. Kerrynzl
    Joined: Jun 20, 2010
    Posts: 2,284


    I’ll keep in the open discussion for everybody’s benefit. As long as it doesn’t turn into a public “Shitfight” [ this is somebody else’s thread which I apologize for it being hijacked ]
    The problem is explaining it all in text, it is easier to show somebody on a blackboard or paper. [ or in my case, chalk on the garage floor ]
  19. Kerrynzl
    Joined: Jun 20, 2010
    Posts: 2,284


    OK, I can see some confusion here!
    Firstly everybody must understand [ or agree ] that the torque comes from the “engine end of the car”

    If we refer to Isaac Newton’s “Principia” here, Newton’s law states everything must be equal and opposite [eg: the torque output of an engine must be supported by equivalent engine mounts ,etc etc ]

    There are losses and some other factors involved but for calculations here, it is easier just keep them at unity

    If you have 1000 lb/ft’s torque at the pinion trying to rotate the ring gear down, then you also have 1000 lb/ft’s torque reaction trying to lift the pinion shaft straight up.
    If this torque reaction is controlled by a 4ft 6in [ 54inch ] long “torque-arm” you will have lift at this pivot point

    1000 x 12” ÷ 54” = 222 lb’s lift at the front pivot [ it also can be calculated a 1000 ft/lb’s ÷ 4.5ft = 222 lb’s ,but multiplying Ft/lb’s x 12 to convert to inches is easier when calculating smaller increments ]

    So the 222 lb’s of lift must have equal and opposite 222 lb’s being planted into the ground by the tyres

    OK if we move on to Wheel torque.

    If we still use the same 1000 ft/lb’s driveshaft torque, and 4.56:1 gears [ a common drag racing ratio ] and 28” diameter tyres [ 14” Radius ].

    This is calculated as 1000 x 4.56 x 12” ÷ 14” [ wheel radius ] = 3908 ft/lb’s of total torque at the tyre footprint.
    So 3908 ft/lb’s of torque at the circumference must have equal and opposite 3908 lb’s thrust at the axle centre line [ This is enough to propel a 3000 lb car at 1.3 G’s acceleration ]

    The thrust at the axle centreline transfer loads into the chassis via upper and lower links .So the upper and lower links share the loads equally.[ 50/50 ]
    This is calculated as 3908 ÷ 2 = 1954 lb’s of combined compression loads on both the upper and the lower links [ by combined loads it can be 1/2 for left and 1/2 for right ]

    The torque reaction from the pinion causes lift, [ not the crownwheel or axle]
    but ..............
    The torque reaction from the tyre footprint causes “thrust” at the axle centreline which accelerates the car.

    Acceleration IS a torque reaction!!!!

    Now back to my previous post [ # 192 ],
    In a hypothetical situation if you have 4 links totally parallel [ no instant centre ], The extra compression forces on the lower links from torque reaction will try to cause the rear to squat
    [ this is a parallelogram effect of all the pivot points ]

    When the links are pointing together at an instant centre, this becomes the theoretical point of lift and acceleration. [ the lift is caused by greater compression forces on the lower links ]

    If the Torque reaction lift from a 54in instant centre was in a car with a 108in wheelbase with a 50/50 weight distribution it should lift the sprung weight of the front and rear equally.

    222 lb’s lift should be 55.5 lb’s at each corner [ the amount of lift in inches depends on the “wheel-rate” of the suspension ]
    If the car had a wheel-rate of 120 lb/in on each corner the car ride height should raise 0.46in at all 4 corners.

    [ note: a 120 lb/in wheel-rate on an early mustang is using 480 lb/in front springs and 120 lb/in rear leaf springs ]

    If this same 108in wheelbase car weighed 3000 lb’s and accelerated at 1.3 G’s as previously suggested but with a 22in centre of gravity height [ CGH ]
    By lowering the instant centre to ground level [ only possible with 4-links ] all the rearward weight transfer during acceleration is by “overturning moment” through the suspension.

    This car should have 1500 lb’s front and 1500 lb’s rear static weight , but 1.3 G’s this car should have 794.4 lb’s weight transfer to the rear
    3000lb’s x 1.3G’s x 22in CGH ÷ 108in Wheelbase = 794.4 lb’s

    794.4 lb’s weight transfered through 120 lb/in wheel-rate = 6.62in compression on the rear suspension minus the 0.46in lift from torque reaction = 6.16in compression at the rear
    and there would also be 6.62in lift on the front suspension plus the 0.46in lift from torque reaction = 7.08in lift at the front.

    With modern drag cars the acceleration is greater than the 1.3G’s previously mentioned, so the instant centre is raised to prevent excessive lift [ weight transfer by overturning moment ] or they would flip over backwards.
    You still get massive amounts of weight transfer from acceleration [ that’s the feeling you get in the back of the seat ]

    Now I’ll hand you all over to professor Paul “Blowby” to explain all his calculations :D

  20. Wow guys, thanks SO much for clearing all this up. ;)

    In a DESPERATE attempt to keep this relevant, I'm defining "torque" generically as rotational force present at ANY point in the system. More specifically at the rear axle centerline. You agree that we still have "torque" flowing through the axle shafts and wheels to the tire contact patch?

    Please allow me to state once again that gear ratio has NOTHING to do with rotational force (torque) as applied to the rear axle housing. We are already done with gear multiplication at that point.

    You stated just above that with a torque arm (upper and lower links rigidly connected at a common intersecting point, right?) we have a lift force present. If I extended the links in the photo I posted in #196, and welded the ends together, I would have a torque arm (also known as a ladder bar), correct?

    If we disconnect the links from each other, we again have the 4 link pictured in post #196. Aren't the forces applied to the links by the axle housing exactly the same? PLEASE do not go back to birdcages again....

    So, to answer my original argument to the statements you made in posts #120 and #138; please state, in 1000 words or less, HOW there can be a lift force present at the end of a torque arm or 4 link if BOTH upper and lower links are in compression? Regardless of how much thrust is applied to the lower link, it cannot generate any chassis lift if it is horizontal, can it?
  21. CutawayAl
    Joined: Aug 3, 2009
    Posts: 2,144

    from MI

    This has already been explained. I doubt another 1000 words would change your mind. Your perceptions are common, so you aren't alone in your confusion.:)

  22. I'll give you that, Al. Pretty much an entire industry shares that "confusion".

    I think in that case someone should notify AFCO, Speedway, Coleman, Howe, et al, that their "pull" bar sprung top link should be renamed the
    "reduced push" bar... and that it doesn't work.,1937.html
    Last edited: May 10, 2013
  23. I haven't been following closely on this but from my skim of the posts between Kerry and Ex it seems Kerry is to prove that Newton's Third law is moot.


    "Right Hand Rules!"
  24. Boones
    Joined: Mar 4, 2001
    Posts: 9,560

    from Kent, Wa

    Kerry, you noted that having a lower triangulated bar is better than having the top bars triangluated. Is this not similar to what Chevy did with the truck trailing arms (minus a two upper parts).

    This has been one of the best threads I have read in years.

  25. Gee, Steve, thanks....I think....:D Hey, does that work opposite in the southern hemisphere??? Maybe that's it!

    Boones, I believe Kerry's reference was to what is commonly known as a "Satchell Link" system. Basically a Chevelle type tri-4 link flipped upside down with the converging points of the links still on the axle end...and if this "mass confusion" is the best thread in years, I feel sorry for the HAMB... :eek::D
  26. Kerrynzl
    Joined: Jun 20, 2010
    Posts: 2,284


    I wrote earlier. [ post # 192 ] The bottom paragraph is the part that needs understanding correctly.

    "So the torque reaction loads get subtracted from the upper compression loads [ but because the compression loads are greater than tension loads, it merely reduces these loads ]
    On the lower link the torque reaction loads get added to the lower compression loads. [ this is what I see in the videos ]

    The difference between the upper compression and lower compression loads still create exactly the same lift at the instant centre [ this is also the centre for the acceleration forces ]"

    It is simple triangulation with a 4-link that has an instant centre.
    A ladder bar behaves exactly the same, the top of the triangle is under compression, the bottom of the triangle is under Greater compression.

    If you had a shackle at the front of a ladder bar you would temporarily "see" the results of compression loads driving it forward [ until it ran out of travel, and started moving the car ]
    Last edited: May 10, 2013
  27. Kerrynzl
    Joined: Jun 20, 2010
    Posts: 2,284


    It's because I'm a "Road-Racer" :D
    Having the triangulation on the lower bars gets the roll-centre lower to the ground.
    A low roll centre is more desirable because lateral acceleration [ G's ] is transferred through "overturning moment" [ bodyroll ]
    This can be controlled with roll stiffness.

    It is also more desirable to have the pivot point fixed at the diff-head ,than to have 2 bars to an instant centre.
    With an instant centre behind the rear-end the roll centre goes up and down with suspension travel.

    This is all academic in road vehicles ,
    But when you're a whisker away from the limit of tyre adhesion, it can make the difference.
  28. I'm at a here's another 12 minutes of boring video. If you can argue that the TOP links attached to these rear axle housings are under compression (of any amount) as the cars accelerate, then we will have to agree to disagree.

    (Nice long straightaways on this one and you can DEFINITELY tell when the driver is on the gas....)
  29. Kerrynzl
    Joined: Jun 20, 2010
    Posts: 2,284


    I firstly would like to apologise to all the other members here for the insults to their intelligence. They don’t actually need a good memory to be insulted here , all they need to do is go back and read previous posts


    This is me trying to get back on track again.


Share This Page

Register now to get rid of these ads!


Copyright © 1995-2020 The Jalopy Journal: Steal our stuff, we'll kick your teeth in. Terms of Service. Privacy Policy.

Atomic Industry
Forum software by XenForo™ ©2010-2014 XenForo Ltd.