[/QUOTE] Mustang II front suspensions have no bump stops from the factory - it is part of the shock absorber, not part of the suspension. I used a '76 Mustang II as the donor for the front suspension on my '53 Chevy. It was all original, all OEM. They have no bump stops on the A-arm or anywhere else from the factory - the shock absorbers have rubber bushings at the top that act as the bump stops.
Mustang II front suspensions have no bump stops from the factory - it is part of the shock absorber, not part of the suspension. I used a '76 Mustang II as the donor for the front suspension on my '53 Chevy. It was all original, all OEM. They have no bump stops on the A-arm or anywhere else from the factory - the shock absorbers have rubber bushings at the top that act as the bump stops.[/quote] The bump stops may not have been on the a-arms but you are admitting that the factory designed incorporated them into the shocks. Which means the original design did have bump stops incorporated, just weren't on the a-arms? The Fatman design has no bump stops located anywhere. Just don't get it. The factory is smart enough to put them somewhere but someone with an engineering degree and 30 years of hot rodding can't take a hint.
The factory is smart enough to put them somewhere but someone with an engineering degree and 30 years of hot rodding can't take a hint. so what is the name of your successful aftermarket HOT ROD suspension company?
I give this guy a hard time but that's because I have a valid reason. I have talked to Brent on way too many occasions to only get the run-around. He knows about the problems I have been having and plays it off saying I am the only customer having problems. It has been proven by this forum and other forums that I am NOT the only customer that has been screwed by Fatman. Also, I do not have a hot rod company and no intentions of doing so. If that answers the question. However, myself and my family are no strangers to doing this kind of work and the engineering that is involved with building such parts. We are not a bunch of country bumpkins. Just because a company has made a profit and is as big as Fatman has become does not mean they reached that point by following respectful morals. Screwing a guy over is screwing a guy over. I'm sad I had to furnish the vaseline for this particular experience. Like I have said, Brent and others at Fatman are aware of multiple failures from their products. Poor fitment, poor quality, poor tech support, etc. I don't expect everything to be perfect, I just want to get what I pay for. I honestly don't see where a reasonable person could argue with that.
"The bump stops may not have been on the a-arms but you are admitting that the factory designed incorporated them into the shocks. Which means the original design did have bump stops incorporated, just weren't on the a-arms? The Fatman design has no bump stops located anywhere. Just don't get it. The factory is smart enough to put them somewhere but someone with an engineering degree and 30 years of hot rodding can't take a hint."[/QUOTE] I'm not "admitting" anything, I'm telling you that the millions of Mustang IIs sold by Ford had no bump stops anywhere in the front suspension, and when you go to Autozone and buy replacement shocks they have a rubber bushing on them that stops the shock from slamming too hard when it reaches the limit of its travel. Any shock that is mounted on a Mustang II front end should also have that bumper on it. Period. If someone sells you a shock that doesn't have it, then they have overlooked this important aspect of the factory design. (And for anyone who has shocks, coilover or otherwise, on their MkII that does not have that bushing, take a rubber bushing, slit it, and slide it onto the shaft of your shock!)
Just an update. I finally got the new parts installed and shipped the old ones back to Fatmans this past week. Anxious to hear the failure analysis,I think I know but I'll wait for the official word. Ron
Andy, I think you are working off of some mis-information. There are plenty of reasons why I don't like the split beam suspension as Fatman is/was producing it. But the need to double the spring rate/load is not one of them. I would really like to hear your engineering principles behind this theory. And maybe I could enlighten you on the subject.
Hunh? I think your theory is flawed. The axle is a lever even before being split. What you are trying to explain just does not make sense. As a matter of fact before it is split it can both push and pull on the spring at the same time.
I had an issue during installation of the grease zerks in the ball joints. The threads in the joint were tweaked somehow and I couldn't get the zerk into it. He sent me a new joint at his cost. I wasn't thrilled with the installation instructions but otherwise I've been happy with the setup. And I have the gussets running from the hats down the outside of the frame.
Ok, not a bad start. Yes the axle can be considered a lever but more of a second class lever to be more specific. But even that is not quite correct because it only addresses one side of the vehicle. Except for one thing. There is leverage to be considered. When considering work loads on any beam axle, your virtual fulcrum is in the middle of the fulcrums, load and work ends. And this is exactly why the spring would not change and ride would not be affected as far as the spring is concerned when using an apparatus like this split beam system. This is where your engineering train falls off the tracks. Let's go back to the second class lever. The dimension between the work (red finger/tire) and the work (black arrow pointing up/spring pressure applied down) never changes no matter what you do to the length of the lever and distance to the fulcrum. So the shorter the lever the more work you could do because you have more leverage against the load (spring). The split axle fulcrum is half the distance of the unsplit axle, it's pivot being the other side where the tire meets the road. Except for what I mentioned above, the virtual fulcrum is in the middle of the axle already because you have to consider both sides of the car to work this out. The pivot center location may change vertically by this but it is still in the middle of the car no matter what so the lever arm is really still the same. Really! I think you would be shocked to find out how close in spring rate and load a coil spring from a 55 Chevy is compared to a Model A front leaf spring when you scale things proportional to the weight being carried. I guarantee you that if you were to connect the lower control arms together on this mythical 55 Chevy, it's ride height would not change because we are not changing the distance between the work (tire) and load (spring). Don't give up, learn.
I want you to work something out in your head to help wrap your brain around this concept that even though a beam axle is one piece it is still considered two levers, tied together... Ready? How is body roll controlled on a beam axle and buggy spring? This is a real simple problem that really does explain what I'm trying to say.
Don't go away defeated, that wasn't the point. I know from your point of view Zman and I must seem like brick headed corksoakers on understanding your point of view. That's why misinformation is so dangerous. I have had 20+ years of this stuff, some concepts are a bit tough on the logic system sometimes. I really hope you take some time to consider what I've said here.
I installed a fatmans Stage I hub to hub kit in my truck back in 2001. I am not a pro welder or fabricator and relied on the instructions immensly to insure a successful installation. My truck had a 351W and AOd trans, all orignal sheetmatel, and I even ran 15x8 1/2 wheels with 235 60/15 tires. Truck weighs abot 4000 lbs and I drive it real hard, daily. I am under my truck at least twice a month inspecting the front suspension because I am more worried about my welds cracking than the anything else. I have yet to have any issue with my front end other than the springs originally sent by FM. My truck was waaaaay to soft with the "v-8" springs they sent, So I went to Napa and bought 87 5.0 Mustang springs (I don't have the PN handy) so now it rides just fine. The hats are still solid pieces, the A-Arms show no signs of flexing or bending and I run no bumpstops.
Moot question Andy. You cannot atribute your test to a direct result of spring rate or load. Found this on the web, spring looks normal and holding up a HEMI with a bit more spring left for the bodywork..
LOL, You really don't grasp the concept. Shame really. Here is what looks like exactly the same frame with the wheels on and still has excessive positive camber. Maybe he's using trick photography to hide this sinister plot from everyone...
Ready for the final nail in your coffin? I can't quote you because you pussed out and deleted all your posts. But for the rest of the crew that are actually taking time to read and learn, let me just sum this up. You stated that by cutting the axle in half, you need to double the spring rate and load capacity. Correct? You also stated something to the effect that the spring rate of an IFS car like a 1955 Chevy had springs that were in fact stronger because of leverage. And that it is this leverage that is missing on a beam axle suspension and by splitting it in half the spring would no longer be strong enough to support the car. To use a vehicle I am very familiar with, my personal truck, a 1963 Ford F100 long bed unibody with a twin turbocharged BB Chevy and Tremec Manual trans. With it's Dodge Dakota suspension, I use a CC770 spring in the front which is a replacement heavy duty spring for stock Dodge Dakotas. I know the truck weights in over 4200 pounds and carries just about 2600 of total weight on the front wheels alone. Consulting my TRW Federal-Mogul Coil spring book from 1997 and the charts wherein. These are the specs on the CC770 spring. bar, install ht., load lbs., spring rate, free height 0.781, 12.00, 1418, 716.0, 14.00 Look at the load lbs. number very closely. If you consider that there are two springs carrying a total load of 2,836. pounds, that is really damned close to what the front vehicle weight actually is. This is overly simplified because on a coil spring you need to factor in the angle of the spring and a few other nit-picky bits but you can clearly see that the spring rate is not doubled as you stated it would be.
We need to bring this dead cat back for the evening crowd. You say that "you can not atribute your test to a direct result of spring rate or load" Intersting. What would you attribute it to?? Here is someting to try to think out. If the spring attached right at the end of the lower a arm(split axle) then the load would be the sprung weight --Agree? If the spring was attached to the inner pivot point on the lower arm, the spring could hold up nothing and would have to be infinitely stiff--Agree?? Logic would hold that the required load on the spring would increase as the spring attach point moved toward the inner pivot point--Agree?? High school phisics shows that a spring half way in the middle would have to push twice as hard as one pushing at the wheel end of the arm. Do you think that the spring location has nothing to do with the force it must exert on the arm???????? I have tried to explain the phisics as calmly as possible. I respect you work and the obvious care you take with what you do. I have never critisized you. Why are you making digging at me your life work? Why use catty snipets as a substitute for honest questions? I am tired of it. The nails,trains,theorys,ect.are stupid. You risk looking like what you attemp to portray me. Please answer my questions.
Oh, And the reason I deleted my posts is because I thought you might delete yours. I know absolutly what I said was correct and I thought you responses did no show you in a good light.
Beam axle suspensions are not independent. You cannot divorce the left from the right. To use your test as a result of spring rate or load you would get a false answer because while you would jack up the spring perch on the right side the left side of the spring would hold down the left side of the suspension. If you were to magically split the car down the middle so the left side would not affect the right, and were to tie the center of the axle to space, then your test result would be more accurate. You would then get the same exact result as an IFS car. I want to establish some terms to make this easier to follow. This is the second class lever used earlier, it demonstrates the situation very well. The red finger above is the tire. The black arrow is the spring. The triangle is the pivot or fulcrum. The distance between the tire(finger) is dimension "A" The distance between the arrow and the fulcrum is dimension "B" Unclear on the question. Are you trying to establish the difference between sprung and unsprung weight? Or are you moving the spring to where the tire is in the diagram, therefore making dimension A null? Yup, that would be true. That would mean that dimension B is reduced to nothing the spring could not hold up the car. Yup is sure does.. If the length of lever arm AB is unchanged and only the spring is moved toward the fulcrum. In otherwords +A -B or at least -B. Nope, never have. I have been arguing that dimension A has always remained a constant. And consequently I have also been arguing that dimension B has also remained a constant which you seem to think does not exists or is irrelevant. The center of mass is still in the same place and still must be supported by both halves of the suspension. Because the positions of the tire, spring and center of mass have not changed and we are dealing with two sides at the same time the fulcrum for supporting the mass has to stay in the middle of the vehicle. Because you are spreading false information. Unintentional maybe, but it is still misinformation. I saw on your profile you say you are an engineer. I expect a better understanding of a concept like this or at least be able to smack me around with real world math related to the concept to prove your point. Are you trying to suggest that all I have done is bait you without answering your questions? Seriously? You haven't adequately shown how your theory works, still. You keep trying to say the leverage is doubled, you still haven't shown your process. Where is your fulcrum if it isn't in the middle of the car? You cannot double the mechanical advantage on a lever without showing where your load, work and fulcrums are to start with and compare them to where they are after splitting the axle. Hopefully I have!
Iwill give a quick try and knock off for the night. Assume two supports on the ground.We will call one a tire and the other a fulcrum. The supports are 2 feet apart. A board lays across them. You stand in the middle of the board. How much weight is on the tire? 1/2 your weight. If you move to stand over the tire how much weight is on it. Your total weight. If you move to the fulcrum end, how much weight is on the tire. None. That is the whole deal. In the case of the lower arm, the inner end is attached to the car. The spring pushes down and loads both ends of the arm. The load on the inner end pulls the car down so the spring must be stiffer to push the car back where it is wanted. Take your diagram an attach the fulcrum to the bottom of a car. The spring works against that pull and so has to be stronger than when the spring is over the wheel.
You keep harping on the fact that the load is changing because somehow the leverage has been increased. The leverage has always been there. Ok, finally you added something to the argument where you do have a point. The position of the fulcrum has changed from a fixed (static) position on the axle, to the mass (dynamic) being supported. Taken quite a bit to get here and I will finally bow down to your theory. But only partially. Why? Because it isn't nearly as simple as you make it. And most certainly it is not doubling the load the spring must bear. Let us just concentrate on the split beam and forget about double wishbone suspensions, it is only confusing things at this point. Lets set some numbers to the game here. Can we assume there is 1000 pounds of sprung weight to support by both halves of the car. The axle is 56" wide on the track and the spring is 34" wide supported in the middle. This makes our dimension A 11" and our dimension B 17" and the fulcrum would be supporting pulling down at 67% of what is at the tire. This would mean there is another 670 pounds of force being added to the weight of the car using just your simple demonstrations. Not quite doubling the load but it is also not what the true added load would be. The force is not linear, we are now dealing with a force vector. Beyond my patience to calculate it, so can I cheat? Can we assume that while there is a force on the fulcrum now attached to the frame where it was otherwise attached to the ground is pulling down but also outward to the wheel. Would this not cut the downward force by nearly 50% cutting the new load to 335 pounds? Now back to the spring itself. It is not a linear rate spring, it's progressive. Almost all multi leaf springs are. And in the context of the vehicles we are dealing with, they use a seven leaf spring that is usually too stiff and we end up pulling a leaf or two out to get the ride quality we look for or to lower the car a bit. By the time we have achieved full ride height the spring rate has risen to well over 400 pounds per inch. This would mean that the static ride would change approximately one inch. Something that could easily be countered by changing the height of the fulcrum point or by putting one or both of the leafs taken out prior, back into the pack.
I think I will just give you a formula for determining the spring force required. Spring Force Requird = Weight(A+B)/B This is from the free body diagram summation of moments. The arm is not accelerating so all forces must balance. The rotational moments must balance. The moment of the tire is the weight on it times the arm (A+B) from your diagram. The moments are being calculated at the fulcrum. The other load is the spring force at distance B. Weight times arm(A+B) must equal Spring force times B. Weight(A+B)=B(Spring) Moving the B to the other side and rearanging gives the equation stated. The loads must also balance so the suck down load at the fulcrum is (Spring load)-(tire load)= Fulcrum load. Does this help?
Actually no. 1000(28)/17 = 1,647.058823 Remarkably, nearly the same number but in equation form that I provided before amazingly enough. Even so at 400 pound per inch average in the leaf spring we are considering here that is only a little over 4" of compression on the spring to total weight. The springs I have loose around here have 6/7" of arch with no load (loose) and that would still leave 2" of arch to the spring. The load did not double, the spring would still work. Again worst case scenario you add a leaf to the pack. Not giving up just yet. Got real work to do.
Look it is small round tubing and could have been altered? They are made for daily driving situations, everyone knows someone who trashes everything they drive and blames someone else. But everyone needs to realize that there is a stress factor involved here. The car could have been altered welding? may have the wrong geometry or suspesion travel? frame could be bent causing excessive stress on one side. (bottoming out) Who knows, this would take too much for all of us to analize here. It does look like it could be from the thicker welding on the side of the tubing. Try this out, Here is a basic demonstration of how the principals of metal properties act against each other. Roll up a piece of paper into a roll about 1 inch O.D. and try to bend it? tough! Now distort it slightly anywhere and bend it, changes outside of tube OD also, not just that it bends easily in the location that was distorted. Same thing probably here also but on a higher scale. Tubing is very strong as long as it is perfectly round.
The Jalopy Journal
