Dual Shocks up front? - Page 2 - Ranger-Forums - The Ultimate Ford Ranger Resource


Suspension Tech General discussion of suspension for the Ford Ranger.

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  #26  
Old 01-17-2006
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bla bla bla is much more "elegant" than what I actually said, lol!
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  #27  
Old 01-17-2006
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That is type of rising rate linkage, commonly used with coilovers on race cars and on the rear suspension of motocross bikes. The deeper you go into the travel, the faster the shock moves relative to wheel travel. So, you get lighter damping near the suspension's static equilibrium point and heavier damping at the suspension moves toward full compression.

A rising rate linkage can work very well or it can be quite mediocre, depending on the geometry and the shock valving.
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  #28  
Old 01-17-2006
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y not just buy one good set of shocks instead of wasting money on two ..
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  #29  
Old 01-17-2006
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Quote:
Originally Posted by Therhinosranger
y not just buy one good set of shocks instead of wasting money on two ..
^^^ the easiest, cheapest and most practical solution. ^^^
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  #30  
Old 01-17-2006
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like the looks of em tho. something different.
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  #31  
Old 01-17-2006
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i have single shocks up front on my Effer and my dad has dual...mine rides smoother and has better feel. his is a bit stiffer and jarring sometimes.

handling wise they but feel the same but i cant take turns like his could...mainy from the 33s.

i say no and invest in a good single shock. ur wasting time and money.
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  #32  
Old 01-17-2006
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Quote:
Originally Posted by Splitfire
Same rate, same stroke.
The force being applied to the bottom shock will be different than the force being applied to the top shock, even though the same force is being applied to the lever arm by the wheel. Torque = Force * Length of Radius Arm. The shocks themselves might have the same stroke and rate, but one would be worked harder than the other. So wouldn't it make sense that you'd want 2 different strokes and rates?

EDIT: I got ahead of myself and Jason and John both said what I said already... D'oh!

I like Bob's explanation better anyways. Still trying to figure all of it out, but it does make sense I think...
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  #33  
Old 01-17-2006
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Ok I understand what you're saying about the rising rate Bob, but how can you tell that just by looking at it? As opposed to one where, say, the rate does not increase with the travel of the wheel and it stays static? Would the cantelever (sp. I know) be closer to a right angle?
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  #34  
Old 01-17-2006
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Quote:
Originally Posted by Gearhead61
Ok I understand what you're saying about the rising rate Bob, but how can you tell that just by looking at it? As opposed to one where, say, the rate does not increase with the travel of the wheel and it stays static? Would the cantelever (sp. I know) be closer to a right angle?
First of all, I'm going to pick some numbers out of the air based on the photo. I'm sure that these are not perfectly accurate but they will do for this discussion. Also, there is no way to tell from the photo if it's fully topped out or if the suspension is sitting at some static equilibrium point. It really doesn't matter - the basic concept is the same.

OK. Recall that the curve of the sine function hits a maximum value at 90 degrees. Angles near 90 degrees have a sine value near 1 and as you move significantly away from 90 degrees, the value begins to decrease more rapidly.

Maximum reactive torque from the bellcrank would be required when the force of the link is at right angles to the arm (tangential). If you look at the arm of the bellcrank that is attached to the suspension link, it is an angle significantly different from 90 (I can't tell what it is exactly, so I'll just call it 120 degrees for convenience). In a well designed rising rate linkage, the angle will approach 90 when it bottoms. Sin90 is 1 and the sin120 is 0.87. Assuming these values, the rate rise would be:

[(1/0.87)-1] x 100% ~= 15%

Then look at the other arm and imagine it moving through its arc. Its angular displacement must be the same 30 degrees as the other arm. It appears that it will start and end at something slightly different from 90 and pass through 90 at about midstroke of the shock. I'll assume, based on the picture, that it moves from 75 to 105 degrees. Angles near 90 have a sine value that is very close to the value at 90 degrees. Sin75 = sin105 = 0.97. That means that there will be only minor variations in the linear displacement of the free end of the shock per unit angular displacement of the bellcrank. Using my fabricated numbers, the rate change is only:

[(1/0.97)-1] x 100% ~= 3%

or 3% higher in the center of travel than at the ends.

So while the arm that is moving the shocks will have a nearly consistent "push" on the shocks, the arm attached to the suspension will provide progressively more resistance as the suspension compresses, assuming that it does not pass below 90 degrees.

The net result is that the effective damping rate is continually rising as the suspension compresses.
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  #35  
Old 01-17-2006
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Aaaaaaaaaaahhhhhhhhhh!!!!!!!!!!!!

Physics!!!!

Woooooohoooooo!!!
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  #36  
Old 01-17-2006
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I still stand by what I said. Don't turn it into a **** fight.
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  #37  
Old 01-17-2006
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I'm gonna stand with trent on this one. When it comes to suspension. He's most knowledgable.
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  #38  
Old 01-17-2006
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Quote:
Originally Posted by Splitfire
I still stand by what I said. Don't turn it into a **** fight.
Ha ha! Well, don't design anything then. If you can't see the physics of that with basic levers brother, you're one hurtin' pup. Why not just admit you're wrong and be done with it? Anyone with elementary physics can see the truth of it?

It doesn't matter who stands with who, by the way. Facts are facts and physics is physics. Force applied to a lever (or bellcrank in this case) is multiplied closer to the pivot (or fulcrum) at the expense of distance.

Now do you see it?

It's not a matter of a "pissing contest" and your language tells me you probably know you are wrong. What it is about is giving people good information and your information was incorrect. Period. No emotion here, man, just cold hard facts.
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  #39  
Old 01-17-2006
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I say just go with the easy one shock, no need for two.
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  #40  
Old 01-17-2006
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so what exactly are you saying john? That one shock has more pressure on it than the other?
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  #41  
Old 01-17-2006
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Yes, and they move different distances. Observe below...

When the lever (straightened here for clarity) moves over a range, shock A moves distance A, and shock B moves distance B.

Furthermore, due to the conservation of total energy involved, the load closer to the fulcrum feels a greater force over a lesser distance, and the load further from the fulcrum feels a lesser force for a greater distance.

The shocks have different effective rates as seen by the force (the suspension pressure) as they travel the different distances.

To repeat: the two shocks do NOT exert equal damping, and they do NOT travel the same distance.

Crappy drawing but it's late. It'll do, lol.

Last edited by n3elz; 01-17-2006 at 11:13 PM.
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  #42  
Old 01-17-2006
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Quote:
Originally Posted by rwenzing


First of all, I'm going to pick some numbers out of the air based on the photo. I'm sure that these are not perfectly accurate but they will do for this discussion. Also, there is no way to tell from the photo if it's fully topped out or if the suspension is sitting at some static equilibrium point. It really doesn't matter - the basic concept is the same.

OK. Recall that the curve of the sine function hits a maximum value at 90 degrees. Angles near 90 degrees have a sine value near 1 and as you move significantly away from 90 degrees, the value begins to decrease more rapidly.

Maximum reactive torque from the bellcrank would be required when the force of the link is at right angles to the arm (tangential). If you look at the arm of the bellcrank that is attached to the suspension link, it is an angle significantly different from 90 (I can't tell what it is exactly, so I'll just call it 120 degrees for convenience). In a well designed rising rate linkage, the angle will approach 90 when it bottoms. Sin90 is 1 and the sin120 is 0.87. Assuming these values, the rate rise would be:

[(1/0.87)-1] x 100% ~= 15%

Then look at the other arm and imagine it moving through its arc. Its angular displacement must be the same 30 degrees as the other arm. It appears that it will start and end at something slightly different from 90 and pass through 90 at about midstroke of the shock. I'll assume, based on the picture, that it moves from 75 to 105 degrees. Angles near 90 have a sine value that is very close to the value at 90 degrees. Sin75 = sin105 = 0.97. That means that there will be only minor variations in the linear displacement of the free end of the shock per unit angular displacement of the bellcrank. Using my fabricated numbers, the rate change is only:

[(1/0.97)-1] x 100% ~= 3%

or 3% higher in the center of travel than at the ends.

So while the arm that is moving the shocks will have a nearly consistent "push" on the shocks, the arm attached to the suspension will provide progressively more resistance as the suspension compresses, assuming that it does not pass below 90 degrees.

The net result is that the effective damping rate is continually rising as the suspension compresses.
Ok I see what you are saying! That makes sense. I was thinking that the suspension travel would allow the bellcrank to travel past the 90* mark and thinking that it would have to increase and then decrease the rate after it passed 90*. I guess this is what I get to look forward to in the coming years. Thanks!
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  #43  
Old 01-17-2006
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Yes, but john. You have to understand this as well. By the time that bottom shock is fully extended either 1) The arm is going to be at full droop and most likely strapped. Now, if you look at the actual angle of that pivot arm, take a real good look at the design of it. When that top shock is bottomed out, the bottom shock will have exactly the same amount of shaft remaining as the top one did at full droop. They exert the same.
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  #44  
Old 01-17-2006
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I don't know if I'm exactly following you Cass.

I may be looking at it wrong, or maybe I am misreading... It looks to me like as the suspension reaches its highest point of travel (compressed) then the bellcrank will bottom out both shocks at the same time. The one on the bottom is longer because it travels a greater distance because the radius from the shock mount to the point of rotation for the bellcrank is greater than that of the top shock. Let me draw a picture up and I'll show it real fast.
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  #45  
Old 01-17-2006
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Notice how in the first picture the bellcrank is vertical, and then as the bellcrank approaches becoming horizontal and it reaches its greatest resistance like Bob explained, the shocks each come to the end of their travel at the same time.
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  #46  
Old 01-18-2006
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Quote:
Originally Posted by FauX
Yes, but john. You have to understand this as well. By the time that bottom shock is fully extended either 1) The arm is going to be at full droop and most likely strapped. Now, if you look at the actual angle of that pivot arm, take a real good look at the design of it. When that top shock is bottomed out, the bottom shock will have exactly the same amount of shaft remaining as the top one did at full droop. They exert the same.
The force exerted by the shock for damping isn't dependent on it's position. Your analysis is still faulty. They do not and CAN'T exert the same.

You guys are trying to be suspension experts and you don't understand basic levers?

Listen, you can know about all the configurations of suspensions made and sold out there and that makes you an expert SALESMAN to my mind. If you get one and drive it well, you're an expert USER.

But if you don't understand how something WORKS -- you are NOT an expert on the technology. If you don't understand how being connected to two different points on a lever changes the force, you understand NOTHING about the necessary mechanics of suspensions, since a lever is mechanics "101".

You guys made a mistake, and now you're arguing out of pride or something. It's ridiculous. Hit the books and get it straight. You're just making yourselves look bad at this point and you don't need that.
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  #47  
Old 01-18-2006
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wow great info but im not gonna get that set up. noooooooooooooooo way! im good! lol
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  #48  
Old 01-18-2006
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hmmm yes physics and engineering .... hell im in college right now, learning all about it



oh and yea...stay with one shock..**** levers and ****! lol
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  #49  
Old 01-18-2006
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Quote:
Originally Posted by n3elz
since a lever is mechanics "101".
ENGR 111 actually...

Atleast by TAMU standards
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  #50  
Old 01-18-2006
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Ha ha! Well, I'm an electrical engineer, what did you expect...
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