Issues With Big Blue!?!?!

[Nothing Special]

Maybe he is thinking of springs for a LD 250 with a Windsor and 5 lug hubs?

Sure as hell the 3560# gawr front end in my truck has stiffer springs than a 350's stack.

Maybe Billy is more about hardcore four wheeling and doesn't get the overlanding thing?

I don't have an F-350 to compare to, but physics says that, assuming the same load rating, the F-250 spring needs to be about 1.5 times stiffer than an F-350 spring (due to the tire being on a lever that's about 1.5 times as long as the lever the spring is on). The F-250 suspension probably has a lower load rating than the F-350 suspension, but not that much less. So the F-250 spring has to be stiffer than the F-350 spring. It's not intuitively obvious until you look at the leverage involved, but it's pretty clear when you do look at the leverage.

Oh, and on the OX install, Gary, if you read my FTE thread on it you might have noticed where I had to grind away some of the shift fork to clear a boss in the diff housing. That was a "complication", but not much of one. And it's one that the OX instructions pointed out (I just didn't pay attention soon enough).

[Gary Lewis]

I don't have an F-350 to compare to, but physics says that, assuming the same load rating, the F-250 spring needs to be about 1.5 times stiffer than an F-350 spring (due to the tire being on a lever that's about 1.5 times as long as the lever the spring is on). The F-250 suspension probably has a lower load rating than the F-350 suspension, but not that much less. So the F-250 spring has to be stiffer than the F-350 spring. It's not intuitively obvious until you look at the leverage involved, but it's pretty clear when you do look at the leverage.

Oh, and on the OX install, Gary, if you read my FTE thread on it you might have noticed where I had to grind away some of the shift fork to clear a boss in the diff housing. That was a "complication", but not much of one. And it's one that the OX instructions pointed out (I just didn't pay attention soon enough).

Jim - From what I've seen at his shop you may well be right - hard core four wheeling, not overlanding. I hadn't thought about that, but I need a suspension that will do everything well. I need it to ride and handle well on the highway so I can get "there". And it needs to have enough articulation/flexibility to handle the rough stuff when on the trails. Plus it needs to handle towing nicely. But I don't want to maximize any one of those things as I suspect that would minimize the others.

Right now the stiffness makes it no fun to drive long distances and it can't be any fun on rough trails as you'll feel every bump. Plus it'll lose traction 'cause one of the front wheels isn't going to be carrying its fair share of the weight. But, it tows nicely.

Bob - I see what you are saying about the physics. The 1986 dealer facts book shows the standard front axle on a 4wd F350 had a capacity of 5000 lbs as opposed to 3850 for the F250HD 4wd. So, if your math is right, the F250's springs would be rated at 5775 lbs. Not a huge difference, but the F350 springs should be softer. And with an RSK the ride would be even better. But wait! There's SD springs available.

Having said that, wouldn't the 250 springs have to be even stiffer since they can only allow 1 1/2" of travel? Perhaps the D44HD axle is the limiting factor and not the springs?

Anyway, I've now talked to Robert at Four Wheel Parts. Basically, they don't do what I am wanting to do. Yes, they could build the D60, but he recommended Chris at Boom's Garage for what I'm looking for. I have a call into Chris, but he's probably gone to lunch so maybe I'll hear soon.

And, Robert suggested another salvage to talk to, so while I'm waiting on Chris I'll call them.......

[85lebaront2]

Jim - From what I've seen at his shop you may well be right - hard core four wheeling, not overlanding. I hadn't thought about that, but I need a suspension that will do everything well. I need it to ride and handle well on the highway so I can get "there". And it needs to have enough articulation/flexibility to handle the rough stuff when on the trails. Plus it needs to handle towing nicely. But I don't want to maximize any one of those things as I suspect that would minimize the others.

Right now the stiffness makes it no fun to drive long distances and it can't be any fun on rough trails as you'll feel every bump. Plus it'll lose traction 'cause one of the front wheels isn't going to be carrying its fair share of the weight. But, it tows nicely.

Bob - I see what you are saying about the physics. The 1986 dealer facts book shows the standard front axle on a 4wd F350 had a capacity of 5000 lbs as opposed to 3850 for the F250HD 4wd. So, if your math is right, the F250's springs would be rated at 5775 lbs. Not a huge difference, but the F350 springs should be softer. And with an RSK the ride would be even better. But wait! There's SD springs available.

Having said that, wouldn't the 250 springs have to be even stiffer since they can only allow 1 1/2" of travel? Perhaps the D44HD axle is the limiting factor and not the springs?

Anyway, I've now talked to Robert at Four Wheel Parts. Basically, they don't do what I am wanting to do. Yes, they could build the D60, but he recommended Chris at Boom's Garage for what I'm looking for. I have a call into Chris, but he's probably gone to lunch so maybe I'll hear soon.

And, Robert suggested another salvage to talk to, so while I'm waiting on Chris I'll call them.......

Gary, you need to remember your simple machines, levers in particular. The springs need to be stiffer on the TTB and Twin I-beam as they are not right at the wheels. On a live axle the effective lever length is from the opposite wheel as a pivot location. Essentially the spring is based directly for the load.

[Steve83]

...physics says that, assuming the same load rating, the F-250 spring needs to be about 1.5 times stiffer than an F-350 spring (due to the tire being on a lever that's about 1.5 times as long as the lever the spring is on).

No, you've got it backward. The length of the lever from the spring to the frame is irrelevant; and the horizontal distance from the contact patch to the spring isn't much more important. What matters is the angle of the line from the spring seat on the axle to the axle pivot point, relative to the line through the spring seats (the spring's line of force). That angle determines how much of the spring's force will bear the truck's weight, regardless of the length of either line.

No matter what the suspension's geometry or layout: the vertical component of the springs' lines of force must support the sprung weight of frame/body/powertrain.

[salans7]

Jim - I understand what you are saying. We are thinking alike.

Shaun - Here is the stuff that David found. Seems complete - right?

Yep, that's everything you need for the swap.

 

[Gary Lewis]

Jim - I understand what you are saying. We are thinking alike.

Shaun - Here is the stuff that David found. Seems complete - right?

Yep, that's everything you need for the swap.

Guys - I'm getting lost in the physics. Which is strange 'cause that's what one of my degrees is in. Anyway, I feel sure that I don't want to use the F250 springs. F350 springs would be better, and SuperDuty springs the best.

Shaun - Thanks for the confirmation.

All - I called the salvage Robert suggested and he had nothing. When I asked the guy at that salvage if he had any suggestions he referred me to the place in OKC which has the one I've called about.

And, I've not heard from Chris. But it is possible that due to the rain and storms he didn't make it in today.

[Nothing Special]

Guys - I'm getting lost in the physics. Which is strange 'cause that's what one of my degrees is in. Anyway, I feel sure that I don't want to use the F250 springs. F350 springs would be better, and SuperDuty springs the best.

Shaun - Thanks for the confirmation.

All - I called the salvage Robert suggested and he had nothing. When I asked the guy at that salvage if he had any suggestions he referred me to the place in OKC which has the one I've called about.

And, I've not heard from Chris. But it is possible that due to the rain and storms he didn't make it in today.

WARNING - BORING GEEK STUFF AHEAD!

Guys - I'm getting lost in the physics....

I started to figure it out more completely and got lost too, so I had to go back to what my Statics TA back in college called "a fwee-ah ah-body ah-diagwam-ah" (or "free body diagram" for those of us to whom English is not our fourth language. And yes, I get that his English was MUCH better than my Chinese, so I'm really not putting him down).

What we're talking about here is how stiff the springs are. Spring stiffness is measured as the compression rate, how much force it takes to deflect the spring a certain amount. But springs are linear, so it's also how much MORE force it takes to deflect the springs a certain amount MORE.

Let's start with the F-350 suspension because it's so simple we don't need a free body diagram. Let's start with the assumptions that the front of the truck weighs 3000 lbs and we want a spring stiffness of 500 lbs/in (meaning it takes 500 lbs to compress the spring 1 inch). I just pulled that number out of the air, but it'll work for a comparison.

With 3000 lbs on the front of an F-350, that's 1500 lbs on each of the front tires, and 1500 lbs supported by each of the front springs.

Now if the springs have a rate of 500 lb/in, that would mean that if we put an additional 1000 lbs on the front of the truck we would be putting 500 lbs more on each spring and it will deflect 1" below wherever it is at ride height. And since this is a simple solid axle suspension, 1" more compression on each spring means each tire moves up (relative to the truck) the same 1".

Now lets take the F-250 front suspension and figure out what the actual spring stiffness needs to be to give an apparent stiffness of 500 lb/in. By that I mean when we put an additional 1000 lbs on the front of the truck we want each front tire to move up 1", the same as we saw in the F-350.

This is where a free body diagram becomes necessary for me. A free body diagram is a rough sketch of a "solid" piece (it doesn't have to actually be solid, or one piece, it just needs to not move with respect to itself). You show what forces are applied in what locations. Newtons law tells us that unless the piece is accelerating, the sum of all of the forces on it must be zero, and the sum of all of the moments on it must also be zero (in physics a "moment" is the torque exerted on an object by applying a force on a lever arm).

Below is the free body diagram for one tire / axle beam assembly of an F-250 front suspension. Fp is the vertical force applied to the swing arm at the pivot. Fs is the force applied to the spring perch by the spring (it's also the force on the spring). And Ft is the force applied to the tire by the ground (or the amount of weight sitting on the front tire). The measurements of 26" from pivot to spring perch and 41" from the pivot to the contact patch are approximate measurements off my '97 F-250. And they are measured parallel to the ground, not in a straight line from the pivot to the contact patch or spring perch.

Since the forces in the vertical direction need to add up to zero, the diagram tells us that Fs = Ft + Fp. Or solving for Fp (we might want to later...), Fp = Fs - Ft.

And the sum of the moments must also equal zero. Moments need to be defined around a certain point. The physics don't care what point we use, but the math will be easier if we pick the "right" point. So we'll use the pivot point. The moment the spring exerts on the body around the pivot point is Fs x 26". The moment the ground exerts on it around the pivot point is Ft x 41. So that means that Fs x 26 = Ft x 41. Or solving for Fs, Fs = Ft x 41 / 26

Let's take the same 3000 lb front end, but we'll only look at one side. The weight on the tire is half the truck weight, so still 1500 lbs, just like the F-350. But now when we do the math, the force on the spring is Ft x 41 / 26, or 1,500 lbs x 41 / 26, or 2,365 lbs! (for what it's worth, Fp = Fs - Ft = 865 lbs).

Now let's put an additional 1,000 lbs on the front of the truck to move the tire up 1", the same as we did for the F-350. Now the new Fs = 2,000 lbs x 41 / 26 = 3,154 lbs. So we've put an additional 3,154 - 2,365 = 788 lbs on the spring.

But how much has it deflected? Since the tire at a 41" radius went up 1", the spring at a 26" radius went up 26 x 1/41 = 0.63".

Spring rate is (additional) force per (additional) deflection, so 788 lbs / 0.63" = 1,243 lbs/in. Just a little stiffer than the 500 lb/in F-350 spring!

This result was so much more than I expected that I didn't believe it. I've checked my equations and my math a couple of times though, so I'm pretty sure it's right. But if anyone wants to prove me wrong I won't be at all offended.

Edit to add: A few posts later I described how the F-250 spring stiffness likely increases significantly as the TTB geometry pulls it sideways. My calculations above are for this overall spring rate. If you put an F-250 spring on a solid axle it would not get reefed on so badly, so the overall spring rate of the same spring in that case would be lower. I'm still sure that the F-250 spring is significantly stiffer than an F-350 spring, even if it were on a solid axle. But likely it isn't 2.5 times stiffer if the pulling sideways isn't taken into account.

[85lebaront2]

WARNING - BORING GEEK STUFF AHEAD!

Guys - I'm getting lost in the physics....

I started to figure it out more completely and got lost too, so I had to go back to what my Statics TA back in college called "a fwee-ah ah-body ah-diagwam-ah" (or "free body diagram" for those of us to whom English is not our fourth language. And yes, I get that his English was MUCH better than my Chinese, so I'm really not putting him down).

What we're talking about here is how stiff the springs are. Spring stiffness is measured as the compression rate, how much force it takes to deflect the spring a certain amount. But springs are linear, so it's also how much MORE force it takes to deflect the springs a certain amount MORE.

Let's start with the F-350 suspension because it's so simple we don't need a free body diagram. Let's start with the assumptions that the front of the truck weighs 3000 lbs and we want a spring stiffness of 500 lbs/in (meaning it takes 500 lbs to compress the spring 1 inch). I just pulled that number out of the air, but it'll work for a comparison.

With 3000 lbs on the front of an F-350, that's 1500 lbs on each of the front tires, and 1500 lbs supported by each of the front springs.

Now if the springs have a rate of 500 lb/in, that would mean that if we put an additional 1000 lbs on the front of the truck we would be putting 500 lbs more on each spring and it will deflect 1" below wherever it is at ride height. And since this is a simple solid axle suspension, 1" more compression on each spring means each tire moves up (relative to the truck) the same 1".

Now lets take the F-250 front suspension and figure out what the actual spring stiffness needs to be to give an apparent stiffness of 500 lb/in. By that I mean when we put an additional 1000 lbs on the front of the truck we want each front tire to move up 1", the same as we saw in the F-350.

This is where a free body diagram becomes necessary for me. A free body diagram is a rough sketch of a "solid" piece (it doesn't have to actually be solid, or one piece, it just needs to not move with respect to itself). You show what forces are applied in what locations. Newtons law tells us that unless the piece is accelerating, the sum of all of the forces on it must be zero, and the sum of all of the moments on it must also be zero (in physics a "moment" is the torque exerted on an object by applying a force on a lever arm).

Below is the free body diagram for one tire / axle beam assembly of an F-250 front suspension. Fp is the vertical force applied to the swing arm at the pivot. Fs is the force applied to the spring perch by the spring (it's also the force on the spring). And Ft is the force applied to the tire by the ground (or the amount of weight sitting on the front tire). The measurements of 26" from pivot to spring perch and 41" from the pivot to the contact patch are approximate measurements off my '97 F-250. And they are measured parallel to the ground, not in a straight line from the pivot to the contact patch or spring perch.

Since the forces in the vertical direction need to add up to zero, the diagram tells us that Fs = Ft + Fp. Or solving for Fp (we might want to later...), Fp = Fs - Ft.

And the sum of the moments must also equal zero. Moments need to be defined around a certain point. The physics don't care what point we use, but the math will be easier if we pick the "right" point. So we'll use the pivot point. The moment the spring exerts on the body around the pivot point is Fs x 26". The moment the ground exerts on it around the pivot point is Ft x 41. So that means that Fs x 26 = Ft x 41. Or solving for Fs, Fs = Ft x 41 / 26

Let's take the same 3000 lb front end, but we'll only look at one side. The weight on the tire is half the truck weight, so still 1500 lbs, just like the F-350. But now when we do the math, the force on the spring is Ft x 41 / 26, or 1,500 lbs x 41 / 26, or 2,365 lbs! (for what it's worth, Fp = Fs - Ft = 865 lbs).

Now let's put an additional 1,000 lbs on the front of the truck to move the tire up 1", the same as we did for the F-350. Now the new Fs = 2,000 lbs x 41 / 26 = 3,154 lbs. So we've put an additional 3,154 - 2,365 = 788 lbs on the spring.

But how much has it deflected? Since the tire at a 41" radius went up 1", the spring at a 26" radius went up 26 x 1/41 = 0.63".

Spring rate is (additional) force per (additional) deflection, so 788 lbs / 0.63" = 1,243 lbs/in. Just a little stiffer than the 500 lb/in F-350 spring!

This result was so much more than I expected that I didn't believe it. I've checked my equations and my math a couple of times though, so I'm pretty sure it's right. But if anyone wants to prove me wrong I won't be at all offended.

Edit to add: A few posts later I described how the F-250 spring stiffness likely increases significantly as the TTB geometry pulls it sideways. My calculations above are for this overall spring rate. If you put an F-250 spring on a solid axle it would not get reefed on so badly, so the overall spring rate of the same spring in that case would be lower. I'm still sure that the F-250 spring is significantly stiffer than an F-350 spring, even if it were on a solid axle. But likely it isn't 2.5 times stiffer if the pulling sideways isn't taken into account.

You just basically proved mathematically what I was referring to earlier in my reference to levers, specifically the three classes of levers and the calculations of force involved. It's been years since I took physics in high school, 55 to be exact, so my math is a little rusty. I was thinking if I were crazy enough to want to convert Darth to 4WD I would want to keep the front coil spring suspension but with a live axle rather than twin traction beams.

[Steve83]

WARNING - BORING GEEK STUFF AHEAD!

Guys - I'm getting lost in the physics....

I started to figure it out more completely and got lost too, so I had to go back to what my Statics TA back in college called "a fwee-ah ah-body ah-diagwam-ah" (or "free body diagram" for those of us to whom English is not our fourth language. And yes, I get that his English was MUCH better than my Chinese, so I'm really not putting him down).

What we're talking about here is how stiff the springs are. Spring stiffness is measured as the compression rate, how much force it takes to deflect the spring a certain amount. But springs are linear, so it's also how much MORE force it takes to deflect the springs a certain amount MORE.

Let's start with the F-350 suspension because it's so simple we don't need a free body diagram. Let's start with the assumptions that the front of the truck weighs 3000 lbs and we want a spring stiffness of 500 lbs/in (meaning it takes 500 lbs to compress the spring 1 inch). I just pulled that number out of the air, but it'll work for a comparison.

With 3000 lbs on the front of an F-350, that's 1500 lbs on each of the front tires, and 1500 lbs supported by each of the front springs.

Now if the springs have a rate of 500 lb/in, that would mean that if we put an additional 1000 lbs on the front of the truck we would be putting 500 lbs more on each spring and it will deflect 1" below wherever it is at ride height. And since this is a simple solid axle suspension, 1" more compression on each spring means each tire moves up (relative to the truck) the same 1".

Now lets take the F-250 front suspension and figure out what the actual spring stiffness needs to be to give an apparent stiffness of 500 lb/in. By that I mean when we put an additional 1000 lbs on the front of the truck we want each front tire to move up 1", the same as we saw in the F-350.

This is where a free body diagram becomes necessary for me. A free body diagram is a rough sketch of a "solid" piece (it doesn't have to actually be solid, or one piece, it just needs to not move with respect to itself). You show what forces are applied in what locations. Newtons law tells us that unless the piece is accelerating, the sum of all of the forces on it must be zero, and the sum of all of the moments on it must also be zero (in physics a "moment" is the torque exerted on an object by applying a force on a lever arm).

Below is the free body diagram for one tire / axle beam assembly of an F-250 front suspension. Fp is the vertical force applied to the swing arm at the pivot. Fs is the force applied to the spring perch by the spring (it's also the force on the spring). And Ft is the force applied to the tire by the ground (or the amount of weight sitting on the front tire). The measurements of 26" from pivot to spring perch and 41" from the pivot to the contact patch are approximate measurements off my '97 F-250. And they are measured parallel to the ground, not in a straight line from the pivot to the contact patch or spring perch.

Since the forces in the vertical direction need to add up to zero, the diagram tells us that Fs = Ft + Fp. Or solving for Fp (we might want to later...), Fp = Fs - Ft.

And the sum of the moments must also equal zero. Moments need to be defined around a certain point. The physics don't care what point we use, but the math will be easier if we pick the "right" point. So we'll use the pivot point. The moment the spring exerts on the body around the pivot point is Fs x 26". The moment the ground exerts on it around the pivot point is Ft x 41. So that means that Fs x 26 = Ft x 41. Or solving for Fs, Fs = Ft x 41 / 26

Let's take the same 3000 lb front end, but we'll only look at one side. The weight on the tire is half the truck weight, so still 1500 lbs, just like the F-350. But now when we do the math, the force on the spring is Ft x 41 / 26, or 1,500 lbs x 41 / 26, or 2,365 lbs! (for what it's worth, Fp = Fs - Ft = 865 lbs).

Now let's put an additional 1,000 lbs on the front of the truck to move the tire up 1", the same as we did for the F-350. Now the new Fs = 2,000 lbs x 41 / 26 = 3,154 lbs. So we've put an additional 3,154 - 2,365 = 788 lbs on the spring.

But how much has it deflected? Since the tire at a 41" radius went up 1", the spring at a 26" radius went up 26 x 1/41 = 0.63".

Spring rate is (additional) force per (additional) deflection, so 788 lbs / 0.63" = 1,243 lbs/in. Just a little stiffer than the 500 lb/in F-350 spring!

This result was so much more than I expected that I didn't believe it. I've checked my equations and my math a couple of times though, so I'm pretty sure it's right. But if anyone wants to prove me wrong I won't be at all offended.

Edit to add: A few posts later I described how the F-250 spring stiffness likely increases significantly as the TTB geometry pulls it sideways. My calculations above are for this overall spring rate. If you put an F-250 spring on a solid axle it would not get reefed on so badly, so the overall spring rate of the same spring in that case would be lower. I'm still sure that the F-250 spring is significantly stiffer than an F-350 spring, even if it were on a solid axle. But likely it isn't 2.5 times stiffer if the pulling sideways isn't taken into account.

You're still creating some confusion the way you worded the moments. The body is not part of your diagram or your calculations, so you're not figuring "...the moment the spring exerts on the body around the pivot..."; you're figuring the moment through the BEAM. A subtle distinction, and it doesn't directly affect the numbers, but it's still worth noting.

And I probably created some earlier... Hinging the suspension off the frame (twin I-beam) DOES change the spring force, vs. supporting one front wheel with the other (monobeam).

[Nothing Special]

You're still creating some confusion the way you worded the moments. The body is not part of your diagram or your calculations, so you're not figuring "...the moment the spring exerts on the body around the pivot..."; you're figuring the moment through the BEAM. A subtle distinction, and it doesn't directly affect the numbers, but it's still worth noting....

Yes, some of my wording might have led to some confusion. I was struggling a little with how much to keep my wording consistent with the methods I was using and how much to use terms that are more commonly used. What I ended up trying to do was use the terms my professors used, but define them here as (I thought) needed. When I said "body" I was talking about the swing arm / tire that I used as the body in my free body diagram, not the sheet metal. So yes, I was figuring the moment on the beam (or actually the beam/tire "assembly"), but that's exactly what I said, just in a term that caused some confusion. And as you said, it doesn't affect the numbers