update on Rich's build

[RICH MUISE]

Interesing and easy to comprehend read on that link, Lynn. Thanks. I'll get that info later, only questionable part will be determining where to stop including/measuring/counting the last (touching) coil....my guess is at the point of contact.

[canadian_ranchero]

here is a copy of the stock front spring information.the b7a-5310-n are stock for a 6 cyl 2door custom.the b7a-5310-e are what my wagon used.the n is taller and the weight at load height is 2125 lbs.the e is shorter and the load height weight is 2025 lbs.but the rates are different 400 for the n and 500 for the e.so the e should put the ride height lower but would be stiffer.i am thinking that the e spring will work better than the spring i put in [i think they are R or a L ] 

[lalessi1]

Here is a link that further explains counting coils. I counted 7 1/2 active coils on my "stock" springs. They are actually from an Edsel.

http://springipedia.com/compression-counting-coils.asp

Based on:  Wire Diameter             .697"
               Mean Coil Diameter     4.764"
               Active Coils                7.5
           CALCULATED SPRING RATE = 409 lbs/inch

The specified compressed height of the spring at standard ride height is 9.6". I measured a free length of 15" which means the load needed to compress the spring from 15" to 9.6" is 5.4 x 409= 2208.6 lbs.

If I cut 2 coils off of the spring the new spring rate jumps up to 558 lbs/inch

Now the "new" spring free length is 11 3/4" (measured). The 2208.6 lbs load will compress the cut spring 4". That gives an installed length at "design load" of 11.75 - 4 = 7.75". That compared to the 9.6" is about 1 7/8" more but that will lower the car more than that based on the lower control arm geometry. I believe the car will be lowered by roughly 1.5 times the additional spring compression. Cutting two coils will lower the car 1 7/8 x 1.5 or around 2 7/8".

This excersize shows that you lower the car by almost the same amount that you shorten the stock spring. For me that is an unexpected result.

                 

[RICH MUISE]

So....I understand(at least following) pretty much what you're doing above, but in the case where you don't know the load ??
In other words, your spring was made to compress to the std ride height with a load of 2208. lbs. In the case of my drivetrain being set back further than normal, I'm not sure that going thru all the calculations is going to be accurate since we don't acuually know anything about the weight on the front of our cars except the difference in drivetrain weights, which does not include factors for weight distribution differences. Also....2208 lb load??? wouldn't that be doubled since there are two springs?? so 4416 lb load on the front when our Customs only weigh 3450ish total? I'm so confused.
I'm really liking the idea now of cut some and try it, lol.
My springs btw:
.70 wire size
coil space (cl to cl) 2 5/8
OL 16 1/8
OD 5.43
8 active coils (I think)

[lalessi1]

I thought you might ask about that load! Basically that load is the load on the spring and NOT the weight of the car on that corner. The lower control arm carries the weight of the car at the centerline of the tire and is acts as a lever to compress the spring. The distance between the "pivot line" of the inboard bushings and spring center is compared to the distance from that "pivot line" to the tire centerline. For example if inner distance to the spring is 6" and the distance to the tire is 12" the weight of the car at the corner is multiplied by 12/6 or 2. In other words a corner weight of 750 lbs becomes a 1500 lb spring force.

The same principle also means that a 1" spring compression corresponds to a 2" suspension movement using that same ratio.

Now since I was trying to demonstrate much cutting a spring would drop the car, I needed a reference load to get close so I used Ford's spring info that standard compressed length is 9.6" as a base value. Since your car is lighter, that compressed spring length would be longer as a starting point but the CHANGE in that  value due to cutting the spring should br pretty close. (I also guessed the control arm ratio to be 1.5 for the calculations I made).

Ok, so now I will try to use closer numbers and your values and actually measure the control arm. THE MADNESS CONTINUES....

[lalessi1]

Here we go.....

YOUR spring rate calculated to 399 lbs/inch. That matches the lighter factory springs.
The 16 1/8 free height is about 1" to 1 1/2" higher than the factory spring. (Depending on which 400 lb spring you compare it to.)

I measured my control arm. I got 8 1/4" from the inner pivot line to the spring centerline. I measured 17 1/2" from the inner pivot point to the center of the tire (tread). The control arm "lever factor" is 2.12.

As a check, looking at the factory "Capacity at Normal Loaded Height" for spring loads, they run from a low of 2075 to 2500 lbs. Using the 2.12 factor above these spring loads translate to 979 to 1179 lbs front LOADED corner weights. That seems very reasonable and correct.

Now lets assume you reduced your front corner weight by 100 lbs over the lightest factory value. You corner weight is 879 lbs and your spring load is 879 x 2.12 or 1864 lbs.

Your spring would compress by 1864/399 or 4.67" from the 16 1/8" free length giving you an installed spring height of 11.45 inches. Compare this to the stock length of 9.6" and your spring is almost 1 7/8" longer than that. That translates to a suspension height of 1 7/8" x 2.12 or 3" HIGHER THAN STOCK.

If you cut 1 coil.... Spring rate         456 lbs/in
                          Free length         13.5 " (using your coil spacing)
                          Installed height    9.4"
                          Suspension drop   4.35"
                         
                         

[RICH MUISE]

Wow.....and the scary part is I'm understanding/following what you did. OK, well, it's obvious I do not want to remove a complete  active coil, I'd end up back where I am now. but the next thing to figure out is exactly what it is I am cutting off. The flat/ground end of the spring has a complete inactive coil on it, but not so clear cut on the bottom end. The coil's spiral angle on that last coil decreases so the last coil is not fully active because the spacing is closer, but not entirely inactive either. I figured a half coil.
So, the new confusion is....the calculations tell you how much cutting an active coil changes things, but how do you figure out how much a partially active coil changes the function?

My current thoughts are, cutting that first full coil off is not the equivalent of cutting a full active coil so removal of that first coil is not going to drop the suspension the same 4.35.
Removing that first complete coil, if it IS considered inactive would reduce the overall lenth of the spring by .950. I'm going to (probably wrongfully) assume here removing an inactive coil does not affect the spring rate, only the physical shortening of the spring.?, so .950x 2.12+approx 2". Now, if I want a 3" drop from where it was, I need an additional 1", an that will come from removing a portion...23%...of the active coil.
If I have followed your math with any comprehension  at all, I'm concluding along with the said assumptions that if I remove the first inactive coil and 1/4 of the next one,
I may be/should be /probably won't be/ close to the 3" drop I need. Which by the way, is pretty darn close to the 1 1/2 I was talking about removing for the first trial and error. 2.625 x .23% =.95=1.55

[lalessi1]

This is my understanding of where you are...

You have 3" drop spindles?
With the the "stock" springs you have (when originally installed) you were sitting higher than stock.
You would like a 3" drop from actual stock.
The Aerostar springs you have now are too low.

If you cut the 16 1/8" spring from where it is now to a free length of 13 1/2" (one coil) it should compress to an installed length of 9.4" or slightly lower than stock, leaving you with a 3" spindle drop. The 4.35" is from where it was.... (which was substantially higher than stock). http://springipedia.com/compression-counting-coils.asp  says that an open ended coil is counted as a full coil. and a ground end is not counted. Now what we have is "close" to a pure open ended coil so cutting a little more might be in order. Best course of action is measure to 13 1/2" free length and see how many coils that actually is. and we can refigure the spring rate. The 8 active coil number you gave me calculated the rate to a Ford book number, so it is probably correct. Do you know the Eaton spring rate for what you have?

Correct any info that I have wrong and I will run the numbers again. Now that we have some geometry numbers the calculation is simpler.

PS, I am probably going to do the same thing on my car eventually.

[RICH MUISE]

You have 3" drop spindles? 2 1/2 actually
With the the "stock" (eaton)springs you have (when originally installed) you were sitting higher than stock. yes, about 1 1/4 higher than oem I believe, even with the dropped spindles
You would like a 3" drop from actual stock. No, 3" from where it was with the Eaton's
The Aerostar springs you have now are too low. yes, an inch to inch and a half too low

[lalessi1]

Got it. I will "run the numbers" tomorrow.

[KYBlueOval]

You have 3" drop spindles? 2 1/2 actually
With the the "stock" (eaton)springs you have (when originally installed) you were sitting higher than stock. yes, about 1 1/4 higher than oem I believe, even with the dropped spindles
You would like a 3" drop from actual stock. No, 3" from where it was with the Eaton's
The Aerostar springs you have now are too low. yes, an inch to inch and a half too low

Rich, I spoke with Gary at Eaton Spring yesterday regarding my springs, looking for confirmation that they would work with the engine/trans combination I've decided to use. Gary was more than willing to exchange my springs, IF I wanted to try a different spring. I'm going to use what I have, as the spring that he would exchange it for, would add an additional 1 inch in drop, to the 2 1/2 I'm getting from the spindle.
Before you start cutting, you might want to call Gary and  inquire as to what he can do. Just a suggestion.
John

[RICH MUISE]

I've talked with them several times......I got mine way too long ago for them to do an exchange. He did tell me they had a spring that would lower my car 2" from where it was, but that's still going to make it an inch too high, so I'd need to pay full $, plus shipping, just to try it. In all likelyhood, that spring would have been better than what I had, but still not right, so I'd end up cutting anyway.
That part number he was talking about was MC1364. "2" or more drop from what you have now" were his exact words. That is the only "stock" spring from eaton's he talked about as a posible fix. So, anything beyond that I guess would be getting into custom made.

[hiball3985]

This whole front spring thing is a fiasco. After all my searching and calculations I didn't find any replacement spring that would work with the 2 1/2 Granada spindle drop and ended up using the OEM springs, I would have proffered new but I just couldn't find what would work off the shelf.

[lalessi1]

Here we go.... again: 

Ok the numbers actually work out very well.

The Eaton springs based on your info would have had an installed compressed height of 11.4". That would have given you 
1 1/4" higher than "stock" even with 2 1/2" dropped spindles. The spring load calculates to 1900 lbs., close to the weight estimates I used earlier. To get a 3" drop below that you would need an installed compressed height of approximately 10". that would give you a drop (with the spindles) of 1 3/4" lower than stock or 3" lower than Eatons with the dropped spindles.

If you cut 3/4 of a coil, the spring rate becomes 441 lbs/inch. The free spring length becomes 16 1/8 - 2 5/8 x .75 = 14". The installed compressed length would be 14 - 4.3 or 9.7"

Cutting 1 coil       Rate 456,    Free length 13.5,  Installed compressed height 9.33

Cutting 1/2 coil     Rate 425,    Free length 14.8,  Installed compressed height 10.3

I would shoot for a free length of 14 1/4 , mark that spot and see how much of a coil(s) to cut off.

[RICH MUISE]

Because of the geometry of that first coil, trimming to free length of 14 1/4 would remove pretty close to 1 1/2 coils. At least that for sure gets it into full open coil for the math
With 'our' previous calculations, I had mentioned I thought 1 1/4 coils would be a good place to start, so I'm good with the -1 1/2 coils/-1 7/8"/14 1/4 as a final estimate.
I'm going to find some 1/2" spacer/osolators to have on hand in case needed, and cut the springs to the 14 1/4. I'm not sure when the actual replacement is going to happen, depends on when Kip has some shop time open...may not be until the 2nd week in March after we get back from Phoenix.
I ordered two new tires yesterday. They were able to order the next size up for me in the same brand. So now my new front tire size will be 215/70/15's. They should be here Friday.
I'm having a heck of a time finding lowering blocks for our cars...most of the blocks available for "2 inch leaf springs" are actually 2 1/2". I see that as an issue with the rubber/sheetmetal isolator pads not fitting, as well as the fact they are not solid, but thin walled, so the outer walls would actually not even contact the spring. Am I missing something here? I thought this would be really easy, but I'm finding most of the speed shops don't even carry them. WTH??