Cutting coil springs to reduce ride height

Or...

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In all seriousness though, as has been discussed the only ones you could do are the fronts since the rear has reduced radius to locate on the center pad. Overall I think it'd be better to just buy the springs you want and re-sell the 4" set.
 
Or...

View attachment 446460

View attachment 446461


In all seriousness though, as has been discussed the only ones you could do are the fronts since the rear has reduced radius to locate on the center pad. Overall I think it'd be better to just buy the springs you want and re-sell the 4" set.

Cut the front springs to get the height you want. Reinstall, then torch the rear springs until level.


(it'll work, but don't do this)
 
By the formula you posted it basically does. Cut length in half and the only thing that changes is N, which will be really close to cut in half.

Are we going the route of one of Zeno's paradoxes? :)
 
When I was road racing motorcycles the front springs where too soft as delivered. The aftermarket was always slow getting suitable weight springs. We ended up cutting coils and adding a longer spacersto get it liveable until the aftermarket caught up. Physics is not a good idea... it's the law.
 
Keep in mind that cutting springs also stiffens them. The spring rate goes up because you are reducing the amount of bar that you are compressing.

Stiffens them? I would think it makes them "softer" because there's less spring to push back


unless you mean once compressed? I can see that, there's less movement available due to the greater compression due to less coils (if that makes sense)
 
Keep in mind that cutting springs also stiffens them. The spring rate goes up because you are reducing the amount of bar that you are compressing.

This is absolutely correct, with one minor correction. You aren't compressing the bar, you're twisting it, as PeteN95 noted.

Yes, a 150 lb/in spring is a 150 lb/in spring. But cut it in half and it is now a 150/0.5 or 300 lb/in spring.

Also correct.

Edit - found it.

Note that the number comes out of this is in SI units (Newtons per meter). To convert to the more familiar lb/in, you have to divide by 175.

psrivats, that is indeed the formula used. However, it does not require a conversion factor depending on units. There are only two units used in the equation - force and distance. As long as consistent units are used, it will yield a correct result. For instance, for US units, use pounds for the force unit and inches for all distance units, and you're good to go. Typically, the value for shear modulus is 11,500,000 pounds per square inch. That's close enough for most steel and steel alloys. Actual values vary a bit, but that value will be within 2-3% of actual values.

Yes it does. A coil spring is simply a torsion bar coiled up. If you shorten a torsion bar, it is stiffer because it is harder to twist a shorter piece of metal.

Correct again.

It does change, but does it change enough to matter?

On a TJ, no, it doesn't change enough to matter (continue reading for more on that.)

At the end of the day, none of it matters. If the OP is going to cut his front coil springs, he will simply cut off a little at a time until he gets the ride height he wants. Then he buys a pair of rears that also provide the ride height he wants.

Agreed, but there is a way to get close on the first cut (continue reading...)

By the formula you posted it basically does. Cut length in half and the only thing that changes is N, which will be really close to cut in half.

Basically true. One could argue that a progressively-wound or multi-rate spring won't follow that rule, but the formula psrivats posted doesn't address helix angle because, in practice, it doesn't matter. The spring won't be stable if the coils are far enough apart for that to matter. It will bow severely when compressed.

K is constant for a given spring. Cut the spring in half and it's no longer the same spring.

True again

I've attached a spreadsheet I put together this evening to help the OP determine how much to cut off. Unfortunately, you'll need Microsoft Excel to open it. I'm not savvy enough to create a Google version of that so that anyone can look at it, so if someone wants to do that, knock yourself out. Note - I don't have data on the Zone springs. I do have my RockJock CE-9132F springs sitting on the floor of my shop, so I measured those and put the values in for them. I then tweaked the amount cut off until I got close to a 14" installed length, which I believe is a 2" lift (someone correct me if I'm wrong, please). It comes out to 2.75" cut off, and the spring rate increases from 136 lb/in to 155 lb/in (a 14% increase). As has been discussed on this forum many times, spring rate on a TJ is just not a factor in ride quality. A 14% increase won't matter. Of course, the RockJock spring is dual rate, and I didn't have time to actually verify that for the spreadsheet calculations, but I don't think it'll matter much; however, I could be wrong, so be careful if you actually use this on a multi-rate or progressive spring like the RockJock springs. I could take a look at how the calculations turn out for a dual-rate spring, but I don't have time tonight. Here's a screen shot of the spreadsheet:

1691028987042.png


The way the spreadsheet works, you enter the yellow highlighted cells, and the other cells get calculated for you. If the OP decides to use this, he'll need to replace all yellow cells with measurements from his springs, with the exception of the Shear Modulus and the Stock Spring Installed Length because those won't vary from spring to spring. Also, spring rates in practice can have wide variations, so I'd suggest that you cut less off at first because it's tough to put part of a coil back once you cut it off!

Final caveat - I have designed too many valve springs and suspension springs in my previous career to count, and I even wrote the specification manual for suspension springs at one company I worked for. However, as a human, I'm prone to mistakes (especially at this age). I can't guarantee that this spreadsheet is correct, so use it at your own risk, please! If you do find an error, let me know and I'll post a correction.
 

Attachments

  • Cut Spring Calculator.xlsx
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Tremendous amount of information from all, thanks to all

my plan will be to cut the front springs
and replace the rear springs after i see the results of the front spring modification
 
This is absolutely correct, with one minor correction. You aren't compressing the bar, you're twisting it, as PeteN95 noted.



Also correct.



psrivats, that is indeed the formula used. However, it does not require a conversion factor depending on units. There are only two units used in the equation - force and distance. As long as consistent units are used, it will yield a correct result. For instance, for US units, use pounds for the force unit and inches for all distance units, and you're good to go. Typically, the value for shear modulus is 11,500,000 pounds per square inch. That's close enough for most steel and steel alloys. Actual values vary a bit, but that value will be within 2-3% of actual values.



Correct again.



On a TJ, no, it doesn't change enough to matter (continue reading for more on that.)



Agreed, but there is a way to get close on the first cut (continue reading...)



Basically true. One could argue that a progressively-wound or multi-rate spring won't follow that rule, but the formula psrivats posted doesn't address helix angle because, in practice, it doesn't matter. The spring won't be stable if the coils are far enough apart for that to matter. It will bow severely when compressed.



True again

I've attached a spreadsheet I put together this evening to help the OP determine how much to cut off. Unfortunately, you'll need Microsoft Excel to open it. I'm not savvy enough to create a Google version of that so that anyone can look at it, so if someone wants to do that, knock yourself out. Note - I don't have data on the Zone springs. I do have my RockJock CE-9132F springs sitting on the floor of my shop, so I measured those and put the values in for them. I then tweaked the amount cut off until I got close to a 14" installed length, which I believe is a 2" lift (someone correct me if I'm wrong, please). It comes out to 2.75" cut off, and the spring rate increases from 136 lb/in to 155 lb/in (a 14% increase). As has been discussed on this forum many times, spring rate on a TJ is just not a factor in ride quality. A 14% increase won't matter. Of course, the RockJock spring is dual rate, and I didn't have time to actually verify that for the spreadsheet calculations, but I don't think it'll matter much; however, I could be wrong, so be careful if you actually use this on a multi-rate or progressive spring like the RockJock springs. I could take a look at how the calculations turn out for a dual-rate spring, but I don't have time tonight. Here's a screen shot of the spreadsheet:

View attachment 446586

The way the spreadsheet works, you enter the yellow highlighted cells, and the other cells get calculated for you. If the OP decides to use this, he'll need to replace all yellow cells with measurements from his springs, with the exception of the Shear Modulus and the Stock Spring Installed Length because those won't vary from spring to spring. Also, spring rates in practice can have wide variations, so I'd suggest that you cut less off at first because it's tough to put part of a coil back once you cut it off!

Final caveat - I have designed too many valve springs and suspension springs in my previous career to count, and I even wrote the specification manual for suspension springs at one company I worked for. However, as a human, I'm prone to mistakes (especially at this age). I can't guarantee that this spreadsheet is correct, so use it at your own risk, please! If you do find an error, let me know and I'll post a correction.

Question about the "amount to cut off" field: I'm guessing that's measured along the length of the spring while free?

If you measure how much to cut while under load, doesn't that give you exactly how much change you'll see when installed? Not an easy measurement to take, though, unless you're doing full coils.

As a practical tool it seems that a measurement along the wire would be more precise.
 
Question about the "amount to cut off" field: I'm guessing that's measured along the length of the spring while free?

Yes, the amount to cut off is measured perpendicular to the axis of the spring coils with the spring under zero load.

If you measure how much to cut while under load, doesn't that give you exactly how much change you'll see when installed? Not an easy measurement to take, though...

Yes, it does, and yes, it is a difficult measurement to perform. That is why I made the calculator. It's much easier to measure, mark, and cut the spring with it out of the vehicle.

As a practical tool it seems that a measurement along the wire would be more precise.

I don't know that I would agree with that. Precisely measuring a round object wound in a helix and installed in a vehicle seems a frustrating task...
 
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