Is a section of a spring broke for front strut on a 2018 VW Tiguan (222k) do we park till we get drive to mechanic or can drive on and get fixed really soon? Once I removed section clank stopped it seems to drive fine.
Broken spring on front strut of a 2018 VW Tiguan
- Thread starter andrew_j
- Start date
It'll be fine. I'd plan to get it fixed but I wouldn't avoid driving the car.
CleanSump
Site Donor 2026
Is this the bottom coil? Common on some BMWs too. You can drive, just don't be that crazy kid.
Crazy way to lower that bad-boy!
It's probably not even that much lower, if at all. As you shorten a spring you increase the spring rate... Half a coil? Could probably just leave it that way.
Just cut the same off the other side.
Great combo with either spring, I have used both on many vehicles and never had a problem. Just look for any colored dots on the original spring in case the new one is offered with different variants.Rock Auto has quick struts for $75 each
OK fine, if you want real parts, they also carry Bilstein shocks/struts and Lesjofors and Suplex springs
How much driving do you need to do before you can get it fixed?
You must make sure the remaining rough edge can't shift and possibly cut into he inner sidewall.
Are you planning on keeping this 222k vehicle? This can effect the "quick strut" vs. a better repair decision.
Are you planning on keeping this 222k vehicle? This can effect the "quick strut" vs. a better repair decision.
Back in my days ricers sometimes CUT their spring for lowering, then says OEM struts sucks.....
I don't think that's how spring rate works. You have the same rate just bottom out a little sooner before you hit the bump stop.It's probably not even that much lower, if at all. As you shorten a spring you increase the spring rate... Half a coil? Could probably just leave it that way.
Just cut the same off the other side.
No, that is how it works.
k=(Gd⁴)/(8ND³), where N is the number of coils.
https://grassrootsmotorsports.com/articles/how-calculate-spring-rateand-how-understand-cuttin/
CleanSump
Site Donor 2026
It gets stiffer as it gets shorter.
However, your distance also reduced.
Let's say you have a spring, cut in half. Your total spring support of the new half length spring would be 1/2 as stiff, travel half as far before bottom out, if the same weight is applied to it.
I used to sit on broken sofa all day and I know how it feels when a spring is broken.
If you don't believe me, next time when you eat steak try to double stack and bite into it. It would be twice as stiff. Thinning the steak in half would not make the steak harder to chew. I think you understand that right?
True.
If you take a linear-rate coil spring and cut in it half, you will end up with two coil springs of a higher linear-rate, meaning it will take more force to compress the cut coil per inch than it would to compress the original coil
A steak is not a coil spring.
Everything that can compress is a spring. A piece of foam is a spring, an eraser is a spring, metal is a spring, as long as they don't went too far into the plastic region and stay inside the elastic region, it will go back to the same position after you release your force / weight, it is a spring.
Go get a scrap spring from a mechanical pencil, and cut it in half. They don't become stiffer and take more force to compress per inch. The number of coil per distance travel hasn't change.
So back to your equation:
Where k=spring rate; d=diameter of the spring wire; G=shear modulus of elasticity of the spring material (11,500,000 for a steel spring); D=mean diameter of the spring's coils; and N=number of active coils.
k=(Gd⁴)/(8ND³)
Your material hasn't change, so G hasn't change. Diameter of your wire hasn't change, so d hasn't change. Diameter of your spring coil hasn't change, so D hasn't change.
In order for your understanding to be correct after you cut the spring you have to PULL IT LONGER back to the original length as well. That way the spring slope sharper, and would be stiffer. But if you didn't pull it longer it would still be the same N.
The N here is number of active coil for the total spring length (total spring length is not in this equation). If you reduce both the number of coil AND the total spring length at the same time, this equation doesn't apply any more. The k here also has the force per total spring length in unit. They cancel each other out. So if you are not keeping the total spring travel length the same as before, by cutting the spring, this won't be stiffer.
The equation should be k (spring rate) = F (force applied) / L (total spring length) = (Gd⁴)/(8ND³) = (Gd⁴)/(8(n / L)D³)
where n = number of total coil turn, L = total length of the spring and N is really number of coil turn / total length of the spring that your link didn't mention.
So if you cut the spring you are having the same number of coil turn per spring length and N doesn't change because you are not changing the slope of the spring (which is what number of coil for the same length really is).
https://engineerfix.com/what-is-spring-rate-and-why-does-it-matter/
So, your total spring "rate" is the same because your distance and number of coil both reduce. However, because your total spring length is reduced, the total force it can support for the same rate is reduced.
Thus, you have a weaker total spring for the same car weight and it will bottom out sooner.
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CleanSump
Site Donor 2026
Someone has it backwards. Engineering should be left to trained Engineers.
What one thinks they see isn't always what is going on.
What one thinks they see isn't always what is going on.
All the coils in a spring experience the same force. Given a force, each coil compresses the same distance. The total distance the spring compresses is the per coil distance times the number of coils. If there are fewer coils it takes more force to move the same distance, i.e. a stiffer spring with higher k.
But if you just break or cut a car spring without stretching the remaining coils out, the car starts lower and there is less distance to work with before the suspension bottoms out. The force needed to bottom out a shortened spring turns out to be the same as the original spring. It is a harsher ride to get there though.
With all that said, the last half coil of a spring is often inactive because it starts out intentionally touching the coil above it. This is done to make the spring sit flat on the plate under it. So now you have a shorter spring but no increase in k.
But if you just break or cut a car spring without stretching the remaining coils out, the car starts lower and there is less distance to work with before the suspension bottoms out. The force needed to bottom out a shortened spring turns out to be the same as the original spring. It is a harsher ride to get there though.
With all that said, the last half coil of a spring is often inactive because it starts out intentionally touching the coil above it. This is done to make the spring sit flat on the plate under it. So now you have a shorter spring but no increase in k.
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