This is relatively simple. If you take a 3" exhaust, I know from basic maths at GCSE level that 2pieR is the going to be the area, so in this case it is around the 9.4 square inches figure. Fair enough....

Now, the point I am struggling with is what happens if you "squeeze" said pipe into more of an oval? I imagine initially area will remain similar but there has to be a point where the area is diminishing until the point where eventually it will be flat shut and there'll be zero square inches.

Tell me I'm not being a bit "fick" please. There's a point to this (in the end )

pi*a*b where a = half length of major axis (horizontal) and b = half length of minor axis (vertical)

pi*a*b where a = half length of major axis (horizontal) and b = half length of minor axis (vertical)

Cheers. After doing some rough sketches I can see how right away the area will diminish, based on estimated 4" across x 2" high dimensions when crushed a bit. Just means if I'm going to have my new exhaust crushed to save on ground clearance, it will need to be much bigger to start with.

2X6, 1X12 :eek:

when moving into an oval it can get a bit more calculus'y rather than straight formula, if you know what width/height combo is available for space post it up and I'll give it a go.

Alternatiivley it can be sketched in AutoCAD or Inventor and it'll work out the areas itself (not that I hate calculus or owt)

eh, am I being dumb. But if you have a circle of said size, then squash it with out adding or removing any metal; It'll still cover the same amount of space but in a different shape ? Your not adding or removing material from the exhaust so how can the area covered shrink ?

Sorry, What I meant is its straight forward to measure the area of a circle, but if you only have a usable area of 2" by 4" then its not as straightforward as pi R^2.

I'm assuming the system will be designed before it is made and wont just be crushing random bits of pipe :P

When you know the size of the pipe required, then the space available you can work out the equation of the curve required (or half the curve and double it)

I think it just does - The vertical dimension can be crushed and it won't keep on going outwards forever. Eventually it will just pancake.

Mr W - wouldn't mind some playing around, starting with a 2.5"/3.0"/3.5" and crushing each pipe down so it's height is half what it was originally to start with (and whatever horizontally it happens to be )

its nothing like that simple as bending/crushing etc pipe will affect its flow rates and resistence, this isn't school maths at all.

its nothing like that simple as bending/crushing etc pipe will affect its flow rates and resistence, this isn't school maths at all.

Excatly. Thats why exhaust makers use mandrel benders which retain the exhausts round profile.

Thats certainly desirable and IMHO thats why you pay more for a system from a large manufacturer like Milltek, H&S, Tubi etc because they have been designed accordingly.

The places you can go into with all the right equipment and they make you an exhaust on the spot are not so good as they are just cobbled together to follow the route requested and in the pipesize required, no thought into the effectivness of the pipe at all.

If I get a min at work today Jason I will have a look but crushing a pipe to half its original diameter will start to cause failure on the 4 'corners' if you know what I mean and would probably have to be hotworked with a former in the middle.

It might not go to half the height. That was just a figure I plucked out the sky. No bends will be crushed either as I'll be using bought in bends from an exhaust supplier - Only straight bits.

What's wrong with the exhaust you've got now?

Stop messing

This is relatively simple. If you take a 3" exhaust, I know from basic maths at GCSE level that 2pieR is the going to be the area, so in this case it is around the 9.4 square inches figure. Fair enough....

Would appear we have all missed the deliberate error, 2piR = Circumferance not area:O

Oh well, lol. Geometry was never my strongest point during A level maths and that was a fair number of years ago.

would appear we have all missed the deliberate error, 2pir = circumferance not area:o

pi r^2.

Can you not just fit a 4x2 box section along the bottom then fillet the round pipe to it at each end.:confused:

But did you pick up on Jasons error:P

Is it even necessary to "crush" the pipe? I've yet to see a system, even using 3" diameter tube, where the widest parts of the system weren't the silencer boxes (or catalytic convertor).

Is it even necessary to "crush" the pipe? I've yet to see a system, even using 3" diameter tube, where the widest parts of the system weren't the silencer boxes (or catalytic convertor).

Well, let's just say Skoda never designed the fabia to have an exhaust going where this one will be going.

pie-r-squared is for area of a circle! Pi*r^2

pie-r-squared is for area of a circle! Pi*r^2

Here's a little diagram I just did in CAD, I mean MS Paint. As well as the bend to the offside, there's also going to have to be 2 kinks to take it down slightly and then back level.... Also going to involve a flexi joint.... Kind of an idea of what it will look like when you get your head down to sill height - Only half of the tail will be obvious, peeking out under the sill. However nothing, nyada, will be visable from a normal viewpoint. It also what it looks like from above (if the car weren't there )

Stainless Steel stockist and supplier UK | National Delivery | Metal Finishing Services would save faffing about with crushing etc, then you know what your CSA will be.

Is this for performance or just to be different? I have a feeling that you might loose too much back pressure with such a short run?

If i'm interpreting the diagram right....... you're going to change from covering the car behind in soot to cover the pedestrian on the pavement in soot instead ? :P:D