Performance Maths Part 1

Were you one of those students who in school loved gazing out of the window during maths lessons? Someone for whom the intricacies of pi and square roots and sines weren’t just of little interest but the whole deal was simply repugnant? Join the club. But that probably stops you now using maths as an extremely useful tool when modifying cars.

In this occasional series we will not cover the purity of maths, the intellectual pursuit so beloved of mathematicians (and often those that teach maths!). We won’t be showing you multiple ways of doing things, or how the equations can be proved. Instead, we’ll just concentrate on using maths to achieve better results in car modification.

A mix of units will be used in this series – the most commonly used units being used for each example. For example, exhaust pipe diameters in inches and petrol tank volumes in litres.

Cross-Sectional Areas

Knowing and comparing cross-sectional areas is vital in car modification. This is the case because the flow of a fluid (like air or exhaust gas or fuel) is largely proportional to the cross-sectional area of the pipe carrying it. So, is a single 3 inch exhaust pipe going to flow more than dual 2 inch pipes? Or, if the rest of the engine intake system is in 2.5 inch round tube, is a mouth that is 2 inches by 1 inch likely to be causing a restriction?

Cross-sectional areas of square and rectangular sections are found simply by multiplying height by width. So if the intake system has a mouth that is 2 inches wide and 1 inch high, the cross-sectional area is (2 x 1) = 2 square inches. If you change the units to centimetres, exactly the same calculation applies except the answer will be in square centimetres.

Cross-sectional areas of circular sections are found by: radius multiplied by itself multiplied by 3.14 (pi). So a round pipe that is 3 inches in diameter has a cross-sectional area that’s calculated by: 1.5 x 1.5 x 3.14 = 7.065 inches square (7.1 inches square when rounded up.) Note that the 1.5 inches used in the calculation is the radius, that is, half the pipe diameter of 3 inches.

Now don’t underestimate this stuff – it’s very good information. Let’s look at the comparison we stated near the beginning of this section: is a single 3 inch exhaust pipe going to flow more than an exhaust using dual 2 inch pipes?

We now know that a 3 inch pipe has a cross-sectional area of 7.1 square inches. What about a 2 inch pipe? 1 x 1 x 3.14 = 3.14 square inches, and if you have two of the pipes, the total internal cross-sectional area is 3.14 + 3.14 = 6.28 inches square, rounded to 6.3 square inches.

So at 7.1 square inches, a single 3 inch exhaust has a bigger cross-sectional area than the cross-sectional area (6.3 square inches) of dual 2-inch exhausts – the single 3 inch pipe will flow more. However, there’s not a helluva lot in it, so if for example it’s easier to package two 2-inch exhausts under the car than a single 3 inch, you won’t be that far behind. (And what if you use dual 2.5 inch exhausts? What’s the comparison then?)

This stuff isn’t theoretical. A proud owner once showed me the performance exhaust he’d just had fitted to his car. Extractors, mufflers, cat converter, 2.5 inch pipe diameter – they all looked fine. And then I saw the exhaust tip. It was rectangular in section and looked to my eyes like it would have a pretty small cross-sectional area. In fact, when I later worked out the cross-sectional areas of the pipe and tip, the tip proved to be just OK, but there wasn’t much in it: if the tip had been even half an inch shorter in height, the shiny chrome bit would have been a restriction to flow...

Sometimes the intake system of a car will include a triangular bit of pipe. The area of a triangular section is worked out by 0.5 multiplied by the width of the base multiplied by the height. So a triangular cross-section that’s 3 inches wide and 1.5 inches high has a cross-sectional area of 0.5 x 3 x 1.5 = 2.25 square inches.

Area of a rectangle or square = width x height Area of a circle = radius x radius x 3.14 Area of a triangle = half base x height

Volumes

Once you know how to calculate cross-sectional area, it’s very easy to calculate volume. If the cross-sectional area is constant, all that you need to do is to multiply the cross-sectional area by the length.

Say we’ve got a cylindrical swirl pot that’s 3.25 inches in diameter and 6 inches long.

Step 1 is to calculate the cross-sectional area. Because we use the radius (not the diameter), we need to divide 3.25 by 2. That equals 1.625 inches. So the calculation of cross-sectional area is 1.625 x 1.625 x 3.14 = 8.29 inches square (or 8.3 inches square when rounded off).

Step 2 is to multiply the cross-sectional area by the length. So 8.3 x 6 = 49.8 cubic inches, or near enough to 50 cubic inches.

To convert from cubic inches to cubic centimetres, multiply by 16.4. 50 x 16.4 = 820 cubic centimetres, or 0.82 litres.

The same approach is taken when calculating the volume of a box – work out the cross-sectional area and then multiply by the length. So if a box is 6 inches wide, 7 inches high and 15 inches long, the calculation becomes 6 x 7 x 15 = 630 cubic inches. (That’s 10.3 litres.)

Working out volumes is important in car modification. The swept volume of each cylinder can be calculated in this way. So can the volume of the inlet manifold plenum chamber and filter airbox - if you’re making your own, it’s good to make quick comparisons with factory ones. Ditto with the internal volume of mufflers - muffler volume is a very important aspect in controlling sound levels.

Volume = cross-sectional area x length (applies to cylinders and boxes)

Mass

Knowing the volume is also useful when calculating the weight of something.

For example, how much will a petrol tank weigh if it has a capacity of 50 litres? (To put this a different way, if you’re building your own car, how much weight are you putting way out the back when the petrol tank is full?) In this case you already know the volume – 50 litres. Now, how much does each litre of petrol weigh? The quick Google answer is about 740 grams per litre (ie 0.74kg per litre). So the contents of the petrol tank weigh 0.74 x 50 = 37kg.

The same sort of calculation can be made if you want to compare the weight of different building approaches. For example, a floor panel might use flat steel sheet that’s 1.6mm in thickness. How much will the panel weigh?

If the panel is 50cm x 75cm, the area is 3750 square centimetres. But what is the volume – how do we take into account the 1.6mm thickness? It is vital that the units remain the same, so 1.6mm needs to be converted to 0.16cm. The area (3750) multiplied by the thickness (0.16) gives a volume of 600 cubic centimetres – 0.6 litres. Now, how much does each litre of steel weigh? The answer is about 8kg per litre. Our 0.6 litres multiplied by 8 kg/litre gives us an answer of 4.8kg.

Now it’s easy to say: why bother with the calculation – why not just weigh the bloody steel? The answer is that different design approaches can be much more easily trialled on paper. If you’re working on a project where every gram is important, many different design approaches can have their weight calculated before a single piece of metal is cut.

Mass = volume in litres x kg/litre of material

Note: The kg/litre measurement of different materials is most often specified as the ‘specific gravity’. Specific gravity numbers are quoted without units but in fact they are in kilograms/litre. Some examples are:

Product	Specific Gravity
Aluminium	2.6
Brass	8.4
Brick	8.7
Copper	8.8
Epoxy	1.8
Glass	2.4
Gold	19.3
Iron cast	7.1
Lead	11.3
Nickel	8.9
Platinum	21.5
Steel	7.8
Tungsten	19.2
Wood min	0.35
Wood max	0.99

Conclusion

A tool kit that includes the ability to calculate cross-sectional area, volume and mass is a very powerful one indeed. I reckon in my own car modification I’d use one or other of these tools maybe 30 or 40 times a year.