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.

CrossSectional Areas
Knowing and comparing crosssectional 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 crosssectional 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?
Crosssectional 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 crosssectional
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.
Crosssectional 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 crosssectional 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
crosssectional 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
crosssectional 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 crosssectional area than the crosssectional area (6.3 square
inches) of dual 2inch 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 2inch 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 crosssectional area. In fact, when I later worked
out the crosssectional 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 crosssection that’s 3 inches wide and 1.5 inches high has a
crosssectional 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 crosssectional
area, it’s very easy to calculate volume. If the crosssectional area is
constant, all that you need to do is to multiply the crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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 crosssectional 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
= crosssectional 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
crosssectional 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.