Were you one of those students who in school loved gazing out of the window
during maths lessons? Me too. But that probably stops you now using maths as an
extremely useful tool when modifying cars....
Last time in Performance Maths Part 1 we covered calculating
cross-sectional areas, volumes and masses. Each of these is incredibly important
if you’re modifying the intake or exhaust, are building a swirl pot – or even
building a whole car. This week we’re going to start looking at triangles, one
of the single most useful mathematical shapes in existence.
Right-Angled Triangles
Something that makes triangles very useful is that they are used so commonly.
We talk about a race car with a ‘triangulated’ frame – that’s a very stiff
shape. But without much thinking about it, we also use imaginary triangles all
the time. Furthermore, they’re usually right-angled triangles, where one of the
three internal angles is 90 degrees. In fact, anything that is measured
against vertical and horizontal lines will form a right-angled triangle.
So why is this useful?
We’ll get into it in more detail in a moment, but picture for example the
steering axis of a car – what used to be called kingpin inclination. As this
diagram indicates, it can be shown by a line (red) passing through the upper and
lower balljoints.
But what angle is the steering axis inclination? To find out, we draw a
vertical reference line, shown here in blue.
And by adding an extension to the blue line, we form a triangle with one 90
degree internal angle – so a right-angled triangle.
So what? Well, if we know the length of any two of the sides, we can
calculate the length of the third. Or, if we know either of the two angles that
aren’t 90 degrees, and we know the length of one of the sides, we can calculate
the length of the others.
Again, so what?
Let’s look at this if you were building or modifying a front suspension. To
get the desired scrub radius (that is, how far across the tyre’s width the
steering axis impinges on the road) you find that (say) a steering axis
inclination of 17.5 degrees is needed. If you measure how far vertically apart
the balljoints are (this is fixed by the wheel’s upright) you can then calculate
how much further towards the centre of the car the top balljoint needs to be.
Or, to take a step back, if you know how much scrub radius you want, and you
know the width of the tyre and its diameter, you can calculate what steering
axis inclination is actually needed to gain that scrub radius.
Or, in a McPherson strut car, how much extra camber you’ll get if you move
the strut top inboard...
The above are complicated examples, but they’re complicated because in real
life, making actual measurements on suspension systems is very difficult – tape
measures need to pass right through solid wheels and difficult things like that!
Calculating the results is far easier and more accurate.
And here’s another suspension example. Let’s say that you have a lower
suspension arm that’s 45cm long. The outboard end of the suspension arm (the
part that connects to the wheel hub) needs to be able to move up and down 20cm.
You’re wondering what type of bush or bearing to use on the inboard end of the
arm and you approach a bearing supplier. The first question asked is: what angle
does the bush need to rotate through? If you sketch two right-angled triangles,
the answer can be easily and quickly calculated (and is done so below).
Identification
Before we can go any further, we need to have a way of identifying the
different sides of the right-angle triangle.
One angle is identified here with an arrow – it’s what we’ll call our ‘angle
of interest’. Directly opposite that angle is a side called ‘Opp’, or opposite.
Next to the angle is a side identified as ‘Adj’, or adjacent. The remaining side
is called ‘Hyp’, or hypotenuse.
You can see that the two sides Opp and Adj are defined in terms of our angle
of interest. Here we’ve moved the angle of interest and Opp and Adj have moved
appropriately. Note that the Hyp doesn’t change.
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Opposite and Adjacent sides are defined in terms of the angle of interest.
The Hypotenuse is fixed.
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Relationship Between Angles and Lengths of Sides
If with a triangle we divide the length of the opposite side by the length of
the adjacent side, a certain number will be created. For example, in the case of
this triangle, that calculation is 12 divided by 24 = 0.5. So in this case the
adjacent side is twice as long as the opposite side. It doesn’t matter how big
or small we draw the triangle, if we keep the same internal angles, the result
of dividing the adjacent by the opposite will always be 0.5.
In other words, the ratio of the lengths of the triangle sides is dictated by
the internal angles (ie the shape) of the triangle.
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There’s a fixed relationship between the internal angles of the triangle and
the ratios of the lengths of its sides.
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Tan
From the above paragraph we know that the number gained when the opposite
side is divided by the adjacent side is strictly related to the angle of
interest, shown by the arrow. In fact, the ratio of opposite divided by the
adjacent is called the tan of that angle. To put this another way:
| Opposite |
Tan of angle = | ------------ |
| Adjacent |
So in this case:
Tan angle = 12 divided by 24 = 0.5
Tan angle = 0.5
Angle = 26.6 degrees
The last step is as easy as punching 0.5 into a calculator and then pressing
the 'inverse' then 'tan' buttons.
Even if you read no further in this series, this is an extremely useful tool
to have at your fingertips. Here are two examples.
1. Rotation of a Bush
Let’s look again at the angle of rotation that a suspension bush needs to
move through. The suspension arm described above was 45cm long and the outboard
end of the suspension arm needed to be able to move up and down 20cm.
This diagram shows the system drawn as two right-angled triangles. The pivot
is at right and the suspension arm is the horizontal line marked as being 45cm
long. The end of the arm can move upwards by 10cm or downwards by 10 cm (giving
a total travel of 20cm).
Let’s just look at the top half. The angle (arrowed) is found by the tan of
the Opp over Adj. That is, tan of angle = 10 divided by 45, or tan of angle =
0.222, or angle = 12.5 degrees.
Now that angle is for movement in only one direction (upwards) so we need to
double it to get the full rotational movement the bush will work through – 25 degrees. In a real life application (say a race car) that’s a large-ish
angular movement for a rubber bush – in this case, it might be better to use a
bearing.
2. Torsion Bar Calculation
If the inner bush is replaced with a torsion bar spring, the same rotational
angle calculation can again be incredibly useful.
If the wheel is supporting 400kg in bump (say 200kg statically) and the lever
arm is 45cm long, a torque is applied to the torsion bar. Torque is worked out
by the force x length, so it’s 400kg x 0.45 metres, or 180 kg/m. (This is
converted to Newton-metres by multiplying by 9.8, giving 1764Nm.)
We now know that for the outer end of the suspension arm to move from full
droop to full bump, the pivot point rotates by 25 degrees. Therefore, an
appropriate torsion bar is one that twists by 25 degrees when 1764Nm of torque
is applied.
And how is that information useful? Here’s why: any spring manufacturer will
immediately be able to tell you the diameter and length of torsion bar you need
to achieve that spring rate without over-stressing the steel.
| Opposite |
Tan of angle = | ------------ |
| Adjacent |
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Finding the Length of the Hypotenuse
If you know the lengths of the two sides that meet in the 90 degree
right-angle, you can calculate the length of the third side – the hypotenuse.
You don’t even need to know any of the internal angles.
So if we know the length of A and the length of B, and want to calculate the
length of H, we need to know of some relationship between the lengths of the
sides. The relationship is:
(A x A) + (B x B) = (H x H)
So in other words, A squared plus B squared equals H squared.
Let’s put some numbers in.
(A x A) + (B x B) = H x H
(12 x 12) + (24 x 24) = (H x H)
(144) + (576) = (H x H)
720 = (H x H)
H = 26.8
And the way we turn (H x H) into just H is to square root 720. The square
root key looks a bit like a tick with a horizontal bar extension. Punch 720 into
the calculator and then press the square root key and the answer is that H is
26.8.
So why would this be of us? The example is again from suspension – in part
because that’s where I have found all this stuff so very useful.
Spring Damper Length Calculation
Imagine you’re using a combined damper/spring assembly that will be angled
from the vertical. At rest, the upper wishbone is parallel to the road and the
spring/damper connects from the upright to the same metalwork that supports the
upper wishbone pivot. It can be seen that here again is a right-angled triangle.
If the upper wishbone (red) is 25cm long and the distance between the upper and
lower outer balljoints on the wheel hub is 20cm, how long does the spring/damper
have to be?
The spring/damper is the hypotenuse of the triangle, so (20 x 20) + (25 x 25)
= length of spring/damper squared.
400 + 625 = length of spring/damper squared
1025 = length of spring/damper squared
Length of spring damper = 32cm
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(Side A x Side A) + (Side B x Side B) = (Hypotenuse x Hypotenuse)
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Internal Angles
The internal angles of a triangle always add up to 180 degrees. Therefore,
when dealing with a right-angled triangle (where we know one angle is 90
degrees), we need to know only one other angle to be able to work out all the
internal angles.
So say we have a right-angled triangle that has one 30 degree internal angle.
The right-angle of 90 degrees plus the known angle of 30 degrees adds up to 120
degrees. Since a triangle must have internal angles that total 180 degrees, the
unknown angle is 180 – 120 = 60 degrees.
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Internal angles of a triangle always add up to 180 degrees
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Conclusion
Many of you will already know basics like a triangle’s internal angles add up
to 180 degrees, than the tangent of an angle is found by opposite divided by
adjacent, and that the length of the side of a triangle can be found from the
other sides.
But this story is not really for you! Instead, it’s for those for whom these
concepts are new, or at least, ring only faint bells.
And if you’re one of those people, read through the story again. Draw some
triangles (or suspension systems!). Do the calculations. Try to picture the
angles and sides clearly in your mind and see how useful the concepts can be...
