Engine Architecture and Mathematical Design

I'm hoping to make this blog appeal to the masses so lets start with the basics and move it on from there. I will add a worked example so you can follow along as I go. However I will use dimensions different from the engine I am developing for obvious reasons;

Basic Circle and Cylinder Theorem;

If we take the cylinder in figure 1.1 we can calculate the area and volume of the bore and cylinder respectively.

Blue Arrow = Diameter(d) = Cylinder Bore(B)

Yellow Arrow = Radius(r) = Cylinder Bore/2

Orange Arrow = Circumference(c)

Red Arrow = Height(h) = Engine Stroke(S)

You will no doubt remember from High School that πd = c, πr^2 = Area of a circle(A) and that

πr^2 x h = Volume of a cylinder (V_sv)

∴ 

Swept Volume per Cylinder = π x (Bcm/2)^2 x Scm
or

Swept Volume per Cylinder = V_sv (cm^2 ) = π/4 x (B^2 mm)/100 x Smm/10

Swept Volume per Engine = V_tsv = n_cyl  x V_sv

So lets take the 2012 Formula 1 regulations as a guide to create an engine. The Bore must not exceed 98mm and the engine must be a 2.4litre V8. That basically means 2400cc/8 = 300cc per cylinder. If we were to use a large bore and a short stroke for high revolutions per minute, although the engine cannot rotate any higher than 18000 RPM, (I will discuss all this in depth later this is merely a basic example of the cylinder formulae as the RPM and mean piston speed are detrimental to the stroke) so with a stroke of 45mm;


πr^2  x S=V_sv
∴
√(V_sv/πh)  x 2 = B
√(300/(πx4.5))  x 2 = B
√(300/(14.137))  x 2 = B
√(21.22)  x 2 = B
4.61 x 2 = 9.21cm = 92.1mm bore size

Therefore the engine would have a stroke of 45mm and bore of 92.1mm
Capacity would therefore be;

V_sv (cm^2 ) = π/4 x (92.1^2 mm)/100 x 45mm/10
 = 0.7854 x (8482.41)/100 x 45mm/10
 = 0.7854 x 84.8241 x 4.5mm
= 299.79cc

and

V_tsv = n_cyl  x V_sv
= 8 x 299.79
= 2398.32cc