Mechanism 01 of 03

Flat-plane crank and piston motion

Joshua Kyan Aalampour

I started by building a single cylinder.

The slider-crank

Rather than trusting the parameter driving it, I measured the piston travel in the model and got 86.0 mm. This matched what I had for stroke exactly.

Textbook slider-crank relation assumes the piston slides along the axis the crank offset is measured in, and in a 90° vee it doesn't, since both banks sit 45° off vertical. So I resolved this by projecting the offset onto the bank axis first. At crank angle $\varphi$ , for a cylinder whose throw sits at base angle $\theta$ , the crankpin offset is

o_y = -r\sin(\theta + \varphi), \qquad o_z = r\cos(\theta + \varphi)

That offset then projects onto the bank axis $\mathbf{n} = (0, n_y, n_z)$ , which is $(0, -\sin 45^\circ, \cos 45^\circ)$ for the left bank and $(0, \sin 45^\circ, \cos 45^\circ)$ for the right.

o_n = o_y n_y + o_z n_z

and the distance from the crank center to the wrist pin, measured along the bank axis, is

d(\varphi) = o_n + \sqrt{\ell^2 - \left(r^2 - o_n^2\right)}

One crank angle drives all eight cylinders through this same expression. Only $\theta$ and the bank normal change between cylinders. The rod angle and the piston orientation fall out of it as well.

One cylinder through two crank revolutions.

Engine speed9,000 rpm

Mean piston speed25.8 m/s

Peak acceleration5,100 g

180°pos 76.7 mmvel 22.4 m/sacc -2,725 g

Rod ratio and the second harmonic

I went with a 142 mm rod on an 86 mm stroke. I chose a long rod deliberately, because once bore and stroke are locked the rod is the only lever left on how symmetric the piston's travel is.

\frac{\ell}{s} = \frac{142}{86} = 1.651 \qquad\text{so}\qquad \frac{r}{\ell} = \frac{43}{142} = 0.303

That puts the rod ratio at 1.651, which is long even among engines built to rev. I also set the 96 mm bore against the 86 mm stroke, a bore-to-stroke ratio of 1.116, so the layout comes out oversquare.

Expanding the exact expression for small $r/\ell$ splits the travel into a fundamental harmonic at crank frequency and a correction term. The correction term is the one that matters here, since it survives differentiation as a harmonic at twice crank frequency.

x(\varphi) \approx r\left(1 - \cos\varphi\right) + \frac{r^2}{2\ell}\sin^2\varphi

a(\varphi) \approx \omega^2 r\left(\cos\varphi + \frac{r}{\ell}\cos 2\varphi\right)

At top dead center both terms are positive and add together, while at bottom dead center they oppose each other, which gives the following:

\frac{a_{\text{TDC}}}{a_{\text{BDC}}} = \frac{1 + 0.303}{1 - 0.303} = 1.87

The piston is accelerated close to twice as hard at the top of its travel as at the bottom, which sets the design case for the rod small end, the wrist pin and the crown. That 1.87 comes from the two-term expansion. The exact expression, which is what the plot computes, gives 1.84 at this rod length.

Piston acceleration through one revolution at 9,000 rpm, with the stroke held at 86 mm (only the rod changes).

Rod ratio1.651

Peak at TDC5,100 g

Peak at BDC2,800 g

2nd harmonic1,200 g

This rodInfinitely long rodSecond harmonicAs built, 142 mm

Connecting rod length142 mm

Peak to peak is fixed at 2ω²r, 7,790 g at any rod length (the rod only moves where zero sits between the peaks). The shaded area is the departure from a pure sine, the second harmonic a flat-plane crank cannot cancel.

Four throws in one plane

Four crankpins carry eight rods, so I had to decide where those four throws would sit around the axis. I put all four in one plane at 0/180/180/0. That is the same arrangement the Ferrari 458 uses, and the same one in Ford's Voodoo (the 5.2 L flat-plane V8 from the Shelby GT350, which in my opinion is kind of the most European sounding Ford engine there is). It contrasts with the 90° increments you commonly see in American V8s. I measured the finished crank to confirm it: all four throws in one plane at zero lateral offset, up, down, down, up.

End-on, this crank collapses to a single vertical plane, pins 1 and 4 up against 2 and 3 down. A cross-plane crank never collapses.

Firing order

With the throws fixed at 0/180/180/0 the crank angles at which each cylinder reaches top dead center are fixed too. That meant the only choice left was which cylinder gets labeled number one. Working through the throw angles for this layout gives 1-8-3-6-4-5-2-7, with cylinders 1 to 4 on the left bank and 5 to 8 on the right, which is the canonical flat-plane V8 order. The badass thing about this is that I never actually typed that sequence in anywhere. It came out as a natural consequence of the throw angles! I was quite happy about this haha!

Firing order 1-8-3-6-4-5-2-7 across the 720° cycle. Every event alternates banks, with no two consecutive firings on the same side.

Left bank, cylinders 1 to 4Right bank, cylinders 5 to 8

0°Cylinder 1Left bankEight events, 90° apart.

Every exhaust pulse leaves from the opposite bank to the one before it, so each manifold receives an evenly spaced train of pulses.

I set the cam lobe phasing from the same top dead center angles the firing order comes from. This way both the valve events and the firing sequence derive from one source.

Secondary imbalance

A flat-plane crank runs far less counterweight, which is why it is lighter and revs harder. However, it comes with more challenges as it relates to balance. The second harmonic from above comes back as a shaking force:

F_2 \;\approx\; m\,r\,\omega^2 \cdot \frac{r}{\ell} \cdot \cos(2\varphi)

where $m$ is reciprocating mass, $r$ the crank radius, $\ell$ the rod length and $\omega$ the angular velocity. In a 90° V8 the primary forces cancel under either crank. The secondaries cancel only on a cross-plane, so a flat-plane carries a net shaking force the mounts have to tolerate. I accepted that trade. The long rod softens it, since the secondary term carries a factor of $r/\ell$ and 0.303 is a small one, at a cost in package height.

Crankshaft and bearings

The crankshaft is a single component made of 17 bodies: five main journals the shaft turns in, four crankpins the rods hang from, and the webs and counterweights joining them.

Rotating assembly

Main journals: 5 × Ø56 mm
Crankpins: 4 × Ø46 mm
Throw radius: 43 mm
Throw angles: 0 / 180 / 180 / 0°
Pin spacing: 106 mm
Crankpin width: 50 mm
Big end bore: Ø50 mm
Small end bore: Ø27 mm
Wrist pin: Ø22 mm

Because two rods share every crankpin, the banks are offset along the crank axis by about 12 mm each so the big ends sit side by side. That stagger shows up end on as the left and right cylinder centers not lining up.

Speed and acceleration at 9,000 rpm

Mean piston speed follows from stroke and engine speed alone:

\bar{v}_p = 2 s n = 2 \cdot 0.086 \cdot \frac{9000}{60} \approx 25.8\ \text{m/s}

Peak acceleration is at top dead center:

a_{\max} \approx \omega^2 r \left(1 + \frac{r}{\ell}\right)

With $\omega = 9000 \cdot 2\pi / 60 = 942.5$ rad/s, $r = 0.043$ m and $r/\ell = 0.303$ , that comes to about 49,800 m/s², or roughly 5,100 g. Both figures follow from geometry once an engine speed is chosen.

Four throws, all in one plane. Viewed end on, the crank is flat.