User:Kevinthenerd/undersquare

This is a sandbox for an article section I'm working on. It will take me some time to complete it. Adding references will probably be the hardest part since most of this is off the top of my head. It may actually be considered "original research" (in the language of Wikipedia) since, in effect, this is a synthesis based on available material and is not indeed a direct paraphrase of the original sources.

Please consider everything here total garbage until I remove this line. Thanks. kevinthenerd (talk) 22:39, 3 March 2008 (UTC)

Torque in an Undersquare Engine
An undersquare engine usually develops more torque than an oversquare engine of equal displacement. However, contrary to a popular belief, the extra leverage created by the crank arm is not responsible for this effect.

Beginning with a discussion of torque, work, and power, a relationship will be drawn between torque and the work developed per cycle, where P=power, N=rev/time, T=torque, and W=work per cycle.

$$Power \equiv \frac{dW}{dt}\, \equiv \dot W \equiv \mathcal{P} \, $$

For constant power, this relationship becomes simpler:

$$\mathcal{P} = \frac{W}{t} \, $$

where W is the total work developed over a time period of length t.

From automotive engineering literature ,

$$\mathcal{P} = 2 \pi N T \, $$

Note that it is also convenient to write $$N=\frac{1}{t}$$ where $$t=time \, $$.

Putting these together,

$$\mathcal{P} = \frac{2 \pi T}{t} = \frac{W}{t} \, $$

Therefore,

$$W = 2 \pi T \, $$

and more importantly

$$T \sim W \, $$

Torque is directly indicative of the work developed per engine cycle. The two are related by a constant factor.

The boundary work in a fluid system can be expressed by the following integration.

$$W \equiv \int F\, dx = \int p\, dV \, $$

Imagine two engines of equal displacement. One is oversquare, and the other is undersquare. They are spinning at the same crankshaft speed (rpm). If they are developing the same internal pressures, the torque will be the same, since torque only depends on in-cylinder pressure $$p$$ and volumetric displacement $$dV$$.

The radius of the crank arm is therefore irrelevant when only considered directly. While a larger crank radius will produce more torque for a given piston force, an undersquare configuration actually develops less piston force because it has less area over which to spread the pressure. These two effects directly cancel.

$$F = p A \, $$

$$dV = A dx \, $$

therefore

$$ F dx = p A \frac{dV}{A} = p dV \, $$

Somehow, the undersquare engine must develop more in-cylinder pressure than the oversqure engine in order to produce more torque. The in-cylinder pressures that drive the piston downward are a direct result of combustion, and combustion requires fuel and air. Assuming a constant air/fuel ratio and constant efficiencies, the undersquare engine must be drawing in more air/fuel mixture.

Air induction in a naturally aspirated engine is strongly related to the speed of the piston. Mean piston speed dictates the camshaft profile, and it largely affects siphoning within the ports. For the undersquare engine, a higher mean piston speed is reached at a slower engine speed. Momentum siphoning is allowed to occur at a slower engine speed, improving volumetric efficiency. By the time the oversquare engine develops a similar mean piston speed, the volumetric flow required to develop the same torque is higher since the engine's air needs are now greater, scaling with the crankshaft speed. Viscous forces within the port and inertial forces within the manifold now dominate, starving the engine of air, producing less torque.

Oversquare engines can usually produce more power in a naturally aspirated engine, however. Torque is almost fixed by the limitations of induction, so to produce more power, the engine should produce the same amount of work in less time. In other words, the same torque should be produced at a higher engine speed. Higher engine speeds are available to an oversquare engine since the intertial forces that want to tear the engine apart are related to the stroke length.

Assuming an infinite rod-to-stroke ratio,

$$F = m a \, $$

$$a = \omega^2 r \, $$

$$F = m \omega^2 r \, $$

where $$\omega = 2 \pi N \, $$

For a fixed deck height, the oversquare engine also benefits from an increased rod-to-stroke ratio, reducing piston acceleration and reducing force on the connecting rod.

An undersquare engine can be exploited for its smaller connecting rod force in a turbocharged or other forced-induction engine since high in-cylinder pressures would produce too much force against a large-area piston. The 7M-GTE is a good example of such engineering. This effect is also exploited in economy car engines since their pressure forces also typically exceed their accelerative forces.

In engineering practice, the bore-to-stroke ratio is mostly determined by the ratio of inertial forces to pressure forces. A highly turbocharged diesel engine, which experiences a great deal of torque at a low speed, would dictate an undersquare design. An air-restricted, naturally aspirated race engine, which can spin to 10000+ rpm but can only produce a small amount of torque, would dictate an oversquare design. One other factor worth mentioning, however, is that a highly oversquare engine has difficulty allowing for high valve lift due to interference with the piston, and it is also limited in its compression ratio for the same reason. An oversquare engine does, however, allow for a large diameter valve, improving valve area. A highly undersquare engine has a limited chamber area for wide valves and is therefore restricted to a smaller valve diameter, but it more comfortably allows a higher compression ratio as seen in diesel engines.