7 replies to this topic

[knighty]

I was having a chat on another forum aboyut typical rod length to stroke ratios for road and race engines, the road derived engines typically have a ratio in the region of 1.6 to 1.8, so I thjought I'd have a look at what the F1 boys are doing.........the asiatech is running 104.89 rod, 46.1 stroke, therefore 2.27.......the ferrari 049 is running 41.4 stroke and about a 115mm rod length, therfore 2.77......holy 5hite thats quite high!......I dare say they are even higher nowdays........I was trying to rack my brains why the newer generation of F1 engines are running such high rod-stroke ratios, then I heard that the longer the rod, the less the side loads on the piston pin forces are.........then the penny dropped........I'm thinking with the higher rod-stroke ratios, this allows a lighter piston design due to much lower side loading from a longer rod, therefore more revs, and more power.......does this sound right to you guys?.......I would be interested to know your thoughts

[WPT]

The maximum piston acceleration is reduced with greater l/r ratio. Perhaps another item to think about. WPT

[desmo]

Dwell time around TDC for combustion efficiency and port velocities are probably also factors...

[shaun979]

I have heard that it is a parameter that is a result of other design targets (taller block heights for the right amount of space for airbox design and runner length and angle), and not that they specifically design to that very high rod ratio. If they could run shorter block heights hence shorter rods, they would, so I hear.

The lower side loads of long rods are good to have, but at a given engine speed, if the overall piston/rod assembly gains mass from the longer rod, then the crank loads will increase - which is a negative.

I think there are many factors to consider and that the degree of benefit or hurt of each needs to be calculated. Some are smaller than we think, others, larger.

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While we are here.. how does one calculate side load from inertial force (assuming engine purely motored)? I have seen spreadsheets for percentage of combustion force on piston top applied as sideload, but never one for pure inertial load. Also, how does inertial and combustion factors in sideload affect each other?

[McGuire]

If packaging is the primary issue, you want as short a rod as possible. Longer rod requires increase in deck height, adds block weight and raises CG height.

[WPT]

Posted by shaun979
While we are here.. how does one calculate side load from inertial force (assuming engine purely motored)? I have seen spreadsheets for percentage of combustion force on piston top applied as sideload, but never one for pure inertial load. Also, how does inertial and combustion factors in sideload affect each other?

Assume crank RPM is constant and that you know r and l. Use the piston acceleration equation {a=-rw^2(cos(Theta)+r/l*cos(2*Theta)} to calculate the piston acceleration for various crank angles. Determine reciprocating mass, then use F=ma to calculate the inertial force. Line of action is along the bore center line. For a given crank angle, you know the inertial force magnitude and line of action, the line of action of the force on the piston due to the connecting rod, and the line of action on the piston due to the cylinder wall. From this info you can determine the magnitude of the other forces.
r=crank throw radius, l=connecting rod length, w=crankshaft angular velocity in radians per second, Theta=crank angle measured from TDC (Theta=0 with piston at TDC)

[shaun979]

Thanks WPT, I'll try it out soon!

[WPT]

shaun979;
I should add more info. For purely motored engine (assuming engine is in a vacuum, so no compession or pumping) the inertial force caused by the reciprocating mass will result in the connecting rod to be in tension around TDC (say within about 80 degrees either side of TDC, it depends on r/l ratio), and in compression around BDC (about 100 degrees either side of BDC). Line of action of the connecting rod force is along the center line of the connecting rod, and the line of action of the thrust force is perpendicular to the bore center line.
If you have data for the combustion pressure as function of crank angle (power stroke), then use that to calculate the force on the piston at a given crank angle due to this pressure (PXBore Area). If you don't have the data, then use PV^k=constant to calculate the pressure (use k=1.4, and say P=1500psi at TDC). You will need to know the comprssion ratio to do it this way.
If you realize that the inertial force subtracts from the pressure force around TDC (power stroke), you are right. What's the deal, then? See if this thought experiment (Gedanken) helps. Imagine we have a single cylinder engine, in vacuum, no valves or ports, zero friction, and there is a flywheel attached to the crank. Say the crank/flywheel is spinning. When the piston reaches TDC (or BDC) the piston has zero kinetic energy, so all the energy in the system is in the flywheel. When the piston starts to accelerate away from TDC it gains kinetic energy from the flywheel (thru the connecting rod). Since total energy must remain constant, the crank/flywheel slows down. When the piston starts to decellerate around BDC it gives up its kinetic energy to the flywheel, so crank/flywheel speeds up. The piston and flywheel are trading energy so as to keep the total energy of the system constant. Now, say at TDC we introduce a burning charge in the cylinder, and this charge burns so that the connecting rod force is zero for 80 degrees of crank angle. During this interval the torque on the crank from the connecting rod is zero, so the crank/flywheel does not slow down, but continues at constant speed. At 80 degrees ATDC we remove the burning charge. The crank/flywheel will gain energy from the piston as the piston decellerates toward BDC. The system has gained energy just equal to the work done by the burning charge on the piston.
I hope this helps in understanding. I make no claim that the connecting rod has zero load during the power stroke in real engines. WPT