I own a Mark I Tiger, B9471136. I have owned it for 31 years and have over those years made and unmade many changes, not the least of which were engine mods. Many modifications are undertaken without any basis of fact or reason. This analysis attempts to address one of the more confusing: Rod length to stroke ratios. Rod ratio or the rod length to stroke is said to increase torque at (pick your RPM range).

Since I am an old retired aerospace engineer and have both the time and inclination, I decided to investigate, analytically, what the effects of a varying rod length for a given crank throw means with respect to torque, rod axial force and piston side wall force. Each of these parameters is important: torque because it is what turns the gears to get you down the road, axial force on the rod because this is partially causes bearing wear, and side force because it translates into friction and hence heat and wear on the cylinder walls.

My model is simple. I used a crank and slider mechanism for the analysis. The model comes from "Kinematic Analysis of Mechanisms", McGraw-Hill, 1969. Because I wanted to develop axial forces in the rod and side forces on the piston, I used a parameter called BMEP or brake mean effective pressure (psig) acting on a piston which had a diameter of 4.00 inches. Since my Tiger is a Mark I with the 260 cubic inch engine I used the published BMEP value of 124 psi (which occurs at 2200 RPM). This gives an axial piston load of 1558 lbf. I held the crank throw constant, at 1.435 inches, which is half the stroke of our 260/289 cubic inch engines, and varied the rod length. Coincidentally, I used our rod length of 5.155 inches as the starting point and used 5.0, 5.5, 6.0, and 4.5 inches as other lengths.

The math model I developed, based on the crank and slider referenced above, looks like this:

               sin(x)*(1-((r/l)sin(x))^2)^1/2  + cos(x)*sin(x)*(r/l)
M(x) =  F * r* -----------------------------------------------------
                               (1-((r/l)sin(x))^2)^1/2

 where:
          M(x) =  torque (ft-lbs)
             F = Piston axial force (lbf)
             x =  crank rotational angle (radians), evaluated from TDC to
                   BDC (0-180 degrees)
             r =  crank throw (inches), stroke/2
             l = rod length (inches) pin centerline to bearing journal
                 center line

As can be seen, it is a reasonably complex trancendental equation. I had originally planned to take the derivative and set it equal to zero to find the maximum angle at which torque occurs based on specified rod ratios, but the complexity got out of hand, at least for me. I would be interested to hear from anyone whose can accomplish the feat (I used Math Cad 4.0 and it, for sure, produced the required derivative, but I did not take the time to evaluate it.). What I did was program the equation, in both Math Cad and Excel 4.0, to see the results. And interesting results they are! The data sheets are far to extensive for me to type in and turn them into tables, but the Excel data plots for the various parameters are shown below as Figures 1 through 5.

Results:
1) Longer rods have lower piston/cylinder side loads, hence less cylinder and piston wear [note1].
2) Short rods have a higher axial loading, hence potentially more bearing wear [note 2].
3) Short rods produce higher torque early, less torque later [note 3].
4) Short rods produce a higher peak torque [note 4].

and finally, an astounding (at least to me)
5) No matter what the rod length, the area under the torque curve is EXACTLY the same for each case! Conservation of energy, perhaps?

note 1. The new 4.6L OHC engine has a very high rod/stroke ratio (2.5+). I suppose this is one of the ways that Ford manages to have an engine which will run for a hundred thousand or more miles without significant attention.

Department of Corrections. Thanks Bill! He said:

"Dear Sir,

In your calculations on the 4.6L Ford V8 you show the rod length as 8.937
inches which happens to be the crankshaft centerline to deck dimension. The
correct connecting rod length (mean) for the 4.6L Ford V8 is 5.9331 inches
which will give you a rod/stroke ratio of 1.6745. That is actually not a
very good ratio if one is looking at durability. The 5.4L and 6.8L is even
worst with a ratio of 1.5984.

Bill Kohn"

note 2. Maybe. Maybe not. Longer rods have higher weight and thusly more inertial loading as they reciprocate. And for an engine which lives at high RPM, this might mean the difference between long or short life.

note 3. I can hear it now! There are some of you out there who will throw in cam timing, valvelifts etc to make the argument that one rod or the other is better. But, that would be changing the problem by changing multiple parameters at a time. I agree that each can be tuned with cam changes, carb or EFI changes to get more performance for any given particular set of constraints.

note 4. This also surprised me a little. I had expected that the torque would be less. But I can rationalize the data by short rods having more axial force which pushes on the crank. But it does happen earlier in the power stroke. As a matter of fact all rods produced torque peaks before the 90 degree rotation point and fell off slower.

As an interesting side bar, I computed the rod ratios for all of the engine types as shown in the Ford Motorsport SVO catalog (1994), see Table 1, below. The high performance engines have rod ratios between 1.65 to 1.73. Interesting to note that the SVO 302 uses the slightly longer rod (5.155 vs 5.092).