Excellent and thank you MR....

I, not being Mechanically literate, always considered torque---obviously functions off horsepower---az something like a......stump pulling thing.

your information of the rpms gives me much more to ponder....

Well, to add some Facts, a bad maintained Chain Drive can cost you more Loss than an average constructed Shaft Drive.

A BMW Engineer talked about the Controversy in a german M/C Mag about a Year ago. He said:

A well maintained and fresh Chain Drive Set has around 4 Percent Loss nowadays. No Lubrication, no Chain Adjustment, Dirt and Rust can drive the Loss up to 15 + easily.

A Shaft Drive normally has a Loss less than 15 Percent, depending on the Construction. A neglected Shaft Drive has as nearly the same Loss, until it fails.

If we look at the Vmax Shaft Drive, you'll see a typical Loss of 13.5 Percent on a 1Gen Vmax with the later final Drives, the 2Gen has a typical Loss around 11.5 Percent.

Good info Squeeze! I love the shaft drive on the 'max! Fluid change once a season (6,000 miles) and that's it!!

Here is another observation. Since the Valkyrie engine crankshaft is in-line with the bike, a 90 degree change of direction, using bevel gears, would be required whether shaft drive or chain drive is being used. Since the majority of the loss from a shaft drive, as opposed to chain drive, is caused by the 90 degree direction changes then, in theory, chain drive would have no advantage over shaft drive for the Valkyrie. There is also an additional very small loss with shaft drive due to the U-joints periodically not being in perfect alignment on both ends but we can ignore that if the U-joint angle is small. However, I still don't see why the shaft drive would produce more real-wheel torque than a chain drive version. In theory, they would produce equal loss at the rear wheel for the Valkyrie. Perhaps this is what your friend is referring to.

So does a "lowered" Gen2 Max have less HORSEPOWER to the ground compared to a stock one? everything else apples to apples? if so...how much? ...if not...nevermind

"I can produce 400 foot-pounds of torque with my own arm exerting 100 pounds of force on a 4-foot lever but hardly any power (less than 1 HP) because I can't turn it very fast".
This example is the best explanation I've heard, really

Remaxed wrote:
The more the U-joints can be kept in alignment with the shaft, the lower the loss. This applies both vertically and horizontally. If the U-joints are in perfect alignment with the shaft then there is no loss from the U-joints in theory.

I have eyeballed the U-joint angles on the VMAX with a protractor and the angles appear to be quite small so any reduction of power loss by lowering would be neglible. Even if lowering could achieve a perfect vertical U-joint alignment, there is still a horizontal misalignment.

If you are really interested, here is the equation that predicts the efficiency for a single U-joint based on bend angle. This equation is only accurate for bend angles less than 15 degrees.

Efficiency = 100 x (1 - (0.003 x Theta))

where Theta is the bend angle in degrees.

Also of interest is that the efficiency of spiral, bevel, spiral bevel, and spur gears is in the range of 98% to 99%.

I was bored one day so I actually added up the estimated losses in the VMAX drive train and I was actually able to predict the loss in power quite closely from the engine crankshaft to the rear wheels. I checked my calculations against the rear wheel dyno numbers that I have seen posted for a stock VMAX (e.g. 174 RWHP from Cycle World). Here are the contributing factors with each introducing a loss:

* Spur gears in gearbox
* Bevel gear at front of shaft
* Bevel gear at rear of shaft
* Front U-joint
* Rear U-joint

When I added it all up I came up with about 11% loss from engine to rear wheel which matches quite closely with the posted dyno numbers (174 RWHP and 197 BHP).

In case you haven't guessed, I am an engineer and such things interest me.