Thread: Federal 604
View Single Post
  #24  
Old 18-06-06, 08:44
Richard Notton
Guest
 
Posts: n/a
Default Re: Re: Federal

Quote:
Originally posted by Alex Blair
They were a massive truck with a 672 Cubic inch 6 cylinder Cummins diesel rated at 57 SAE Horse power....
Ah, I think you have become confused with the generally misunderstood RAC HP rating, this engine is SAE rated at 150HP and probably enough torque to put a knot in the propshaft.

I think we've been here before in another thread, but here goes. . . . . .

The RAC rating is meaningless in engineering terms and is just a hypothetical sum to arrive at a figure for British govt annual vehicle taxation, it was dropped after WWII but because you paid a pound a year road tax per RAC HP this, together with highly taxed fuel, led to the British propensity for small engined cars and the number in the names which referred to the HP rating, eg. Morris 8, Austin 7, Standard 10 etc.

Before WWII the military had to register and pay tax on their vehicles and the RAC rating, for vehicles supplied to the British, was therefore included in the documentation. Even so the RAC rating was fixed in the Brirish minds as an indication of engine size and continued to be used as a descriptive term, there was always a possibility the "Old Ways" would return after WWII of course.

The RAC rating was derived for the Road Traffic Act of 1921 as a tax yardstick and made three fixed assumptions which were accurate at the time but were being well exceeded by the mid 30's:

Mechanical efficiency at 75%.
Mean Effective Pressure at 90psi.
Mean Piston Speed at 1000ft/min.

Assuming these fixed figures you then apply the following formula to get the RAC Rating:

HP = (D² x n) ÷ 2.5

Where D is the bore in inches and n is the number of cylinders.

No account is taken of the stroke as the piston speed is fixed.


Working backwards from your figures then, the Cummins HB600 engine will have a bore of 4.87" to give a RAC Rating of 57HP.

Incidentally having the total capacity stated, a stroke of 6.013" can be calculated.

R.
Reply With Quote