Subject: Gear box torque calculations.
From: rotaryeng
Date: 7/6/2013, 8:16 AM
To: AAA-rotaryeng


Here are some interesting concepts to help get a handle on this ,,, just my 2c

Thanks for all the interesting posts on this amazing little engine.

In order to make the concept clearer, here are five examples of real-world systems which
produce 300 hp.

Example 1: How much torque is required to produce 300 hp at 2,700 rpm? Since hp = torque x
rpm · 5,252, then rearranging the equation to solve for torque produces: torque = hp x
5,252 · rpm.
Answer: torque = 300 x 5,252 · 2,700 = 584 pounds-feet.

Example 2: How much torque is required to produce 300 hp at 4,600 rpm?
Answer: torque = 300 x 5,252 · 4,600 = 343 pounds-feet.

Example 3: How much torque is required to produce 300 hp at 8,000 rpm?
Answer: torque = 300 x 5,252 · 8,000 = 197 pounds-feet.

Example 4: How much torque does the 41,000-rpm turbine section of a 300-hp gas turbine
engine produce?
Answer: torque = 300 x 5,252 · 41,000 = 38.4 pounds-feet.

Example 5: The 300-hp engine in Example 4 drives a gearbox which turns a propeller at
1,591 rpm. How much torque is applied to that shaft?
Answer: torque = 300 x 5,252 · 1,591 = 991 pounds-feet (ignoring losses in the gearbox, of
course).

The point to be taken from those numbers is that a given amount of horsepower can be made
from an infinite number of combinations of torque and rpm.

Think of it another way: In cars of equal weight, a 2-liter twin-cam engine that makes 300
hp at 8,000 rpm and 400 hp at 10,000 rpm will get you out of a corner just as well as a
5-liter engine that makes 300 hp at 4,000 rpm and 400 hp at 5,000 rpm.


Bradly Banks

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