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« Previous Page Table of Contents Next Page »On the Propeller-Shaft Formula Constant

(This Issue’s Question, Page 17)

**
The Know It All questions and correct answers are important design tips for students as well as other marine professionals. We suggest that you file them away for future reference.
**

**The Question Was: **

The standard formula for propeller shaft diameter in inches, as found in ABYC P-6, in
* Propeller Handbook,*
and in
* Boat Mechanical Systems Handbook,*
is:

Where:

SHP = shaft horsepower SF = safety factor

St = shear stress, psi RPM = revolutions per minute

Can you explain where the constant 321,000 comes from?

**The Winners are: **

Naval architect and past contributor to
*The Masthead*
, Eric Sponberg, and Elements of Technical Boat Design graduate, Cynthia Cygan both submitted the correct answer to the March 2011 Know It All question. Clearly, each of our win-ners is too smart for their own good. Indeed, Cynthia is now a two-time Know It All winner—a clear indication of being too too smart. In honor of their now proven status as brainiacs, each of our winners is forthwith officially a Know It All, and should henceforth be addressed as such. Naturally, West-lawn T-shirts, caps and Know It All certificates are on their way to both winners.

**And the Solution Is: **

Over the years, I’ve had many people ask about the constant 321,000 in the shaft-diameter formula. It seems a rather large arbitrary number, as if it was conjured up out of thin air to make the formula work. In fact, the 321,000 is a prod-uct of the standard engineering equations for the shear stress in a rotating shaft as developed for the inches, pounds, horsepower, and rpm units commonly used. Here’s how that works out:

Rotating shafts experience shear stress and the maximum shear stress in a shaft is:

Where:

St = shear stress, psi

T = torque, pound-inches (lb.in.) J = polar moment of inertia, in. 4

From any standard engineering handbook:

Where:

≈ 3.14159

If we know the allowable shear strength of the shaft mat rial we’re going to use, we can find the required diameter based on torque (T) from:

This is the standard formula for shaft diameter based on torque. We now need to modify this to find diameter bas on power and rpm.

Power is work done per unit time and work is a force tim distance. Velocity gives the distance and time. For a rotat circular shaft the distance per revolution is 2r—the circu ference of the shaft (where r is the radius), so velocity is:

V = 2 r x RPM

Where:

v = velocity, in./min.

≈ 3.14159

r = radius of the shaft, in. RPM = revolutions per minute

Force x r (radius) = T (torque)

So power is:

Force x v = 2 RPM x T

Mechanical power in English units is horsepower, which i 33,000 pound-feet per minute, and we’re dealing with th

321, 000 x SHP x SF 3 Dia., in. =

St x RPM

T x Dia. St =

2J Or 2J x St T = Dia.

4 Dia. J =

32

4 3

3

3

2 x St x Dia. St x Dia. T = =

16 32 x Dia. So

16T Dia. =

St Which gives

16T Dia. =

St

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