Why is there no r^2 term in the formula for the swing weight of a golfclub?

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The swing weight of a golf club is calculated using the formula Swing Weight = m * (r - 14), where m is the mass in grams and r is the distance from the top of the shaft to the center of mass in inches. The absence of an r^2 term in this formula raises questions about its rigorous foundation in physics, as it contrasts with the angular acceleration formula, which includes r^2. Swing weight is often viewed as an informal measurement that helps golfers understand how hard it is to swing a club, rather than a precise dynamic measurement. It is based on the club's weight distribution relative to a fixed fulcrum point, typically 14 inches from the grip end. Despite its lack of formal physics backing, swing weight remains a widely accepted concept among golfers.
Lars278
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The Swing Weight of a golfclub = m * (r - 14)
m = mass of the club in grams
r = distance from the top of the shaft to the center of mass of the club in inches
6050 equals the swing weight D-0
But the formula for the angular acceleration of any object spinning around an axis is:
ωdot = τ / I
In the case of a thin rod spinning around one of it's endpoints this will be:
ωdot = 3 * τ / (m * r2)

m, and r are now in SI units though. (kg and meters)

But in the formula for swing weight there is no r2 only an r. Why is this?
 
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Why are you comparing swing ‘weight’ with angular acceleration? They are not the same, and their units are not the same.
 
It's not clear that the formula you show has a rigorous foundation in physics. It may just be an informal unit of measurement with which people identify clubs.
 
The linear acceleration of the club head would be related to the torque at the wrists by something similar to this "swing weight".
 
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FactChecker said:
It's not clear that the formula you show has a rigorous foundation in physics. It may just be an informal unit of measurement with which people identify clubs.
Maybe not in physics but millions of golfers rely on it. It's supposed to describe how hard it is to swing a golfclub. How hard it is to perform angular acceleration.
 
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Lars278 said:
Maybe not in physics but millions of golfers rely on it. It's supposed to describe how hard it is to swing a golfclub. How hard it is to perform angular acceleration.
I can find no reference which defines it in terms of angular acceleration for a fixed torque. Most of what I can Google up is very dumbed down and talks about feel. The references which speak of measurement or definitions allude to equipment which does a static balance (proportional to r, not r squared).

For instance:
  • A club's weight distribution relative to a fixed fulcrum point, typically 14 inches from the butt (grip end) of the club. Swingweight is measured in alpha-numeric units such as D-1, D-2. The letters used are A through G, with numerals from 0-9 (up to 10 for G).
    The higher a club’s letter-number unit, the closer its balance point is to the head, and the heavier it will feel when swung. The lightest possible swingweight is A-0; the heaviest is G-10. For men, the standard swingweights are D-0 and D-1; for women, C-5 to C-7.
or

It is really not a dynamic measurement as the name implies but rather a simple balance system what in mechanics is called First Moments. This is where the distance of the weight from the fulcrum multiplied by the weight itself must be the same on both sides for the beam to balance.
 
Lars278 said:
Maybe not in physics but millions of golfers rely on it. It's supposed to describe how hard it is to swing a golfclub. How hard it is to perform angular acceleration.
That makes sense. And it fits in with @jbriggs444 idea that it is related to the force needed to accelerate the swing rotation the desired amount. In that case, it can be related to a formal physics derivation.
 
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