Bodybuilding Articles

Determining Natural Bodybuilding Potential

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Can you determine natural bodybuilding potential? Is it possible to determine if a lifter is a natural or steroid user? The answer is generally yes to both questions.

The formula. The following equation is derived scientifically, studying approximately 300 winning drug-free strength training athletes and bodybuilders from 1947-2007. For more information please read this article, Your Maximum Muscular Bodyweight and Measurements.

H = Height in inches

W = Wrist circumference located at the hand side of the bony lump on the wrist (known as the styloid process)

A = Ankle circumference at the smallest measurement

BF% = The bodyfat percentage at which the ankle and wrist circumferences were taken

natural muscle potential

Potential Variance. The natural bodyweight potentials listed below are derived using a wrist circumference of 7.5 inches, and an ankle circumference of 9.5 inches. These numbers are reasonable for the heaviest of natural lifters. (Note: the author’s wrist circumference is 8.0 inches, and ankle circumference is 10.0 inches at a bodyweight of 280 pounds. So you can see that for a natural lifter under 220 pounds, 7.5 and 9.5 are very reasonable numbers)

For a 0.5 differential in BOTH numbers (a lifter’s wrist is 8.0 inches and ankle is 10.0 inches instead of 7.5 and 9.5), the weight variance using the formula would be an additional 5.2 pounds of lean mass. In simple terms, for every additional inch of circumference (from either ankle, wrist, or both combined) above the potentials listed below, a lifter could have an additional 5.2 pounds of lean mass.

Again, keep in mind that the numbers used to derive natural potential (7.5 wrist circumference and 9.5 ankle circumference) are at the high end of normal for a natural bodybuilder with under 20% bodyfat. Therefore, a reasonable maximum potential variance for natural lifters from the below numbers would be plus 3 pounds. For smaller wrist/ankle boned lifters, the numbers could be smaller by up to 10 pounds.

The numbers. Using the above formula, and inserting 7.5 wrist inches and 9.5 ankle inches, we derive the following natural bodyweight lean body mass potentials for a 6% bodyfat percentage. Again, the derived numbers below are lean bodyweight, which means total bodyweight less fat. It is not total competition bodyweight including the 6% bodyfat.

The reduced formula with wrist and ankle circumferences and a 6% bodyfat percentage is…

H^1.5 (0.31037632)

Height, 66 inches = 166.4 lean body mass potential. Competition weight = 177.0 pounds

Height, 67 inches = 170.2 lean body mass potential. Competition weight = 181.1 pounds

Height, 68 inches = 174.0 lean body mass potential. Competition weight = 185.1 pounds

Height, 69 inches = 177.9 lean body mass potential. Competition weight = 189.3 pounds

Height, 70 inches = 181.8 lean body mass potential. Competition weight = 193.4 pounds

Height, 71 inches = 185.7 lean body mass potential. Competition weight = 197.6 pounds

Height, 72 inches = 189.6 lean body mass potential. Competition weight = 201.7 pounds

Height, 73 inches = 193.6 lean body mass potential. Competition weight = 206.0 pounds

Height, 74 inches = 197.6 lean body mass potential. Competition weight = 210.2 pounds

Height, 75 inches = 201.6 lean body mass potential. Competition weight = 214.5 pounds

Height, 76 inches = 205.6 lean body mass potential. Competition weight = 218.7 pounds

Bottom line. It is possible for a natural lifter to exceed these numbers by a very small percentage. But, it is virtually impossible for a natural lifter to exceed their potential by more then a few pounds. Based on ankle and wrist circumference, I think it is safe to say that a lifter who exceeds their natural potential by more than 5-10 pounds lean mass is certainly suspect.

Steve Shaw

Steve Shaw is the primary content manager for Muscle and Brawn.

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