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Proportionate Barbie (or, Barbie is not skinny)

Julian D. A. Wiseman

Abstract: scaling Barbie such that the ratio of weight to strength is preserved suggests that Barbie is not skinny.

Publication history: only at www.jdawiseman.com/papers/barbie/proportionate_barbie.html. Usual disclaimer and copyright terms apply.


Barbie is a fashion doll manufactured by Mattel and marketed to young girls. There has been much negative comment about Barbie’s proportions, particularly her waist (see, for example, the BBC’s What would a real life Barbie look like?, 6th March 2009). But the calculations underpinning these complaints contain a basic error, and fixing that error reveals that Barbie is not skinny.

Two of the Barbies in the author’s household have heights of 30cm and 29cm, and waists of 9.5cm and 9cm, and the calculations that follow are based on these numbers. Let us start by repeating the simple scaling used by Barbie’s detractors.

The DollsHeight30cm29cm
Waist9.5cm9.0cm
6:1 scalingHeight180cm ≈ 5′11″174cm ≈ 5′8½″
Waist57cm ≈ 22½″54cm ≈ 21¼″

And indeed a waist of about 22″ is very small. But that’s because the scaling is wrong.

Imagine a garden spider that is, say, 5cm ≈ 2″ across. Scale this spider by a factor of 100, so it becomes fully 5m ≈ 16′ across. This would have increased the spider’s weight by a factor of 100³ = 1,000,000. (Imagine a stack of boxes 100 across, 100 deep, and 100 high. The number of boxes in the stack would be 100×100×100 = one million, and hence the weight of the stack would be a million times the weight of a box.) But the weight-bearing strength of one of the spider’s legs is approximately proportional to the surface area of its cross section, which has grown by a factor of 100² = 10,000. So the spider’s weight has increased by a factor of 100 more than the increase in the the strength of its legs. Such a giant spider couldn’t stand—its legs would snap. Large things are not just larger versions of small things: the change in scale compels a change in proportions.

Honeybee’s narrow waist

This can be rephrased less mathematically. Imagine seeing models, of size about one foot, of a spider and of an elephant. Would one know, instinctively, which was the large creature and which the small? Of course: something with legs as thin as a spider’s must be a small creature, no larger than a human hand. But the thickness of the legs of the elephant make obvious that this creature is as large as, well, as large as an elephant. If one were to enlarge a spider to the size of an elephant, its legs must be thickened.

Alternatively, observe the small size of the ‘waist’ of the honeybee in the picture on the right. This creature can’t be five feet long: its waist would snap. And if it is to be scaled such that it becomes five feet long, its waist would have to be thickened.

So consider scaling Barbie by a factor of 6, to become of human height, whilst preserving the ratio of weight to strength. To support Barbie’s weight the legs and waist must thicken by a factor of something like √6 ≈ 2.4. This gives new dimensions as follows.

The DollsHeight30cm29cm
Waist9.5cm9.0cm
6:1 height scaling, preserving the ratio of weight to strengthHeight180cm ≈ 5′11″174cm ≈ 5′8½″
Waist140cm ≈ 55½″132cm ≈ 52″

Preserving the ratio of weight to strength, a real-life Barbie would have a waist chunkily over four feet! Of course, this is also a simplification (Barbie isn’t that much of a tub of lard): the waist could hardly need to grow wider than the rest of the torso. A richer model is needed: the arms thickened; flow of weight and stress through the torso computed; torso thickened; and then the legs thickened and feet enlarged to hold the extra weight of the scaling and of the thickenings. But some parts of the model are simpler: increasing the size of the bust by a factor of 6 increases its weight by a factor of 6³ = 216, and increases the protrusion distance by a factor of 6. The torque thus becomes 1296× larger, requiring a similar increase in the strength of the supporting skin. Relative to a plain linear scaling, the bust would need to flatten. (So weight increases as a third power; torque as a fourth power. Different structures would need to be thickened by differently disproportionate factors.)

In summary, for the same reason that a spider can’t be scaled to the size of an elephant, Barbie is not skinny. Indeed, for an adult one foot high, Barbie is sensibly proportioned.

Julian D. A. Wiseman, New York, 29th March 2009


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