Posted in Geometry

If One World Trade Center is a prism and not an antiprism, would it be less in volume?

One World Trade Center, more simply known as 1 WTC and previously known as the Freedom Tower, is the lead building of the new World Trade Center complex in Lower Manhattan, New York City. The supertall skyscraper is 104 storey  and is being constructed in the northwest corner of the 16-acre World Trade Center site. The image at the right shows the design as of May 2012.

One World Trade Center is an example of an antiprism. The square edges of the world trade centre tower’s cubic base are chamfered back, transforming the building’s shape into an elongated square antiprism with eight tall isosceles triangles—four in upright position and another 4 in upside down direction. Near its middle, the tower forms a perfect octagon, and then culminates in a glass parapet whose shape is a square oriented 45 degrees from the base. My question is Is this bigger than if it were a square prism? How about in terms of surface area?

Antiprisms are similar to prisms except the bases are twisted relative to each other, and that the side faces are triangles, rather than quadrilaterals. Here’s a model of a square antiprism.

square antiprism

The surface area of the antiprism may be bigger than the prism because of the additional faces but I’m not quite sure. How about the volumes? Cavalieri’s principle does not apply because the cross sections do not have the same area. Anyone wants to resolve and explain this? There is a formula for volumes and surface areas of antiprisms but I’d appreciate a more intuitive solution.

Note: All information and images about 1 WTC is from Wikipedia. Square antiprism model from eusebeia.dyndns.org.

2 thoughts on “If One World Trade Center is a prism and not an antiprism, would it be less in volume?

  1. In the 1WTC prism, the top square is smaller than the bottom square instead of the basic square antiprism which has identical top and bottom square. This the ‘rotated’ top square fits exactly within the bottom square so that, in plan view, the corners of the top square meet the mid-point along each side of the bottom square. Thus the area of the top square is only half the area of the bottom square. This means that the upward pointing triangular sides are vertical and the downward pointing triangles slope inwards towards the top of the building. As a result the surface area of the facades must be less than it would be for a simple square prism of equal height and base dimensions. If the new 1WTC building had been a true antiprism with equal sized base and cap, the upward pointing sides would have sloped outwards and the total surface area of the facades would have been approximately the same as the equivalent square prism.

  2. So I’m *guessing* here a bit, but my initial reaction is:
    1. since each traingluar face’s width varies linearly from top to bottom the perimeter on each floor must remain constant: as one triangle gets narrower it’s neighbour gets wider by the same amount. This means that the surface area is equal whether it’s a prism or an antiprism.
    2. Since the area of a regular octagon exceeds that of the corresponding square with the same perimeter (it’s more like a circle) then the volume of the middle floors of a antiprism tower exceeds that of the middle floors of a prism tower.
    Now I’m off to look at a formula to see if my intuition is sound or whether this hunch is a howler!!

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