Q: What is the terminal velocity of a 24-105 f/4 L?

[wickerprints]

I calculate that the terminal velocity of an EF 24-105/4L IS USM under these circumstances is roughly 72 miles per hour, or 32.23 m/s. My calculation is based on the following assumptions:

1. The lens may be modeled as a short cylinder with negligible surface features.
2. The projected (cross-sectional) area of the lens at any time is given by its diameter times its length--i.e., it did not extend.
3. The lens at the moment of impact, falls through air at 20 degrees Celsius at a pressure of 1 atmosphere.

The relevant formula is then given by:
<p style="padding-left: 30px;"]V_t = Sqrt[2 m g / (rho A C_d)],


where:

• m = 0.69456 kg is the mass of the lens + hood (Source: TDP)
• g = 9.80665 m / s^2 is the nominal average acceleration due to gravity at sea level (Source: en.wikipedia.org/.../Standard_gravity )
• rho = 1.20 kg / m^3 is the approximate density of air at 20 C at 1 atm (Source: en.wikipedia.org/.../Density_of_air )
• A = 0.0095014 m^2 is the cross-sectional area of the lens with the barrel fully retracted (Source: TDP)
• C_d = 1.15 is the drag coefficient of a short cylinder (Source: en.wikipedia.org/.../Drag_coefficient )

The resultant calculation yields V_t = 32.2327 m / s = 72 mi / hour. This is fast, but not as fast as some of the estimates mentioned in the media, which put the terminal velocity as high as a few hundred mph.


That said, the calculation DOES change if we assume that the lens hood is mounted to the lens, and the lens extends upon freefall because air resistance "pulls" it open. There are more complexities involved as well, due to the fact that the lens is not actually a smooth cylinder. If we modify our first two assumptions accordingly; i.e.,

1. The lens may be modeled as a long cylinder.
2. The projected area of the lens at any time is given by its diameter times its extended length.

Then the last two variables change:

• A' = 0.01221 m^2 is the cross-sectional area of the lens with the barrel fully extended
• C'_d = 0.82 is the drag coefficient of a long cylinder.

This gives a terminal velocity of V'_t = 33.67 m / s = 75.3 mi / hour, slightly faster than the previous estimate. Finally, if we suppose that the drag on the lens is as minimal as that of a perfect sphere (which I think is the lowest reasonable value), then V_t(sphere) = 50.42 m / s = 113 mi / hour. This last value provides what I consider to be an absolute upper bound on the terminal velocity of this lens. Its true value should be considerably slower, given that the lens is not a smooth sphere, but has numerous ridges and even a lens hood that would significantly increase drag when mounted in nearly any position. While it is true that the terminal velocity is greater at high altitude, by the time the lens hits the roof, the relevant variables correspond to the atmospheric conditions at low altitude.


In summary, I would put the terminal velocity of the lens to be somewhere in the neighborhood of 65 - 80 mph.

[Daniel Browning]

+1 wickerprints!

[Rocco]

Ha!