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*From*: fletcher@physics.usyd.edu.au*Date*: Thu, 09 Feb 2012 15:54:52 +1100

Hi Mike

Try giving the authors of

http://onlinelibrary.wiley.com/doi/10.1017/S0952836904005813/abstract

an email.

Cheers Fletch

Quoting "M. Horton" <scitch@verizon.net>:

I'm trying to help a young student with the calculations for her science

fair project. She read that some of the lift generated by a hummingbird is

created by swirling votices coming off of the wings. She wants to know what

percentage of the lift comes from traditional downward acceleration of air

and what percentage comes from other effects during hovering. So, she got a

hummingbird to land on a hummingbird feeder that was on a digital scale and

got a reasonable mass (4.2 grams). She then used high speed video to

determine how long it takes for a wing to go back and forth and got a

reasonable answer (7/300 of a second). She then used a life-size photograph

to build a model of the wing beating and determined its volume to estimate

the volume of air moved by the wings. She then used the equation F=mv/t

(equation 1 on this site: http://jeb.biologists.org/content/207/8/1345.full)

from to calculate the force applied to the air. She used the middle of the

wing since it starts at 0 at the shoulder and a maximum at the tip. She

used the average velocity over the entire stroke. The answer came out about

an order of magnitude smaller than I expected. I expected for the upward

force to be at least as large, if not larger than the downward force due to

less than 100% efficiency.

Here are the calculations. Did we make a mistake?

Force down = weight of bird

W=mg = .0042 kg (9.8m/s^2) = 0.041 Newtons

Force up = force applied to air by the wings

volume of air moved = 37 mL in one back and forth stroke (measured by

building a model and using water displacement)

mass of air = (37 mL)(.0013 g/mL) = .0481 g or .0000481 kg

Distance traveled by the center of the wing (as measured on the model) is

9.6 cm or 0.096 m

One back and forth motion takes place in 7 frames of 1/300 second each on a

high speed camera or .023 seconds

So, the average velocity during the motion is .096/.023 or 4.2 m/s

So, the force generated is equal to (4.2m/s)(0.0000481 kg)/(0.023 seconds) =

.0088 Newtons

According to this calculation, only 21% of the bird's lift is provided by

forcing air down. This seems pretty small. Did we miss a decimal somewhere

or make an invalid assumption?

Thanks in advance,

Mike

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