wingspan33 wrote:What a great piece of outreach equipment!
Thanks Scott!!
Margie (and a whole lot of other people) deserve a lot of credit for helping with the project.
wingspan33 wrote:I wonder what wind speeds Little Hawk would need to develop enough lift to "fly" a typical adult pilot. Less than Jet Stream speeds I hope.
OK, let me put on my aeronautical engineering hat along with my 2 years of wind tunnel experience ...
The lift generated by a wing is represented as follows:
Lift = q x CL x SW
where CL is the Coefficient of Lift (a dimensionless quantity that captures all the nasty aerodynamic details)
where SW is the Wing Surface (in square feet, for example)
and where q is the "Dynamic Pressure" which is equal to 1/2 x rho x velocity x velocity
In this last equation...
rho is the density of the air
velocity x velocity is "v squared" (which I cannot easily represent in this forum).
OK, now we're ready to figure it out. Let's say that this thing is going to be flying me, and so the lift that it generates must equal the same lift that my Falcon 195 generates (since I weigh the same regardless of which glider I'm flying). But Little Hawk is 65 square feet while my Falcon is 195 square feet, so that's a factor of 3 times smaller. So let's write the two equations for the two gliders.
For my Falcon: Bob's weight = Lift = qFalcon x CLFalcon x SWFalcon
For Little Hawk: Bob's weight = Lift = qLittleHawk x CLLittleHawk x SWLittleHawk
Since Bob's weight is the same in either case, we can equate those two equations (ignoring the difference in the weight of the gliders):
qFalcon x CLFalcon x SWFalcon = qLittleHawk x CLLittleHawk x SWLittleHawk
Now if we assume that they both have the same Coefficient of Lift (which they would if they were scaled versions of each other rather than being a single surface compared to a double surface) then we can just use CL for both sides of the equation:
qFalcon x CL x SWFalcon = qLittleHawk x CL x SWLittleHawk
Since CL is known to be non-zero, we can divide both sides by CL to get:
qFalcon x SWFalcon = qLittleHawk x SWLittleHawk
But we know that the surface of the Falcon is 3 times the surface of Little Hawk, so we can substitute as follows:
qFalcon x 3 x SWLittleHawk = qLittleHawk x SWLittleHawk
Again, since we know that SWLittleHawk is non-zero, we can divide both sides by that value to get:
qFalcon x 3 = qLittleHawk
Now we can plug in the equation for Dynamic Pressure ("q") for each wing to get:
1/2 x rho x velocityFalcon x velocityFalcon x 3 = 1/2 x rho x velocityLittleHawk x velocityLittleHawk
Of course, we can divide out the 1/2 and the "rho" since both are non-zero to get:
velocityFalcon x velocityFalcon x 3 = velocityLittleHawk x velocityLittleHawk
Now we can switch both sides (since we're looking for velocityLittleHawk) to get:
velocityLittleHawk x velocityLittleHawk = 3 x velocityFalcon x velocityFalcon
Now we can take the square root of each side to get:
velocityLittleHawk = squareroot(3) x velocityFalcon
The square root of 3 is about 1.73, so that means that the speeds for Little Hawk would be about 1.73 times the speeds of my Falcon.
So the real key to remember here is that the lifting capacity of a wing goes up as the SQUARE of the velocity. That's why a 20mph wind on launch is 4 times more difficult to deal with than a 10mph wind. The forces go up as the SQUARE of the velocity. But the forces only go up linearly with the wing surface. So if you cut the wing surface in half, the speed would go up as the square root of 2. But if you cut the wing surface by 3, then the speed goes up as the square root of 3.
I hope that answers the question. Sorry for all the math.
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