Bird strike(s)

Jim Logajan wrote:
If I recall correctly, damage is roughly proportional to energy of
impact, not momentum. (Based on the theory of spring deflection, I
believe: Suppose the object (goose or large weight) strikes a
compression spring. The spring would compress to about the same amount
because the spring equation, E_spring = k_spring_constant *
X_deflection, shows the linear proportionality between energy and
compression.)

Oops! What I wrote here is wrong. The equation E = k*X is only true for a
rare breed of springs known as constant force springs[*]. For conventional
Hook's law springs (F = k*X), the equation is of course E = 0.5*k*X^2.

So if E_kinetic = 0.5*m*V^2 and E_spring = 0.5*k*X^2, and the two energies
are set equal, after a little algebra the deflection is found:

X = V*sqrt(m/k)

So by the spring theory, damage WOULD be linearly propotional to the speed
while proportional to the square root of the mass - i.e. doesn't rise as
fast. Given the earlier example:

X_goose = 120*sqrt(14/k) ~= 449 * sqrt(1/k)
X_wt = 15*sqrt(1000/k) ~= 474 * sqrt(1/k)

Hmmm - interesting that they are still comparable with this selection of
weights and speeds!
[*] A spring loaded measuring tape is the most commonly known household
example of an item that has a constant-force spring in it. The restoring
force is the same no matter how far you pull the tape out.

"Jim Logajan" wrote in message
.. .
So by the spring theory, damage WOULD be linearly propotional to the speed
while proportional to the square root of the mass - i.e. doesn't rise as
fast. Given the earlier example:

X_goose = 120*sqrt(14/k) ~= 449 * sqrt(1/k)
X_wt = 15*sqrt(1000/k) ~= 474 * sqrt(1/k)

Hmmm - interesting that they are still comparable with this selection of
weights and speeds!

The new numbers are just the square roots of double the old numbers, so
they're pretty much guaranteed to still be comparable. :-)

--Gary