Bird strike(s)

"Jay Honeck" wrote in message
news:rVHbf.520936$x96.436058@attbi_s72...
Paraphrased from Sport Aviation this month:

Hitting a Canada Goose exerts the same force as dropping a 1000 pound
weight 10 feet.

Yikes! That would make mince-meat out of ANY GA aircraft.

It would if the claim were true. But a little high-school physics shows it's
not. (And knowing that it's not could bear on important choices you make
while flying--if the claim were true, you'd want to choose almost any
alternative to such a collision.)

Assuming the same compressibility, the forces of the two collisions would be
proportionate to the colliding objects' respective momenta. After dropping
ten feet, an object has a velocity of about 15 knots; hence, a 1000-pound
weight has a momentum of 15,000 knot-pounds. A Canada Goose weighs up to 14
pounds; hence, at (say) 120 knots, its momentum is at most 1,680
knot-pounds--about an order of magnitude less than what's asserted above. If
the 1000-pound weight is harder (less compressible) than the goose, then the
asserted comparison is wrong by an even greater factor.

--Gary

"Gary Drescher" wrote:
Assuming the same compressibility, the forces of the two collisions
would be proportionate to the colliding objects' respective momenta.
After dropping ten feet, an object has a velocity of about 15 knots;
hence, a 1000-pound weight has a momentum of 15,000 knot-pounds. A
Canada Goose weighs up to 14 pounds; hence, at (say) 120 knots, its
momentum is at most 1,680 knot-pounds--about an order of magnitude
less than what's asserted above. If the 1000-pound weight is harder
(less compressible) than the goose, then the asserted comparison is
wrong by an even greater factor.

The goose and falling weight do, however, strike with comparable kinetic
energies:

E_goose = 0.5*14*120*120
= 100,800

E_wt = 0.5*1000*15*15
= 112,500

So maybe this is why Sport Aviation claims the strikes are comparable. 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.)

"Jim Logajan" wrote in message
.. .
The goose and falling weight do, however, strike with comparable kinetic
energies:

E_goose = 0.5*14*120*120
= 100,800

E_wt = 0.5*1000*15*15
= 112,500

So maybe this is why Sport Aviation claims the strikes are comparable. 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.)

Yup, good point. If the goose exerted the same force as the falling weight,
the goose's energy would be much greater than the falling weight's; instead,
the goose exerts far less force, but its energy is comparable to the falling
weight's.

--Gary

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