Originally posted by vorticon
can a floatplane take off going up a fast moving river? and why is this different?
A float plane and a plane on wheels are not the same.
Plane on wheels.The force exerted on the bottom of the wheel is caused by rolling friction which DOES NOT depend on velocity. The force of rolling friction is ...
T = C * W
T - force of friction
W - the weight of the plane
C - constant coefficient dependent only on materials of the wheel and the surface on which the wheel is rolling
There are two sources of rolling friction. The rolling friction between the bottom of the wheel and the ground (or the conveyor), and the other is the rolling friction of the wheel bearing . The former is designed to be large. You do not want wheels to lose traction and slip, so you make the tire out of soft rubber. The later source of friction is designed to be as small as possible (thus you use hard and polished steel balls rolling on the hard and polished steel surface). Both friction points are subject to the same equation (F = C * W), but they have significantly different coefficients C. One purposely large the other one purposely small.
Now, imagine the wheel (forget the plane for a moment) weighing W sitting on the ground. The rolling friction is as we said T = C * W, but remember this is not a force yet.
Now, this is very important to understand. Friction is a tricky thing. It is not a force in itself. It is "an ability" of the coupling between two surfaces to resist the tangential force P. When P is zero, T will be zero as well. As P slowly increases, T will match it exactly and as a result the there is no movement between the surfaces. This continues until P reaches C * W. This is the MAXIMUM resistance to movement coupled surfaces can produce. As soon as P exceeds C * W, the wheel starts rolling (or in a case of static friction the surfaces start slipping).
Until now, we had equilibrium. For all values of force P between 0 and C * W, P was always matched by equal and opposite T, and therefore no movement.
As P increases even more, the equilibrium is no more. The coupling can only produce a maximum of C * W, the rest of the force P becomes unbalanced, so the movement must result. The wheel starts rolling and if P > C * W the wheel rotation will accelerate. If P = C * W, the wheel will not accelerate but it will maintain its state. If it was stationary, it will remain so, but if it was rolling with the velocity of V it will remain doing so.
This is why if you want to start taxiing, you must apply more thrust to get the plane rolling, but then you can lower the power to sustain a constant roll.
The moral from this story is that there is a limit to how much force the ground (or conveyor) can exert on the plane. There are two friction couplings between them, and the amount of force transfer is determined by the weaker of the two. Normally the wheel/ground coupling is stronger (it was designed to be so) and the wheel/axle coupling is weaker. The wheel will not slip on the ground, the wheel will roll. When the engines push the plane, the plane moves forward because the wheel/axle friction is too small to stop it.
The situation changes when you apply brakes. Now you just made C of the wheel/axle coupling very large. Now, since the wheel/ground coupling is weaker of the two, the outcome depends only on its coefficient C. If this coupling is strong enough to transfer a force equal to engines thrust, the plane remains stationary (this is what we want), but if the coupling gives, the wheels slip (no rotation) and the plane moves forward. Put the plane on ice and even with brakes applied the full thrust most likely will move it forward.
Unless the brakes are applied, you can tug the bottom of the wheel forward or backward all you want. You can use any value of the force you want. Only a small fraction of this force (defined by C of the wheel/axle friction) will be transferred to the plane. The rest will produce an accelerated angular velocity of the wheel.
The angular momentum of the wheel does not affect plane's movement forward even if it is very large. It will only affect the plane if it turns left or right (gyroscopic effect), which we don’t care about in this case.
Float plane.The force exerted on the pontoons by the flowing stream is NOT friction. It is a force produced by the pressure of water (drag) which is proportional to the square of the velocity of the stream.
A simple case, a torpedo. To simplify the case let’s assume that instead of the domed bow, its bow is flat. In other words, our torpedo is a cylinder.
What kind of forces act on the torpedo. The thrust pushes it forward, and two forces push back (let’s ignore cavitation, turbulance and other gunk). The first is the pressure of water in front of the torpedo being compressed by its movement. Let’s call this force P. The other one is a viscotic friction acting on the sides of the torpedo as it moves forward. Let’s call it Q.
Q depends on the velocity V, but unless the torpedo is moving extremely fast in molassas, it will be negligible in comparison with P.
Let’s start with the Newton’s law. A familiar form is …
F = m * a
or
F = m * V / t
F * t = m * V
… but this is a static case where neither of the components changes.
Let’s consider an infinitely small interval of time dt and virtual movement. The torpedo wants to move forward a distance dx which with the constant velocity V will be equal to V * dt. The mass of water displaced by this movement dm is equal to the volume of water displaced W times water’s density R.
So, we star with …
F * t = m * V
… for a virtual movement
F * dt = dm * V
Both F and V are constant (do not depend on time). The mass of water displaced does depend on time. More time elapsed more water is displaced.
The volume of water displaced in time dt is equal to the surface of the torpedo’s cross section S times the distance traveled dx
dm = S * dx * R
since
dx = V * dt
and
S = pi * D^2 / 4
dm = pi * D^2 / 4 * V * dt * R
so
F * dt = pi * D^2 / 4 * V * dt * R * V
F = (pi * D^2/4 * R) * V^2
F = const * V^2
This was a simple solution for a flat surface. If the torpedo’s bow is conical or semi spherical, the solution is the same but the const will change.
The point is that for the sea plane moving against the current, the force resiting the movement IS NOT FRICTION. It is a water pressure which not only depends on the velocity, but the dependency is pretty strong (V squared).
There is no friction coupling between the pontoon and the plane. The connection is rigid, so the entire force applied to the pontoon by the water will be transfered to the plane.
For the plane moving on wheels, the force resisting the movement is friction and it DOES NOT depend on velocity.
There is a friction coupling between the wheel and the plane, therefore only a portion of the force applied to the wheel by the ground/conveyor will be transfered to the plane.