Terminal Velocity of Falling Objects Question

[TEShaw]

I've got try in here.

Wind would have no effect on the downward velocity of anything falling unless the wind was blowing UP.

Everything is 32 ft/sec/sec

Call me Galileo.

[Kermit de frog]

If you set wind layer from 2-20k for 100mph from the West.
You level your bombers at 21k heading North.
Will your bombs hit the target or veer to the side?

[VonMessa]

I've got try in here.

Wind would have no effect on the downward velocity of anything falling unless the wind was blowing UP.

Everything is 32 ft/sec/sec

Call me Galileo.

As long as you are on Earth.     

[boomerlu]

Oneway,

While I understand the logic behind your assessment, there are several things to consider...

First, the force due to drag is not linear with relative airspeed. In fact, it is proportional to the square of the relative airspeed.

Second, the velocity gained or lost whilst in a given wind layer is Integral[ drag force dt ]. If the time changes, then the amount of velocity gained in a layer will change. This factor only washes out if we're at terminal velocity.

Now, consider three wind layers, one still (0 mph) 10-15k, one west (25 mph) 5-10k, one east (25 mph) 0-5k, and you are traveling due north. Assuming the bomb reaches terminal velocity when it hits the west-layer, the velocity gained is given by

dv/dt = - k (25 - v(t))^2

or in terms of position

dx^2/dt^2 = - k (25 - dx/dt)^2

from the wind. This is a second order nonlinear differential equation, which in general does not have a solution which you can write down nicely, let alone a solution which is linear.

The next layer will have another of the same type of equation. In real life, it is virtually impossible that these effects will "wash out" if you are heading due north since the force is proportional to the square of the velocity.

In this perfect scenario, there should be two drop headings which give a true overall trajectory i.e., you've picked out a heading so that these will wash out. HOWEVER this is due to an inherent symmetry in the system - you do not care about the whether a quantity is left or right wrt to the airplane so long as it cancels out, so with only two crosswinds, we have a symmetry to exploit. With no such symmetry, there should only be one solution.

[oneway]

If you set wind layer from 2-20k for 100mph from the West.
You level your bombers at 21k heading North.
Will your bombs hit the target or veer to the side?

I have completely assumed that all layers will effect bombs as they pass through them...a previous poster contends that is not the case.

In all of my testing when dropping into the wind bombs fall short, and dropping downwind bombs fall long...in these cases winds were above 5k as a single layer and it was dead calm below that...

Obviously as the bomb falls into the dead zone it no longer has the net relative head or tail wind component and thus will either fall in front of or over the target. In this case it is demonstrably clear that the bomb is indeed effected by more than 1 layer.

Its very simple to perform that test in the offline TA...

[Shuffler]

Terminal velocity is required to calculate the effect of the various wind layers as the bomb passes through them.

We would need to know how long it took to reach terminal velocity and what that velocity is to solve the puzzle.

The bottom line of what is needed is the time the object spends in each layer in order to calculate the cross wind components relative to the initial release trajectory

Not to mention that the atmosphere is thinner the higher you go. So terminal would also be higher. Now..... will the bomb ever actually be at terminal velocity? As the bomb falls through the atmosphere and the barometric pressure increases there is more drag. Slowing the bomb down. Since the bomb is slowing it is not "at" terminal velocity but above it as it attemps to neutralize. However it never neutralizes as the pressure is ever increasing as it the bomb falls.

[Bosco123]

Everything in this game I would thing would be constant, making this so much easier than you guys are making this. I would like to know were we will start at, I personally will start out at 10000ft., and that would be easy enough to find out what the final velocity it will reach before it hit's the ground.

Vi(Velocity initial)=0m/s
A(Acelleration)= 9.0 m/s2 (squared) (and that is due to gravity) and since we are going with gravity, it will a positive number
Di(Distance initial)= 3048m
Df(Distance final)= 0m (asuming that you hit somthing at ground level)

I am going of memory, my physics notebook is in my locker, and there is an equation that you have to know for this, but I'm pretty sure I know it.

Vf2(squared)= Vi2(squared)+A(Di) Plug in the numbers!
Vf2(squared)= (0m/s2)+(9.80m/s2)(3048m)
Vf2=29870.4
Vf=172.835m/s Becasue there is this wonderful thing in science called significant figures (which is way to hard explain) The real answer would be 173m/s
Finding the terimal Velocity of somthing is going to be fun. The Coefficent is going to be equal to the force of gravity because both gravity and wind resistance have reached equalibrium.

When I come back tonight, I'll get my physics notebook out and give you guys the equations and i'll probably even solve it lol

<S>

[oneway]

Not to mention that the atmosphere is thinner the higher you go. So terminal would also be higher. Now..... will the bomb ever actually be at terminal velocity? As the bomb falls through the atmosphere and the barometric pressure increases there is more drag. Slowing the bomb down. Since the bomb is slowing it is not "at" terminal velocity but above it as it attemps to neutralize. However it never neutralizes as the pressure is ever increasing as it the bomb falls.

Good point...

The Golden Fleece is being able to figure out how much time the bomb spends in each layer...its actual velocity at any given point in time is irrelevant to the broader problem...what heading neutralizes all cross wind component vectors?

Oneway

[boomerlu]

Oneway,

Basically, this is a freaaaakin complicated problem. To solve this for a given wind layer pattern and airplane heading/velocity, you're going to have to program a numerical differential equation solver or find one. I have one in C++ I made for a physics class once, solving for projectile motion with drag (the artillery problem), but it only uses two spatial dimensions.

[boomerlu]

Everything in this game I would thing would be constant, making this so much easier than you guys are making this.

Sorry, but the atmospheric density varies with altitude, else we wouldn't have realistic IAS vs TAS. Also, this problem is not simple kinematics. Since we've introduced wind layers (and thus implicitly drag) into the problem, we're no longer in the realm of the four kinematical equations, but rather must resort to differential equations which are quite a deal harder to solve.

[Shuffler]

Based on what I said before ...... you'd only have to calculate for the last 5 feet. Assuming the bomb were up to speed. Mass and drag at that point will tell you. As long as it is dropped high enough to reach terminal velocity, from then on it is actually not at terminal because it will slow all the way to the target.   

[boomerlu]

Good point...

The Golden Fleece is being able to figure out how much time the bomb spends in each layer...its actual velocity at any given point in time is irrelevant to the broader problem...what heading neutralizes all cross wind component vectors?

Oneway

The vector DOES matter because drag is proportional to the square of the RELATIVE velocity of the object and the surrounding fluid (i.e., the wind layer). In other words, the drag is proportional to the square of the difference between the object velocity and the atmospheric velocity. And again, it is a differential equation, so things do not wash out nicely.

[Shuffler]

I'm beginning to think this bomb will explode before we reach a conclusion.

Did we drop this or are we on the receiving end?

[Kermit de frog]

I still don't think it will take off!
 

[Ghosth]

HT explained this one at the con.

The short answer is that the  Bombsight allows for crosswinds when you calibrate.