Aces High Bulletin Board
General Forums => Aces High General Discussion => Topic started by: oneway on October 23, 2009, 01:28:32 PM
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All,
Does anyone know the terminal velocity of falling bombs in Aces High?
Oneway
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Which bomb?
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Which bomb?
I don't think it would matter.
Would it? :headscratch:
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Which bomb?
All of them. Any of them.
I understand that each bomb has its own unique drag coefficient that would effect the final trajectory...but for the sake of this discussion lets say a 500lb bomb.
?
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indeed as would a host of other factors (wind obviously not being one in this game)
but things that DO matter.........
the density of the atmosphere (clouds, etc) that the bombs are falling through would
the drag coefficient
the area of the object
the mass of the bomb
the gravity
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All of them. Any of them.
I understand that each bomb has its own unique drag coefficient that would effect the final trajectory...but for the sake of this discussion lets say a 500lb bomb.
?
Plug in your variables.............
(http://i239.photobucket.com/albums/ff107/tymekeepyr/TERMVELOC.png)
Vt = terminal velocity,
m = mass of the falling object,
g = acceleration due to gravity,
Cd = drag coefficient,
ρ = density of the fluid through which the object is falling
A = projected area of the object.
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I honestly do not think that most bombs dropped in the course of "normal" MA game play would even have the time to reach their terminal velocity.
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I don't think it would matter.
Would it? :headscratch:
I enjoy mental puzzles.
Consider this:
Wind 0-5k 0 @ 0kts
Wind 5-10k 225@10kts
Wind 10-15k 180@15kts
Wind 15-30k 090@25kts
Bomber drops bombs from 23k
Hypothesis:
For each and every unique arena wind layering set up, there exists only 2 solutions in terms of heading where the sums of the cross wind components cancel themselves out, yielding a trajectory that in essence would have the same effect/result as bombing either straight into or straight down wind in a single layer wind model.
Unless the terminal velocity and either the time it takes the bomb to reach terminal velocity or the bombs drag coefficient are known...the above cannot be proven.
Oneway
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That's because wind does not effect the bomb while falling.
Only the aircraft dropping it....
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That's because wind does not effect the bomb while falling.
Only the aircraft dropping it....
Wrong on both counts
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Wrong on both counts
There is no wind in the Main Arena
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There is no wind in the Main Arena
The velocity of my 38 only becomes terminal at the point of auger.
It is one of the first things you learn upon entering S.A.P.P.
The second being to save the blender.
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I don't think it would matter.
Would it? :headscratch:
Depends on the game engine and whether or not atmospheric variations are modelled.
In a vacuum, all objects accelerate at the same rate, dictated by gravity.
On planet Earth, this is not the case because of atmospheric variations; temperature, air density, humidity, etc, etc. In Aces High, which is a "virtual Earth," I dunno whether or not these things are modelled because Im not a coader.
Good question.
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Expecting a "Holy Grail" quote at any moment.
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Wrong on both counts
Proof instead of a simple statement without data?
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The only wind level that effects the bomb trajectory is the level in which you release them.
I just did a test off-line to verify this, if you disagree perform the following test.
0-2K- no wind
2-12k- 100mph west to east
12-14k- 20mph north to south
I upped a B-25C and dropped from 13K flying due north (20mph head wind 0 cross)
Calibrated ground speed = 270
E6B ground speed= 270
Lined up on target hanger (course and heading were due north) and hit it from 13k, dropping through a 100mph crosswind. If the lower wind levels were effecting the bombs I would have had to flown a different course and heading.
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Proof instead of a simple statement without data?
Everything in the atmosphere is effected by the atmosphere...including those things falling through the atmosphere.
Skydivers drift with the wind both in freefall and after opening their parachutes. A skydiver that freefalls for 60 seconds will drift 1320 feet in a 15mph wind. (There will be other factors that come into effect as well such as forward throw from the aircraft at exit and any maneuvering the jumper does. The 1320 feet number does not take into effect forward throw and assumes the jumper falls "straight" down.
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The only wind level that effects the bomb trajectory is the level in which you release them.
I just did a test off-line to verify this, if you disagree perform the following test.
0-2K- no wind
2-12k- 100mph west to east
12-14k- 20mph north to south
I upped a B-25C and dropped from 13K flying due north (20mph head wind 0 cross)
Calibrated ground speed = 270
E6B ground speed= 270
Lined up on target hanger (course and heading were due north) and hit it from 13k, dropping through a 100mph crosswind. If the lower wind levels were effecting the bombs I would have had to flown a different course and heading.
I thought we were talking about terminal velocity. The point at which a falling object stops accelerating. :headscratch:
Stop (terminal)
Acceleration (velocity)
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That's because wind does not effect the bomb while falling.
Only the aircraft dropping it....
There have been a lot of skydivers having to find a ride back to the DZ because they thought the same thing. <G>
A bomb is a bomb, only difference is some are meat and some are Comp B.
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I enjoy mental puzzles.
Consider this:
Wind 0-5k 0 @ 0kts
Wind 5-10k 225@10kts
Wind 10-15k 180@15kts
Wind 15-30k 090@25kts
Bomber drops bombs from 23k
Hypothesis:
For each and every unique arena wind layering set up, there exists only 2 solutions in terms of heading where the sums of the cross wind components cancel themselves out, yielding a trajectory that in essence would have the same effect/result as bombing either straight into or straight down wind in a single layer wind model.
SKUZZY! I WANT LaTeX!
In the end, we want the final velocity vector to equal the velocity which the bomb would have had assuming no wind at all.
x(T) = x0(T) where x0 = integral[v0 dt] and v0 solves dv/dt = - k v^2 initial condition v = vinitial, x = xinitial and xinitial is fixed. This part is the no-wind model (substitute for your single wind layer). It suffices since we can always shift our frame of reference so that there is no wind, resulting only in a change in the velocity vector v, which is what we have control over in the first place.
To solve the multiple wind layer rigorously, we have to consider forces. Knowing the drag equation (force is proportional to the square of velocity) and knowing that force is the first derivative of velocity:
dv / dt = - k (n1 - v)^2 in wind layer 1
dv / dt = - k (n2 - v)^2 in wind layer 2
... etc
where k is the constant appearing in the drag equation and the n1, n2, ... etc are the wind speeds in each layer. To get the final velocity vector, we have to solve the first equation and find the final velocity, then take that velocity as the initial velocity for the second equation... and so forth until we have a velocity function v(t) over the whole range. We then have to integrate to get x(t) = integral[v(t) dt]. We then have to choose vinitial so that
x(T) = x0(T) where x0 = integral[v0 dt] and v0 solves dv/dt = - k v^2 initial condition v = vinitial, x = xinitial
These being nonlinear diffeqs, "all bets are off" mathematically.
However, my intuition says that there should only be one velocity heading vector at which this could work since we only have one variable under our control (vinitial) to solve for one variable we want (x(T)).
Welcome to classical mechanics, where we can't solve the sun-moon-Earth problem without a computer!
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I must be confused with the OP's question with all answers in this thread.
He wanted to know the terminal velocity, not drift, trajectory or anything else.
Unless he was asking the wrong question.
Terminal velocity is the point at which the net force of a falling object is 0. Whereas the (downward) force of gravity and the (upward) force of drag equal each other out and said object stops gaining speed.
Depending on how far they fall, objects reach a certain speed where they STOP getting faster, and fall at a constant speed.
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Von, he was asking about this as well..
"Hypothesis: For each and every unique arena wind layering set up, there exists only 2 solutions in terms of heading where the sums of the cross wind components cancel themselves out, yielding a trajectory that in essence would have the same effect/result as bombing either straight into or straight down wind in a single layer wind model."
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Von, he was asking about this as well..
"Hypothesis: For each and every unique arena wind layering set up, there exists only 2 solutions in terms of heading where the sums of the cross wind components cancel themselves out, yielding a trajectory that in essence would have the same effect/result as bombing either straight into or straight down wind in a single layer wind model."
But it is Dale's world.
Assuming that g = 32' per second (squared) we still don't know any of the other variables such as the density o the atmosphere or the Cd, etc.
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But it is Dale's world.
Assuming that g = 32' per second (squared) we still don't know any of the other variables such as the density o the atmosphere or the Cd, etc.
True enough. We also don't know what the bombing model is, the wind model... etc.
I was just trying to give an answer for "real life".
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Would that be an african bomb or a european bomb?
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Would that be an african bomb or a european bomb?
I don't know that........ AAAAAAARRRRRRRGGGGGGGGHHHHHHH H!!!!!!!!!!!!
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OK Holy Grail.....
Everyone knows that terminal velocity is zero. All terminals I have seen are situated in concrete.
(http://www.hayneswhaley.com/project_images/IAHConcourse.jpg)
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SKUZZY! I WANT LaTeX!
In the end, we want the final velocity vector to equal the velocity which the bomb would have had assuming no wind at all.
x(T) = x0(T) where x0 = integral[v0 dt] and v0 solves dv/dt = - k v^2 initial condition v = vinitial, x = xinitial and xinitial is fixed. Note this is a no-wind model, but it suffices since we can always shift our frame of reference so that there is no wind, resulting only in a change in the velocity vector v, which is what we have control over in the first place.
To solve this rigorously, we have to consider forces. Knowing the drag equation (force is proportional to the square of velocity) and knowing that force is the first derivative of velocity:
dv / dt = - k (n1 - v)^2 in wind layer 1
dv / dt = - k (n2 - v)^2 in wind layer 2
... etc
where k is the constant appearing in the drag equation and the n1, n2, ... etc are the wind speeds in each layer. To get the final velocity vector, we have to solve the first equation and find the final velocity, then take that velocity as the initial velocity for the second equation... and so forth until we have a velocity function v(t) over the whole range. We then have to integrate to get x(t) = integral[v(t) dt]. We then have to choose vinitial so that
x(T) = x0(T) where x0 = integral[v0 dt] and v0 solves dv/dt = - k v^2 initial condition v = vinitial, x = xinitial
These being nonlinear diffeqs, "all bets are off" mathematically.
However, my intuition says that there should only be one velocity heading vector at which this could work since we only have one variable under our control (vinitial) to solve for one variable we want (x(T)).
Welcome to classical mechanics, where we can't solve the sun-moon-Earth problem without a computer!
Actually 2 Is my hunch, one representing net up wind and the other net downwind...
Layer 1 = 0>5 20kts 360
Layer 2 = 10>15 0kts 180
Layer 3 = 15>20 10kts 270
Layer 4 = 0>5 10kts 090
Layer 5 = 20>25 10kts 360
Lets look at a simplistic example. Assume the bomb is at terminal velocity the moment it is released from 25k, the aircraft is traveling directly into the wind (layer 5), and the wind layers are all of the same magnitude, 5k separations, and their vectors represent the 4 cardinals of the compass, with the last (bottom layer) being the same direction as the initial layer.
Assume: The bomb sight is perfectly calibrated to the ground speed, such that if wind was consistent from drop alt to target, the bombardier need simply place his cross hairs on the target and pickle off bombs. The aircraft is flying due North into the wind with zero cross wind component relative to its trajectory. The bomb sight is a self compensating mechanism for head and tail wind components of any wind vector.
As the bomb falls through the first layer (20>25k...aka layer 5)...the bomb tracks perfectly to target.
When the bomb falls through layers 3 and 4, it drifts left then right the same distance; both vectors cancel themselves out during our bombs journey.
Upon entering layer 2, the bomb picks up what in essence is a 10kt tail wind. As the bomb enters the final layer, it encounters a 10kt relative head wind.
This bomb having transited through 5 different wind layers will still impact the target perfectly, as each of the four lower layers will have canceled themselves out, and the bomb sight was perfectly calibrated for the layer in which it dropped.
If the aircraft were flying directly down wind, the effect and result would be identical, hence the 2 solution theory.
Now consider this scenario:
Wind 0-5k 0 @ 0kts
Wind 5-10k 225@10kts
Wind 10-15k 180@15kts
Wind 15-30k 090@25kts
I propose that their exists 2 unique headings whereby the cross wind of this layering scheme is washed out, leaving only head or tail wind variables on the table to consider. Bomber pilots who are adept at bombing in the wind will recognize the benefit of calculating those two headings...
Of course without some form of drag value to calculate the speed of the bomb as it passes through various layers, its impossible to calculate.
Writing a computer program that does this would be simple if the values were known.
Oneway
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Proof instead of a simple statement without data?
If you study the whole thread you will see proof that wind does indeed effect the trajectory of a bomb relative to a fixed land target, which of course is proof that "ONLY the speed of the airplane effects the bomb" is Wrong...
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I thought we were talking about terminal velocity. The point at which a falling object stops accelerating. :headscratch:
Stop (terminal)
Acceleration (velocity)
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
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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.
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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?
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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. :D
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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.
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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...
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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.
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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>
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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
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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.
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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.
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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. :D
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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.
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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?
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I still don't think it will take off!
:D
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HT explained this one at the con.
The short answer is that the Bombsight allows for crosswinds when you calibrate.
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HT explained this one at the con.
The short answer is that the Bombsight allows for crosswinds when you calibrate.
What's the long version?
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HT explained this one at the con.
The short answer is that the Bombsight allows for crosswinds when you calibrate.
:huh
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HT explained this one at the con.
The short answer is that the Bombsight allows for crosswinds when you calibrate.
HOLY CRAP that's awesome.
Is that how things were historically?
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Wrong on both counts
In AH, sir, you would be misinformed.....
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Von,
My response was to answer the third post by the OP. The initial question was about terminal velocity, but the more complete question was about crosswinds. That is what I was answering, and the test I ran is at odds with what Hitech stated at the CON, just like Ghosth posted.
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oneway,
When dropping bombs with a headwind or tailwind the key to hitting the target, is to match the calibrated speed to the ground speed, not the true air speed.
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In case of interest (worked out several years ago for the Stalin's Fourth scenario):
http://electraforge.com/brooke/flightsims/aces_high/levelBombing/bombingInWind.html
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Dang Brooke, give up all our secrets why don't you, he was thinking that every layer effects bomb trajectory.
That would have kept him busy for days.....
:D
Strip
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In case of interest (worked out several years ago for the Stalin's Fourth scenario):
http://electraforge.com/brooke/flightsims/aces_high/levelBombing/bombingInWind.html
Wow you used perturbation theory.... last time I used that was for quantum. Looks like a very good analysis right there. Should be on the trainer site... if it isn't already.
Out of curiosity, about how fast must the crosswind be to have a noticeable effect on the impact site assuming no bomber drift?
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Out of curiosity, about how fast must the crosswind be to have a noticeable effect on the impact site assuming no bomber drift?
You mean where the bomber is in a wind layer of 0 wind, but under it is a wind layer of some velocity? I haven't worked out what the speed would need to be, but it would be much higher than any realistic wind. Even in cases with the bomber in a calm layer (zero wind) at 10k, and under it all the way to the ground is a 30 mph crosswind, it didn't have any significant effect on side deviation of bomb-impact point. That's the third picture of the drop from the Ju 88.
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Yes, that's what I meant. Thanks for the info... I guess that's what I get for not working out the numbers... I go off on a hugearse tangent about differential equations and nonlinearity and the inability to actually solve the equation when the actual change due to wind layer drag is very small.
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Can I assume that the bomb falls 45 degrees?
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oneway,
When dropping bombs with a headwind or tailwind the key to hitting the target, is to match the calibrated speed to the ground speed, not the true air speed.
Of course....
Did I mis-speak?
Oneway
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Can I assume that the bomb falls 45 degrees?
No
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I sure at one microsecond in time you could....
:D
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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.
Absolutely correct. It would accelerate at a constant 9.8 m/s2.
Mass has no bearing. A 1000kg bomb will hit the ground at the same time as a 5kg one will, when dropped from the same height, and with the same drag co-efficient.
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same drag co-efficient.
Being the key.
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Absolutely correct. It would accelerate at a constant 9.8 m/s2.
Mass has no bearing. A 1000kg bomb will hit the ground at the same time as a 5kg one will, when dropped from the same height, and with the same drag co-efficient.
So the real question is, how fast is it going to get to Terminal Velocity, is what your saying.
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It's more like this:
You don't need a weatherman to know which way the wind blows.
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Expecting a "Holy Grail" quote at any moment.
"Your mother was a hamster, and you're father smelled of eldeberries!"
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"Your mother was a hamster, and you're father smelled of eldeberries!"
hehe.. hooray for Monty Python
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hehe.. hooray for Monty Python
Haven't you done enough hijacking for one night?
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Haven't you done enough hijacking for one night?
I was told the last one was not High enough.
I have since refilled my blender.
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I was told the last one was not High enough.
I have since refilled my blender.
My blender was backordered. Until then I'll settle on sugar rushes from Monsters
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My blender was backordered. Until then I'll settle on sugar rushes from Monsters
The Skytrains came in friday night. You need to check.
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D'oh. It's apparently been on my roof since then. Thanks for making me go check. Now I can really auger!
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D'oh. It's apparently been on my roof since then. Thanks for making me go check. Now I can really auger!
Painlessly :D
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Painlessly :D
Well, it all depends on the velocity I reach during my auger...
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now that is textbook hijacking. :aok
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now that is textbook hijacking. :aok
I put the conversation back on topic :devil
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Are we still talking about the original question?
All objects fall at the same rate. No wind in the MA. But terminal velocity also depends on mass, shape and Atmospheric Density.
I am unsure if HTC has modeled atmospheric pressure in relation to falling objects.
NASA has a simplified explanation:
http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html
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Its probably a good bet that HTC used 59F and a standard atmospheric pressure (29.921 in Hg) so that the density of air would come out in the standard unit for engineering as 0.002378 slugs per cubic foot. And to be absolutely precise the acceleration in American units would be 32.174 ft per sec squared (just to help tidy things up).
Carry on! :salute
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Wow reading some of these post is making my head hurt haha! You know when you were in school sitting in say a physics class thinking to yourself " I'll never use this information in my lifetime" here's your chance to use that useless information! Dang I need a beer! :lol
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Just remember the most important factor. Terminal velocity is severely affected upon impact with the target. Or was it the target is severely affected upon impact at terminal velocity?
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Are we still talking about the original question?
All objects fall at the same rate. No wind in the MA. But terminal velocity also depends on mass, shape and Atmospheric Density.
I am unsure if HTC has modeled atmospheric pressure in relation to falling objects.
NASA has a simplified explanation:
http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html
Incorrect. If I drop a lead ball and a feather, which one hits the ground first?
They accelerate at the same rate until they reach terminal velocity, where the drag of the object falling through the air is equal to its weight.
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the density of air would come out in the standard unit for engineering as 0.002378 slugs per cubic foot.
What kind of an engineer measures density in units of mass of slimy snails per volume of a deformed foot? :rolleyes:
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What kind of an engineer measures density in units of mass of slimy snails per volume of a deformed foot? :rolleyes:
You know my whole life I have waited for the first stupid question and I think you must have worked all of two minutes on that one! :neener:
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is it about 120mph for a falling object from a static drop? So something to due with airspeed and drop angle calculated with normal TV of earths atmosphere.
But i was wondering if the fastest the bomb would be is when it is dropped, as this may be as high as 500mph. So is the bomb decelrating the whole time? (i didnt read the whole thread so forgive me if its been suggested.)
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But i was wondering if the fastest the bomb would be is when it is dropped, as this may be as high as 500mph. So is the bomb decelrating the whole time? (i didnt read the whole thread so forgive me if its been suggested.)
It really depends on the design of the bomb. For most, making a big crater is not the desired effect and if they employ an impact detonation, a very high speed on impact is not desirable. Bombs designed for penetration are designed with a higher terminal speed. The 7-10 ton concussion bombs carried by the lancs were meant to dig deep into the ground before exploding. For this reason they were dropped from enough altitude to reach the terminal velocity which was very high (I am sorry I, don't want to quote wrong numbers).
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is it about 120mph for a falling object from a static drop? So something to due with airspeed and drop angle calculated with normal TV of earths atmosphere.
But i was wondering if the fastest the bomb would be is when it is dropped, as this may be as high as 500mph. So is the bomb decelrating the whole time? (i didnt read the whole thread so forgive me if its been suggested.)
No. The bomb would be at it's slowest when it is first dropped
The bomb would be accelerating the whole time until it reaches its <drum roll, please> Terminal Velocity. :)
That is, of course, if the bomb in question is dropped from a high enough altitude for it to reach its potential terminal velocity.
The following equation (and the values of the variables) are all that is needed to determine terminal velocity. Nothing less, nothing more.
(http://i239.photobucket.com/albums/ff107/tymekeepyr/TERMVELOC.png)
The only thing that will be affected by any other variables not included in the above equation such as angle of drop, etc, is the
TIME IT WILL TAKE to REACH it's terminal velocity.