Recovery Drift.

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Alby

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If the wind is blowing at 0-5/mph and your rocket ejects its parachute
at 2,250/ft and its descent rate is 3.5 m/s, about how far would you estimate
it will drift?
 
Hmmmm, maybe I answered my own question. lol Guess I should have pulled
out the calculator.

I'm getting around 1,400/ft away.


3.5 m/s = 11.5/ft per sec
2250/11.5 = 196/seconds (3.26/mins) From Deployment to Ground
--
5/mph * 5280ft-per-mi = 26,400/ft
26400/60secs = 440/ft the rocket drifts per minute
440ft * 3.26mins = 1434.4/ft


Hopefully my basic math is correct. :)
 
A simple way to think about wind drift is to scale the problem to a 10 mph wind and the standard descent rate of 15 fps.

10 mph is 15 fps, so if you descend at 15 fps in a 10 mph wind, your rocket drifts 1' sideway for every foot of altitude lost. For example, if you apogee at 1 kft, take a 1 kft walk.

With a 5 mph wind, the drift is half, and with the max 20 mph wind it doubles.

Bob
 
Thanks... I'll keep that rule of thumb on file. :)
 
My rule of thumb is to time how long it takes for rockets to leave the field. At a club launch, I can watch other rockets and accurately figure this out with a stop watch. Then, I can fly rockets with shorter flight durations.
 
This is certainly a complex question, but it would appear that the greater the surface area to mass ratio, the higher the "drift sensitivity factor" for a given amount of cross-wind and the same rate of descent.

This would certainly be an interesting study to come up with some empirical data.

Greg
 
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This is certainly a complex question, but it would appear that the greater the surface area to mass ratio, the higher the "drift sensitivity factor" for a given amount of cross-wind and the same rate of descent.

No. It's really as simple as noted in the previous posts.

-- Roger
 
So...if the rocket lands at the top of a 50 ft tree, I can save 50 ft of walking :rolleyes:
Absolutely correct, however you need to add 50' of climbing, a 50' recovery pole or a chain saw to accomplish a successful recovery. :blush:

Bob
 
This is certainly a complex question, but it would appear that the greater the surface area to mass ratio, the higher the "drift sensitivity factor" for a given amount of cross-wind and the same rate of descent.

This would certainly be an interesting study to come up with some empirical data.

Greg
Unless you are actively changing the angle of attack and L/D ratio of the chute in an attempt to steer the rocket during its descent, simple vector addition rules apply.


Bob

PS Sorry, I initially edited your orignal post by accident when writing this reply. Pressed edit instead of quote.
 
Unless you are actively changing the angle of attack and L/D ratio of the chute in an attempt to steer the rocket during its descent, simple vector addition rules apply.


Bob

PS Sorry, I initially edited your orignal post by accident when writing this reply. Pressed edit instead of quote.


I would agree that is a generally holds true for standard 3FNC/4FNC's, but for something like the Estes Orbital Transport, it is a little more complex. There is a lot of surface area that can be pushed against, and if it rotating... Plus we are making assumptions that the winds are constant at all altitudes (speed and direction). So while "yes" the math may be the easy part, estimating where it will actually land because of the drift is the complex part. So, if someone has a formula to calculate the CEP for the LZ, I'd like to use it. :)

Greg
 
I would agree that is a generally holds true for standard 3FNC/4FNC's, but for something like the Estes Orbital Transport, it is a little more complex. There is a lot of surface area that can be pushed against, and if it rotating... Plus we are making assumptions that the winds are constant at all altitudes (speed and direction). So while "yes" the math may be the easy part, estimating where it will actually land because of the drift is the complex part. So, if someone has a formula to calculate the CEP for the LZ, I'd like to use it. :)

The surface area doesn't matter. Think of a small helium balloon released into the air. It will drift with the wind at the same speed as the wind. So will a giant hot air balloon.

-- Roger
 
I would agree that is a generally holds true for standard 3FNC/4FNC's, but for something like the Estes Orbital Transport, it is a little more complex. There is a lot of surface area that can be pushed against, and if it rotating... Plus we are making assumptions that the winds are constant at all altitudes (speed and direction). So while "yes" the math may be the easy part, estimating where it will actually land because of the drift is the complex part. So, if someone has a formula to calculate the CEP for the LZ, I'd like to use it. :)

Greg

I agree, if there is more surface area for the wind to push....moar drift.

There is probability more factors than that to look at though, its kinda like when you launch the rocket and it weather cocks.
 
I am thinking this:

Let's say you have a hockey puck on the ice in a skating rink, and a shoebox on the rink with the same mass as the puck. Let's say you start a leaf blower about 10 feet away. My guess is that the box will move more than the puck will because of the higher Cd. But that's just a guess.

Greg
 
I am thinking this:

Let's say you have a hockey puck on the ice in a skating rink, and a shoebox on the rink with the same mass as the puck. Let's say you start a leaf blower about 10 feet away. My guess is that the box will move more than the puck will because of the higher Cd. But that's just a guess.

Greg

Or even moar drastic a piece of paper and a thumb tack.
 
I am thinking this:

Let's say you have a hockey puck on the ice in a skating rink, and a shoebox on the rink with the same mass as the puck. Let's say you start a leaf blower about 10 feet away. My guess is that the box will move more than the puck will because of the higher Cd. But that's just a guess.

Your analogy is wrong. Items sitting on ice have to overcome friction with the ice to move. That's not the same as a rocket drifting under a parachute where there is nothing connecting it with the ground. And the items you chose wouldn't fall at the same rate if dropped.

When the rocket is drifting under the parachute there is no wind from it's reference point. It's moving with the air. There's no wind pushing on it, so it makes no difference how large the parachute or rocket is.

The only reason you feel wind on the ground is because you are standing still in relation to the ground while the wind is moving in relation to both you and the ground. A parachute, however, will move along with the air, like a toy boat floating downstream on the top of the water.

The boat will move at the same speed as the water. Two different size boats will move at the same speed. For one were to move slower, as you suggest with hockey puck/shoe box analogy, the slower moving boat would have to be propelling itself upstream against the current. That's impossible for two reasons. First, there's no energy available to propel it upstream. Second, there's no way for it to know which way is upstream.

Another analogy to explain it would be to think of two different size and weight rockets falling from the same height under parachutes of differing sizes on a day with no wind. The sizes of the 'chutes have been chosen so that the rockets fall at exactly the same rate. You'd obviously expect them to reach the ground at the same time and they will.

Now, instead of standing still under them, you stand on a moving treadmill. You'll feel a wind blowing as you move. And, from your vantage point the rockets will seem to be drifting instead of falling straight down. But, they'll still reach the ground at the same time. This situation is equivalent to the rockets falling in "wind." If it helps, you can imagine the treadmill to be really large, as large as the earth, and circular. The earth (and you) moving while the air stays still is equivalent to the air moving while you and the earth remain still.

-- Roger
 
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[POW]Eagle159;250870 said:
Or even moar drastic a piece of paper and a thumb tack.

Dropped from a height, your piece of paper would, of course, drift less than the thumbtack which would fall to the ground much faster. So, your analogy is incorrect.

Balloons make the best analogy because two neutrally buoyant balloons could be drastically different in size, but they would "fall" at the same rate (i.e. they would fall at a rate of 0 feet-per-second which is another way of saying that they would always float at the same altitude). It should be obvious that both the large and small balloon would drift at the same speed in any wind.

-- Roger
 
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Roger and Bob are, of course, correct. From the recovering rocket's point of reference there is NO wind (unless the 'chute is gliding as Bob noted in his second post). It has no other means to create any relative wind.
 
Roger and Bob are, of course, correct. From the recovering rocket's point of reference there is NO wind (unless the 'chute is gliding as Bob noted in his second post). It has no other means to create any relative wind.

Thanks.

The parachute does create a "relative wind" by falling (though I'm not sure "relative wind" is the correct term). If you were in the rocket, you'd feel air flowing past you, but it would feel like it's blowing up from below. It wouldn't give you any clue as to which way the wind is blowing relative to someone on the ground.

If the parachute does generate lift (or your rocket is a boost glider or helicopter recovered model), it may drift more or less distance during each specific flight, but, unless it is guided (by radio control for example), the average distance it drifts will follow the same simple formula.

-- Roger
 
But wouldn't a rocket with a large surface area (fins) have that act as a sail and be blown farther/faster than a sleek small fin/ performance rocket of the same weight and chute size?
 
[POW]Eagle159;251129 said:
But wouldn't a rocket with a large surface area (fins) have that act as a sail and be blown farther/faster than a sleek small fin/ performance rocket of the same weight and chute size?

No.

If the smaller rocket in your example were to move slower, it would be moving slower than the wind relative to the ground. Therefore it would have to resist the air blowing against it. It would have to push itself upwind. There's no source of power it could use to move upwind and there's no way for it to even "know" which direction is upwind.

-- Roger
 
I thought though, that the wind is not 100% efficient...so if the wind blows 10 mph than the rocket would go only 7-8mph. I also thought this in your boat example, the water might move at 20 but the (unpowered)boat only 15.
 
[POW]Eagle159;251163 said:
I thought though, that the wind is not 100% efficient...so if the wind blows 10 mph than the rocket would go only 7-8mph. I also thought this in your boat example, the water might move at 20 but the (unpowered)boat only 15.

The boat is only part in the water. The air/wind has some effect.
A sunken boat would move at the same speed as the water
The rocket on a chute will move at the same speed as the air mass after it has reached equilibrium with the wind.

If you are in a free balloon there is no wind, if there is a gust there is a bit of wind at first until the balloon accelerates to match the wind but then when the gust subsides the opposite happens as it slows back down. All averages out.

Mark
 
The boat is only part in the water. The air/wind has some effect.
A sunken boat would move at the same speed as the water
The rocket on a chute will move at the same speed as the air mass after it has reached equilibrium with the wind.

If you are in a free balloon there is no wind, if there is a gust there is a bit of wind at first until the balloon accelerates to match the wind but then when the gust subsides the opposite happens as it slows back down. All averages out.

Mark



That makes sense.. The wind is relative to your speed in the wind.
If you and the wind are moving at 10mph, there is effectively no wind
from your perspective. Slow down to 0/mph and the wind is hitting you
at 10/mph.
 
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