For example, if you apogee at 1 kft, take a 1 kft walk.
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.
Absolutely correct, however you need to add 50' of climbing, a 50' recovery pole or a chain saw to accomplish a successful recovery. :blush:So...if the rocket lands at the top of a 50 ft tree, I can save 50 ft of walking
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.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.
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.
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 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.
[POW]Eagle159;250870 said:Or even moar drastic a piece of paper and a thumb tack.
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.
[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?
[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
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