# question and answer

• posted
Lets start at 3000 ft with a 1lb object.....assume a drag coefficent of .75
...assume a descent rate is 50 ft/s and assume the wind speed is 15
ft/s...how far will it drift? or the recovery distance?
3000ft / 50/ft/s = 60 seconds so it will be in the air 1 minute
it has a 2"x108" 1-2 mil mylar streamer for visibility more than recovery...
thats 216 sq in...
Using figures from aboves example, can you determine the actual descent
rate?
please show any mathematical formulas....
shockie B)
• posted
I think a good first approximation is to assume the model under streamer will attain the relative wind speed fairly quickly after deployment. For a typical model headed slightly into the wind at deployment, the model will have to slow to zero horizontal velocity, then be accelerated by the wind back in the opposite direction. You could assume (for no good reason other than simplicity) that the two effects cancel and compute drift from the deployment point.
For a 60 second drift with a 15 foot per second cross wind, you'd get a 900 foot leeward drift from the point of deployment.
I believe descent rate IS the vertical component of motion, which you gave as 50 feet per second.
• posted
You forgot the K -- How was the streamer folded...
And the K1(K,shock cord length,carcass mass) factor--The effective area of the 216 inches as it whips around.
And the F(up) factor - Thermal effects although it's probably close to zero with a one pound craft.
And the R factor - Reliability of the thin streamer not breaking apart (happens with my mach busters all the time)
so...
Rate = (Down(k(R),mass) - Fup(Delta temp,k(R)^1.1,k1(R)^0.33)
The 1.1 and 0.33 power functions are guesstimates somewhere between first and third order...
Am I close?
Where's Alan...
Andy

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