Post by thread:[OT]: Terminal Velocity of raindrops and other int

> > I've heard (our TV weatherman) that raindrops have a maximum velocity > > of 25 km/hr. If this is true, make your spacecraft into a raindrop shape > > and avoid most of the technology! :-) > I think terminal velocity is a function of sectional density and drag. > Metal has a higher density than water so it's terminal velocity would be > a lot faster. Intuition allows a fairly good understanding here (which doesn't always happen in applied Physics) . The expression for mass for a regular body can be recast as M = A x L x Density This can be a useful alternative in formulae but we'll leave it alone here. I'll deal directly with mass and frontal area. Terminal velocity occurs when the falling objects weight is balanced by drag. Slower than this weight exceeds drag and we accelerate downwards. Faster than this and drag exceeds weight and we decelerate (or accelerate upwards). Weight is a function of mass (funny that) but for our purposes is better thought of as being a function of volume and density or for a given shape frontal area and density. For a regular body - Weight = mg = mK1 Drag is a function of frontal area, velocity, gas density (and a few zillion other things which we will ignore). (K1, K2, K3 are suitable constants used to wrap together all the unchanging stuff in each formula). Order of correct formula oft used for drag is Drag = 0.5 x Area x Cd x Rho x V^2 = Area x V^2 x K2 [[Yes, yes - this is a brute force approximation and the REAL estimation of drag is complex and a function of way too many thing s - but this will do fine for this purpose]]. (Rho = air density, A = frontal area, ...) When these two forces (weight and drag) balance we have reached terminal velocity m x K1 = Area x V^2 x K2 or Vterminal ^ 2 = K3 x m / A If you must, unsubstituting and rearranging gives - Vterminal = sqrt((2 x m x g )/(Cd x Rho x A)) Simplified V = 4 x sqrt ( m / Cd / A ) E&OE A = frontal area m = body mass g = gravitational constant Cd = coefficient of drag Rho = air density 2 = 1 + 1 :-) Let's try this for a sky diver, head down, tucked in A = 0.2 m^2 (my guesstimate) m = 100 kg say g = 10 (diving low over Lake Asphaltitas) Rho = 1.3 kg/m^3 Cd = 1 I reckon V = sqrt((2 x 100 x 10) /(1 x 1.3 x .2)) = 28 m/s = 91 f/s = 62 mph = 100 kph This is a bit slow AFAIK - I think about 70% of actual value. Not bad for a back of envelope "design". Flushed with success, let's try that rain drop. How big is a rain drop???? 40 drops from my kitchen tap = 10cc = E-4 m^3 or 0.25cc = 2.5E-6 m^3/drop. Don't read the next line if your head hurts :-) D = 2 x ((3 x V)/(4 x Pi))^0.333 D = 8mm That's a mighty large drop !!! Back to tap - I find the "drops" are large enough to allow several drops to be formed from them when dripped into and then out of a tea spoon. Lets go with a 0.1 cc = 1E-6 m^3 drop. m= E-6 kg dia = 0.0058 m Assume full sphere for now A = 26E-6 m^2 Cd = 1 again Vterminal = sqrt((2 x m x g )/(Cd x Rho x A)) = sqrt ((2 x E-6 x 10)/(1 x 1.3 x 26E-6) = sqrt(15.4) = 3.9 m/s = 14 kph = 12.8 f/s = 8.7 mph This is about half the claimed figure. Again, not bad for "back of envelope" (with a little help from a spread sheet for the annoying powers of 10 in the middle :-) ) Allow Cd to be a little lower or allow the drop to elongate slightly or both and the velocity rises happily towards the suggested 25 kph or so. Not knowing when to quit - what about a thin long heavy dart??? See if I can do figures out of my head. Say a 1kg mass with a 1 cm^2 frontal area and a Cd of 0.4 Vterminal = sqrt((2 x m x g )/(Cd x Rho x A)) = sqrt(2 x 1 x 10)/(0.4 x 1.3 x E-4) = 620 m/s = Supersonic Clearly "other effects" come into play long before this. I'll leave someone else to work out the figure for a bowling ball. Russell McMahon -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

Russell McMahon <spam_OUTPICLISTTakeThisOuTmitvma.mit.edu> wrote: > Let's try this for a sky diver, head down, tucked in > > A = 0.2 m^2 (my guesstimate) > m = 100 kg say > g = 10 (diving low over Lake Asphaltitas) > Rho = 1.3 kg/m^3 > Cd = 1 I reckon Russell: That Cd is pretty high; I'd guess that the real Cd is somewhere closer to 0.7. Regardless, your math appears to be in error: > V = sqrt((2 x 100 x 10) /(1 x 1.3 x .2)) > > = 28 m/s = 91 f/s = 62 mph = 100 kph sqrt (2000/0.26) = sqrt (7692) = 88 m/s = 289 f/s = 196 mph = 316 kph That number's pretty high, probably because your guesstimate of A is small by a factor of 2 or 3 (remember, the skydiver's carrying a big parachute on his back). {Quote hidden}> Flushed with success, let's try that rain drop. > .... > m= E-6 kg > dia = 0.0058 m > > Assume full sphere for now > A = 26E-6 m^2 > > Cd = 1 again > > Vterminal = > .... > = sqrt ((2 x E-6 x 10)/(1 x 1.3 x 26E-6) > = sqrt(15.4) = 3.9 m/s > = 14 kph > = 12.8 f/s > = 8.7 mph > > This is about half the claimed figure. That's because the Cd for a sphere is always 0.5, and the Cd for a raindrop or "teardrop" shape is much lower. The shape of a raindrop, in fact, is generally the ideal toward which designers of low-drag subsonic bodies aspire. It makes sense if you think about it: The raindrop begins as a sphere but is then shaped by the air as it falls; it naturally conforms to the lowest-drag shape possible. > Again, not bad for "back of envelope" Agreed. -Andy === Andrew Warren --- .....aiwKILLspam@spam@cypress.com === IPD Systems Engineering, CYSD === Cypress Semiconductor Corporation === === Opinions expressed above do not === necessarily represent those of === Cypress Semiconductor Corporation -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

Several people have pointed out my skydiver results are wrong. A look at the sum involved shows I divided by 2.6 rather than 0.26 so the answer was low by a factor of root 10 = 3.16 so my result should have been 300 kph or 190 mph which is almost exactly the figure Tom Watson gives from his practical experience. Adjust Cd and frontal area as desired. I used Cd=1 for the rain drop and it has been pointed out that a figure of 0.5 is appropriate for a sphere, which the water drop approximates. This brings the result up by root 2 to about 21 kph which compares well with the 25 kph reported. Experiments in the shower this morning :-) indicate that a drop size of under the 0.1cc that I used may be more appropriate - this would produce a lower terminal velocity as the drop mass/area scales as cube/square respectively of diameter. Haven't tried catching rain drops yet. Russell McMahon _____________________________ {Quote hidden}> That Cd is pretty high; I'd guess that the real Cd is somewhere > closer to 0.7. > > Regardless, your math appears to be in error: > sqrt (2000/0.26) = sqrt (7692) > = 88 m/s > = 289 f/s > = 196 mph > = 316 kph > > That number's pretty high, probably because your guesstimate of > A is small by a factor of 2 or 3 (remember, the skydiver's > carrying a big parachute on his back). -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

While you have your envelope out .... > > Drag = 0.5 x Area x Cd x Rho x V^2 > = Area x V^2 x K2 What about in a fluid environment ? ( I tried to find something on the net about this the other night, but didn't get too far. - Nice of someone to bring this up while I'm still pondering). Intuition tells me that viscosity would be a handy thing to throw in there when moving a solid body through a fluid. eg. Oil is less dense but more viscous. So now that we've done raindrops, here's an even more useless thing to calculate - When I throw my spherical 500g sinker (density = 11.3g/cm^3) over the side of the boat, how long does it take to get to the bottom ? Assume I forgot to tie it to the line and I'm in 10m of seawater (density=1025kg/m^3, viscosity=1.3cp). The next question is if I take a cork (say density=900kg/m^3) down to the bottom, how fast will it come up ? Sometimes I hate having a curious mind ? Steve. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics