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'[OT]: Terminal Velocity of raindrops and other int'
2001\05\17@084055 by Russell McMahon

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> >  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

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2001\05\17@150915 by Andrew Warren

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Russell McMahon <spam_OUTPICLISTTakeThisOuTspammitvma.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}

   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 --- .....aiwKILLspamspam@spam@cypress.com
=== IPD Systems Engineering, CYSD
=== Cypress Semiconductor Corporation
===
=== Opinions expressed above do not
=== necessarily represent those of
=== Cypress Semiconductor Corporation

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2001\05\17@214323 by Jim Korman

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Russell McMahon wrote:
>
<great deal of math snipped>

Just for some 'raw' data that I got out an
Intro to Meteorology text.

Size mm           rate of fall
               ft/min    m/sec

7.000                  abt 10.0      About maximum size, droplets break
                                    into smaller drops.
5.000           1,750       8.9      Large raindrop
1.000             790       4.0      small raindrop
.500             555       2.8      fine rain/large drizzle
.200             300       1.5      drizzle
.100              59       0.3      large cloud droplet

Jim Korman

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2001\05\17@221242 by Russell McMahon

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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
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2001\05\17@225123 by David VanHorn

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>
>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.


How are you measuring this?

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2001\05\17@230608 by James Newton

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Russell has a calibrated tongue. Next time it rains, he will be in the yard
with his head tipped back...

James Newton, PICList Admin #3
jamesnewtonspamKILLspampiclist.com
1-619-652-0593 phone
http://www.piclist.com

{Original Message removed}

2001\05\18@005748 by steve

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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.

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