Re: drag coef. and falling bullets

[brian whatcott <inet@INTELLISYS.NET>]

At 13:48 1/21/00 -0500, David Bowman wrote:
> ///
> If we use the info reported by John D. from Wegner's book:
>
> > Using figure 7.3 on page 89 of Wegener, we have
> >   0.47 -- sphere -- representative of point-first flight
> >   1.17 -- cylinder -- representative of broadside flight
>
> and take the 1.17 value for the drag coefficient and use the weight and
> dimensions for the bullet reported by Brian W. (i.e. 0.073N and
> 9 mm x 27 mm) and take the air density to be 1.29 kg/m^3 we get a value
> of V = 20 m/s.

Checking ..er.. confirming Professor Bowman's method
 And you thought hubris was foreign to my nature...   :-)

Fd = Q A Cd
Q dynamic pressure  (1/2 rho v squared)
A representative area
Cd Coeff of Drag

For the Luger round:
 0.073 = 0.5 1.29 V^2  2.43E-4  1.17

V^2 = 398
V = 20 m/s as David noted.

Returning the compliment...

Jumper:
803 = 0.5 1.29 V^2  1.11 1.17
so V^2 = 959
V = 31 m/s  (?) or 69 mph

This is certainly too slow - we are quite certain the jumper
will settle at 120 mph on up... (200 mph if he makes a tight ball)

It *could* be that the jumper is a much smoother surface
than an average cylinder such as a bullet so we are using a Cd
that's too high.

It *could* be that a jumper subtends much lower cross section

or

It *could* be that the variation in drag with scaling is often
taken out by varying the Cd with size....

..or did I mess up the calculation?

Recalling the simple model I used earlier:

Fd = k A V^2 gave 0.25 for k

Comparing with the usual model
Fd = Q A Cd    or   Fd = 0.5 rho V^2 A Cd

we can see that I hid several terms in the portmanteau constant k

K is 0.5 rho Cd    (I made it 0.25 you recall)

So the implication is that Cd for the jumper is
0.25/(0.5 * rho) = 0.43

If we applied this to the bullet it should come up with a
terminal velocity like I estimated earlier, shouldn't it?

Actually no.
One of the earliest findings of the ACA was that thinner
cylinders have disproportionately more drag than their area
suggests.

Before too long, the scaling rule developed by Professor
Reynolds on his Mancunian water pipes was hauled in to make
the scaling problem more tractable.

So I wouldn't want David to dismiss John D's interesting
stairwell idea (It's cheap, and it's available)

Perhaps if we chose a 0.177 (airgun) pellet, #4, and #8 shot
(to continue the huntin', shootin', fishin' theme) our height
requirement would be less severe. Recall that the Tuscan marble
campanile that inhabits the student imagination is only
55 meters (179 ft) high more or less - though that represents a
more heroic eight stories than the usual floor height would give.


There is another issue that David touches on: data fitting using
logarithmic functions of the input data, as follows:

> ....  So, if we define
> x = ln(g*t) - ln(V) and define y = ln(g*z) - 2*ln(V)
> these variables  will obey the universal equation:
>
> y = ln(ln(cosh(exp(x))).
>
> The effect of fitting the data to the universal curve involves
> horizontally translating x by some fixed amount and vertically
> translating y by twice that amount until the data line up with the
> universal curve./snip/ Something useful or
> interesting is bound to be discovered.
//
> David Bowman

This is a classical prescription, no doubt about it - but in my
innocence, I make so bold as to mention that a least squares
fit done in this way is in no way a least squares fit.
Need to run the mean squares on the raw data. (like I do...)

brian whatcott <inet@intellisys.net>
Altus OK