Calculating speed of a free falling iron ball?

From a height of 1km meters maybe...
Does a 1cm diameter ball (from ballbearing) from an airliner flying in the
sky have enough energy
to destroy a BMW engine?

How do I calculate that?

Will the shape of the ball affect the speed? Let's say I make its four
surfaces all triangular...

Jia Yu Cun wrote:
in the

The ball will accelerate until the rate at which it dissipates energy
by stirring/heating the air (air resistance or friction) equals the
rate at which gravity works on the ball by giving it kinetic energy.

At the point the maximum kinetic energy is reached, the ball will fall
no faster. The speed of the ball is called its "terminal velocity" -
V.

V depends on the weight W, the cross-sectional area A, the density rho
and the drag coeffiecient C(d):

V = sqrt(2*W/(C(d)*rho*A)

The drag coefficient of a sphere can have a range of values
[Only registered users see links. ]

which, for a sphere, depend on a property called the Reynolds number Re
[Only registered users see links. ]

which in turn depends on the velocity V, the density rho, a
'characteristic length' l of the sphere, and the viscosity mu of the
medium (in this case, air):

Re = V * rho * l / mu


Use your 1 cm diameter,
a density for the steel ball of about 7.85 g.cm^3
[Only registered users see links. ]
a specific gravity of 1 (SG rel air) for the surrounding medium
and a viscosity of 0.0000173 N-s/m^2
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in the applet on this page:
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I got a value of 43.4 m/s (142.5 ft/s) - about 97 mph. I've driven
faster on a Harley. A good slingshot could launch your ball bearing as
fast.

While this *might* perforate the hood/bonnet of a Beamer and maybe even
damage an engine part or three, I doubt it could *destroy* the engine.
I certainly would not want it to hit me.

four

It will affect the drag coefficient - making it slower. See
[Only registered users see links. ]
Tom Davidson
Richmond, VA

> [Only registered users see links. ]

Thanks for your reply.

From this formula, the velocity is proportional to the square root of the
diameterof the
ball, thus the energy is power(diameter, 4). I guess a 2cm diameter steel
ball will make
so severe damage as to disable my BMW...

How can I find out the maximum height that will reach the terminal velocity?
What will be the
temperature it will reach if I drop it from a 10km high airliner? Will this
temperature severe the
damage?

If I design my object to be of a spear shape with a dent on one side to
break the balance, when
it drops, will it be in an almost vertical position and allow the pointer to
dig deeper into my BMW?

Thanks.

Jia Yu Cun wrote:
bearing as
of the
steel

The applet I cited before gives a terminal velocity of about 61.4 m/s
for a 2 cm steel ball in air. Thats about 41% faster. Combine that
with the fact that the 2 cm ball has 8x the mass of a 1 cm ball and you
get 16 times as much energy at terminal velocity, but only 4 times the
area.

The definitely adds up to more damage.

velocity?

First, there is no maximum height. The terminal velocity is an upper
limit, so that no matter how high the ball is whan you *drop* it, it
can never go faster than its terminal velocity - it is 'terminal'
because the velocity cannot grow any more once it gets that large.

Finding out how high it must be to reach a given velocity (below
terminal velocity) by freely falling is an even trickier problem. It
involves solving a non-linear differential equation because the
resistance changes with the velocity. I am not prepared to try that
here. Maybe later, when I have access to my personal library (I am at
work now)...if you are really interested?


I really can't answer that.


Probably not. The damage will be caused by the kinetic energy of the
ball:

K.E. = (1/2)*m*v^2

The greatest contribution to damage from the heat of the ball would be
if it punctured the fuel tank and was hot enough to ignite the fuel.

to
pointer to

If you look at the shapes on
[Only registered users see links. ]
you will see that the one that has the lowest drag coefficient (and
therefore the highest terminal velocity) is the airfoil shape. If you
look at it as a 3-D object, it becomes the classic "teardrop" shape
that was so popular in 1940's automotive designs.

The highest velocity will give you the highest kinetic energy for a
body of a given mass (as the equation above implies). The highest
kinetic energy per cross-sectional area will give you the most
penetration.

Tom Davidson
Richmond, VA

Yeah! I am really interested in the mathematics, be it calculus or ODE or
what...

I am sure spear shape is better than teardrop shape...as the fastest
aircrafts all
have pointed heads...e.g. Concorde...Slow ones like Boeing have teardrops...

Can you help me posting here your libraries?

Thanks.





"tadchem" <[Only registered users see links. ]> wrote in message
news:1104410543.685269.238840@z14g2000cwz.googlegr oups.com...

Fat Cat wrote:
ODE or
teardrops...

The teardrop shape shown on this link
[Only registered users see links. ]
is the shape that has the lowest Cd
Note that the flow is from the left, so that the blunt end of the
teardrop is in front.

The Concorde and all other supersonic craft are in powered flight.
Remember the stipulation of 'free falling'? It's in the subject header
for this thread.

The leading point is a specific requirement for supersonic flight.

Terminal velocity is not an absolute upper speed limit. It represents
a *final* value for a free falling body. Projectiles that are moving
at speeds above their terminal velocity (bullets, artillery shells,
meteors) will gradually slow down until they reach terminal velocity -
assuming an unimpeded fall.


My favorite library is
[Only registered users see links. ]

Beyond that I have recently reduced my personal collection of dead tree
editions to a few thousand volumes (about three rooms of my current
home). Most of the overage was donated to a small municipal library in
Colorado.

What in particular were you interested in? I believe I have a couple
shelves of math textbooks.

Tom Davidson
Richmond, VA

Dear Fat Cat:

"Fat Cat" <[Only registered users see links. ]> wrote in message
news:41db9aac$[Only registered users see links. ].sg...

Just to add a couple of words to Tom Davidson's excellent response...
The teardrop shape is affected by rain as it falls. It is a balance
between gravity, higher pressure air in front, lower pressure air in back,
and the affinity of water for itself ("surface tension"). Note that a wing
is almost always shaped like a stretched teardrop... in cross section

Most aircraft are shaped like straightened bananas. You are correct the
tapered front of an air plane is important, as this helps minimize
"boundary layer separation". This does not limit how wide the body can
get... only the number of passengers and amount of fuel do this. The
tapered back is important also, as this maximizes the amount of pressure
the air can deliver to the rear of the body.

Phrases in quotes can be used in Tom's favorite library...

David A. Smith

Well, thank you for your reply.

I think you corrected my misconception that the drag on an aircraft is
mainly caused
the area of the cross section head. Okie, assume it be true. Here is my
question:

In my childhood, we kids play all sorts of swords. They were of Maru type
but the blade
is much sharper(3 degrees and thickness is 5mm). When I cut water with it,
the strength
needed was much less than chopping water with the rounded back. I am sure,
the same
phenomenon will be expected with the wings of an aircraft...
http://home.earthlink.net/~steinrl/terms/mune.gif

How does your pressure theory explain this phenomenon?

Thank you.




"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:[Only registered users see links. ]> wrote in
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or
wing

Plus, you sink into water slower when you dive on your head than feet.





"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:[Only registered users see links. ]> wrote in
message news:adSCd.21661$CH5.17472@fed1read01...
or
wing

Thank you Tom. The Google advanced search is helpful.

If your collection is of hardcopy type, then give it a miss. I am sure I can
find some materials online, though it will take me some time to find the
correct
chapter of any fluid mechanics book as I am not mechanical engineer...

Thanks anyway.








"tadchem" <[Only registered users see links. ]> wrote in message
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