Re: KE = ½ mv^2 is disproved in a new falling object impact test.

On Oct 2, 6:58 pm, NoEinstein <noeinst...@xxxxxxxxxxxxx> wrote:
On Oct 1, 9:51 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:



On Sep 29, 9:48 pm, NoEinstein <noeinst...@xxxxxxxxxxxxx> wrote:

On Sep 25, 12:14 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Sep 24, 11:10 pm, NoEinstein <noeinst...@xxxxxxxxxxxxx> wrote:

Dear Dwight:  The terminal velocity of PTFE is about 220 miles per
hour.  A 12 foot drop is nowhere hear that fast!  — NoEinstein —

Where on earth did you get that number?

Dear PD: The terminal velocity of a human body, SG = 1, is 110 mph.
So the terminal velocity of a ball of SG = 2 is twice as much, or 220
mph.  — NoEinstein —

Here, let me help you. In terms of the specific gravity, the terminal
velocity is given by this relation:
[Terminal velocity] = sqrt{2 [specific gravity] [volume] g / [drag
coeff] [projected area]}
which we'll abbreviate this way:
Vt = sqrt{(2*d*V*g)/(Cd*A)}
So, if you want to find the ratio of terminal velocities between a
PTFE ball and a human body, here is how you find it:
Vt(ball)/Vt(human) = sqrt[d(ball)/d(human)] * sqrt[V(ball)/V(human)] *
sqrt [Cd(human)/Cd(ball)] * sqrt[A(human)/A(ball)]
Now, what you said you know is that d(ball)/d(human) = 2. So, by
itself this would make Vt(ball)/Vt(human) = 1.41.
However you haven't included what sqrt[V(ball)/V(human)] is.
And you haven't included what sqrt [Cd(human)/Cd(ball)] is.
And you haven't included what sqrt[A(human)/A(ball)] is.

Would you like to try your calculations again?

PD

Dear PD:  To get a "grip" on science, throw away your textbooks, and
use your one-neuron brain to figure things out.  — NoEinstein —

No, this isn't textbooks. This is from experiment. You're missing a
few factors. Reason won't suffice if it doesn't match experiment.

PD
.

Relevant Pages
Loading