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

On Oct 5, 4:10 pm, NoEinstein <noeinst...@xxxxxxxxxxxxx> wrote:
On Oct 3, 3:21 pm, "dkel...@xxxxxxxxxxx" <dkel...@xxxxxxxxxxx> wrote:



On Oct 2, 4:58 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 —

Hi
Use one of your many neurons and find where rain falls and 100 mph.
It is not even close.
Dwight- Hide quoted text -

- Show quoted text -

Dear Dwight: Rain which condenses closer to the Earth would be
slower. I used to hitchhike home from college in any weather. Some
raindrops do sting, but because its mass hits gradually, the high
speed isn't too apparent. To sense how speed affects your pain, stick
your hand out of a car window and let the car cause side impacts of
your hand to the raindrops. At 60 plus mph, you can definitely feel
the impacts, but you won't have to go to the doctor. — NoEinstein —


Hi
Most rain condenses in the clouds. Most conensation it several
thousand
feed above ground. It has more than enough distance to reach terminal
velocity.
This is one experiment that can be verified by most anyone with a
car.
drive through some rain while there is no wind or drive perpendicular
to the wind. Have an observer look out the window and note the angle
of the rain drops. 45 degrees means you are going as fast as the
drops.
An angle less than that relative to the ground means you are going
faster.
Face it, you are wrong and nothing you state will make it true. I
never
said that you'd need medical attention for 100 mph rain, only that
it would sting( and it would ). I have never had any rain, not driven
by wind, hit me that ever came close to hurting, ever. Nothing,
even close to 60 mph that wasn't wind driven.
Proving you wrong in this case is so easy, anyone on the web can
find your thinking incorrect. Did you even bother to go to the web
page that
Jerry posted of it your mind so closed that you won't except the truth
when it comes from different sources.
Rain does not fall at 100 mph. Do the car experiment again and note
the angle of the falling rain.
I'm sure you won't run this experiment, just as you refuse to run the
experiemnt with different sized balls or different heights. You are
so obviously wrong. You don't want to see anything that might change
your
thinking.
Dwight
.

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