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From: jmorriss on 19 Jul 2010 18:52
On Jul 19, 10:00 am, NoEinstein <noeinst...(a)bellsouth.net> wrote:
> On Jul 18, 11:53 pm, "jmorr...(a)idirect.com" <jmorr...(a)idirect.com>
> wrote:
>
> Dear jmorr:  By my saying... if the suit springs a leak, I wasn't
> excluding having the leak occur in the hose to the surface.  Thanks
> for your comments!  — NoEinstein —
>

More specifically, the most dangerous leak occurs in the hose to the
surface AT THE SURFACE.

After all, the air coming down the hose IS leaking out of the suit,
through a vent. But the leak is balanced by the flow of air down the
hose, and the pressure throughout the hose and in the helmet, matches
the water pressure around the diver.

Cut the hose at the suit, and the diver drowns or suffocatesl no
pressure effect.

Cut the hose about 30 feet above the diver, and he'll feel a one
atmosphere squish. For every 30 feet up, another atmosphere, more or
less.

A diver can also get squished by dropping too quickly, say while
working on the sloping side of a sunken ship. If a diver falls 30
feet down, too fast for the air compressor to keep up, he's in one
atmosphere trouble again.

Good luck with the New Kinematics...
From: Cwatters on 20 Jul 2010 03:09

"NoEinstein" <noeinstein(a)bellsouth.net> wrote in message
news:743db362-fc90-447f-9873-aae6dc348296(a)q35g2000yqn.googlegroups.com...
On Jul 19, 7:34 am, "Cwatters"
<colin.wattersNOS...(a)TurnersOakNOSPAM.plus.com> wrote:
>
Dear Cwatters: The SUPPOSED 'gs' being experienced by pilots are all
calculated using the now-disproved (by yours truly) 1830 Coriolis
equation, KE = 1/2mv^2. The actual 'survivable' 'gs' are much lower.
� NoEinstein �

They aren't calculated at all. They are measured.


From: Androcles on 20 Jul 2010 05:31

"Cwatters" <colin.wattersNOSPAM(a)TurnersOakNOSPAM.plus.com> wrote in message
news:0-6dnTDMJ_QS1tjRnZ2dnUVZ7q2dnZ2d(a)brightview.co.uk...
|
| "NoEinstein" <noeinstein(a)bellsouth.net> wrote in message
| news:743db362-fc90-447f-9873-aae6dc348296(a)q35g2000yqn.googlegroups.com...
| On Jul 19, 7:34 am, "Cwatters"
| <colin.wattersNOS...(a)TurnersOakNOSPAM.plus.com> wrote:
| >
| Dear Cwatters: The SUPPOSED 'gs' being experienced by pilots are all
| calculated using the now-disproved (by yours truly) 1830 Coriolis
| equation, KE = 1/2mv^2. The actual 'survivable' 'gs' are much lower.
|  NoEinstein 
|
| They aren't calculated at all. They are measured.
|
Not only that but it isn't possible to disprove a definition, whatever the
aetherialistic freak says.

From: PD on 20 Jul 2010 10:23
On Jul 18, 1:25 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote:

>
> (2.)  Mythbusters stated that a person (or “Buster”) dropping 20 feet
> experienced close to 12 ‘gs’ hitting an air bag, and 9 ’gs’ hitting
> bags of foam rubber.  Both of which they claimed were… survivable.
> ‘Accelerometers’ were used to graph changes in velocity, and the
> latter allowed a computer program to determine the… ‘g’ forces.
> Unfortunately, those forces were and are being wrongly calculated
> using Coriolis’s 1830 kinetic energy equation, KE = 1/2mv^2, which I
> have disproved up, down, and sideways!  As is the case for Einstein’s
> SR equation, Coriolis’s violates the Law of the Conservation of
> Energy, as well as a number of other realities.
>
> If a human being weighing 160 pounds experienced 12 ‘gs’, that would
> be like placing a 1,920 pound slab of concrete on the person.
> Survivable?  Probably not.  In actuality, the KE increases LINEARLY
> with respect to either the time of fall or the velocity, provided the
> benchmark velocity is 32.174 feet per second.  The simplest way to
> approximate the ‘g’ forces is to say: The accelerating fall over 20
> feet gets countered by a deceleration over (I estimate) five feet.
> Since the deceleration occurs in 1/4th the time and distance, the
> correct ‘g’ force in only 4, not 9.  If one wishes to confirm this
> mathematically, the correct formula for kinetic energy is my own, KE =
> a / g (m) + v / 32.174 (m).

There's no kinetic energy calculation required. When you experience
"g"s, that is a measure of your acceleration (a negative acceleration
is sometimes called deceleration, but it's an acceleration
nonetheless). If you start at rest when you jump and end at rest when
you land, then there is a simple ratio:
[acceleration]/g = [distance landing pit compresses]/[distance in free
fall].
This ratio is easy to derive from the relation v^2 = v0^2 + 2*a*d.
Thus you are right that if the distance the pit compresses is 1/4th
the distance fallen, then the *average* acceleration on landing is 4
"g"s. There may of course be less acceleration at the point of impact,
and higher acceleration as the pit compresses, which would be the case
for just about any elastic behavior in the pit.

>
> MythBusters is too stuck up to perform, on their show, the simple ball
> drop experiment (at larger scales, of course) explained below.  Like
> too many in “science”, they don’t relish being shown to have ever been
> wrong.  Didn’t they state, recently, that the FORCE needed to break a
> duct tape auto barrier was… 100,000 “foot-pounds”?  Ha, ha, HA!
>
> Respectfully submitted,
>
>    — NoEinstein —
>
> Real name: John A. Armistead
>
> Dropping Einstein Like a Stonehttp://groups.google.com/group/sci.physics/browse_thread/thread/989e1...

From: PD on 20 Jul 2010 10:24
On Jul 20, 2:09 am, "Cwatters"
<colin.wattersNOS...(a)TurnersOakNOSPAM.plus.com> wrote:
> "NoEinstein" <noeinst...(a)bellsouth.net> wrote in message
>
> news:743db362-fc90-447f-9873-aae6dc348296(a)q35g2000yqn.googlegroups.com...
> On Jul 19, 7:34 am, "Cwatters"<colin.wattersNOS...(a)TurnersOakNOSPAM.plus.com> wrote:
>
> Dear Cwatters:  The SUPPOSED 'gs' being experienced by pilots are all
> calculated using the now-disproved (by yours truly) 1830 Coriolis
> equation, KE = 1/2mv^2.  The actual 'survivable' 'gs' are much lower.
> — NoEinstein —
>
> They aren't calculated at all. They are measured.

By an accelerometer. Which you can make with a protractor and a string
and a fishing weight, or a plastic tube and a rubber band and a
fishing weight.
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