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Why Groundspeed?

 

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Premier billvon  (D 16479)
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Apr 11, 2012, 10:02 AM
Post #51 of 87 (857 views)
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Re: [dpreguy] Why Groundspeed? [In reply to] Can't Post

>Nor does a sailboat ever reach the speed of the wind on a downwind. Not
>even close to it.

Agreed. That's because the water's slowing it down; friction and all. If it was hovering above the water it would be going at the speed of the wind.

Question for you - how fast does a hot air balloon go with respect to the wind? If the winds are 80 knots, how fast will that balloon be going?


dthames  (B 37674)

Apr 11, 2012, 10:39 AM
Post #52 of 87 (852 views)
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Re: [billvon] Why Groundspeed? [In reply to] Can't Post

In reply to:
>Nor does a sailboat ever reach the speed of the wind on a downwind. Not
>even close to it.

Agreed. That's because the water's slowing it down; friction and all. If it was hovering above the water it would be going at the speed of the wind.

Question for you - how fast does a hot air balloon go with respect to the wind? If the winds are 80 knots, how fast will that balloon be going?

I think you will find that if you drop (from altitude) 2 objects with similar drag coefficients and have the 2 objects at different weights, you will find they land with the lighter object further downwind than the heavier object. That fact should shed some light on this specific point. (if an agreement can be reached on that happening) You can easily test that one.


kallend  (D 23151)

Apr 11, 2012, 10:43 AM
Post #53 of 87 (850 views)
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Re: [wildcard451] Why Groundspeed? [In reply to] Can't Post

In reply to:
In reply to:
In reply to:
"6-8 seconds to reach equalibrium"

Nope. Not at all true. False assumptiion.

WRONG.

You are out of your depth on this one.

Arguing physics with a physics professor.....

Kallend, going to regale us rigging knowledge?

No, I leave that to the riggers.


kallend  (D 23151)

Apr 11, 2012, 10:48 AM
Post #54 of 87 (847 views)
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Re: [dthames] Why Groundspeed? [In reply to] Can't Post

In reply to:
In reply to:
>Nor does a sailboat ever reach the speed of the wind on a downwind. Not
>even close to it.

Agreed. That's because the water's slowing it down; friction and all. If it was hovering above the water it would be going at the speed of the wind.

Question for you - how fast does a hot air balloon go with respect to the wind? If the winds are 80 knots, how fast will that balloon be going?

I think you will find that if you drop (from altitude) 2 objects with similar drag coefficients and have the 2 objects at different weights, you will find they land with the lighter object further downwind than the heavier object. That fact should shed some light on this specific point. (if an agreement can be reached on that happening) You can easily test that one.



If dropped from a balloon (per Bill's point above) do you STILL think the lighter one will end up farther downwind?

You can play with various scenarios on my web site:

http://lensmoor.org/cgi-bin/chute.cgi

and http://mypages.iit.edu/~kallend/skydive


(This post was edited by kallend on Apr 11, 2012, 10:52 AM)


skr  (D 981)

Apr 11, 2012, 3:28 PM
Post #55 of 87 (819 views)
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Re: [strop45] Why Groundspeed? [In reply to] Can't Post

> Why should the position over the ground have any significance at all? I really don't get it.

You're right. As a question on a physics test,
tossing spherical, isotropic skydivers off the
tailgate and watching their horizontal separation
evolve over time, the ground is irrelevant.

And actually, in the rec.skydiving discussions
several people kept pointing this out. When a new
example or theory was being put forth, they would
wait until the jumpers were in freefall and then
bring out this giant yellow Caterpillar tractor
and start hauling the dropzone around under the
jumpers.

At the time I found this annoying because I was still
trying to understand how both airspeed and groundspeed
arguments could seem so convincing.

You would have been one of those guys :-) :-)

I finally pulled back from trying to follow other
people's arguments and sat down to think it through
from a standing start for myself.

I found that if I imagined standing on the ground,
a couple miles off to one side, and watched a jumprun
unfold, I could see all the parts in a way that I knew
the physics was correct.

The exit points on the ground, the corresponding points
in the space above the ground, the opening points, the
patches of air in the exit layer where people exited,
the way those patches moved across the sky with that
layer of wind, the patches where they opened and the
way they moved relative to the ground, and so on.

Once I had that I could listen as others layed out
their current theory and I could follow it, and I could
see if I agreed or exactly where I might disagree.

That's when I saw that the second freefall trajectory
was exactly the same as the first, just displaced upwind,
and you could separate opening points by separating
exit points.

So my answer was, when the first person goes, look
down at the ground and see where they got out, go
a certain distance across the ground, and go.

It was neither airspeed nor groundspeed, it was all
spatial distance.

So I posted about that and advocated that for several
years. But I gradually saw that it's not practical.
I had spotted a lot in the past, but these days lots
of people never get the chance to spot any.

--------

Once I had that I could see how, when the first jumper
get's out, the plane flies upwind from that point in
space, and the first jumper blows downwind from that
point in space.

And the second jumper sees a large distance to the
first jumper at exit, while at the same time separating
exit points by a smaller distance.

That excess distance is what disappears from the moving
layers effect.

--------

(I guess I'm actually now responding to later posts in
(the thread, not just to strop45.

Some time back John asked why we're still talking about
this since Newton's laws haven't changed in several hundred
years, and I said because the current problem is psychological
and social rather than physics.

Almost everybody's initial intuition for leaving separation
is to look out the door and leave some room between jumpers.

Now you can say "Use groundspeed", and some people are happy
with that. That's all they want to know.

But! They won't know why, and they won't know the conditions
under which it's appropriate.

And there are plenty of people around who don't accept proof
by authority, they want to think for themselves and know how
it works.

And for those people we need to explain the genesis and limits
of that initial intuition, and some ways to see the more complicated
situation of winds, and the approximation and limits of groundspeed.

That's what the initial post in this thread was about.

I think John and Bill and others are doing a good job of
simplifying a complicated situation.

----

Billvon, it's a good thing your 80 knot example wasn't 120 mph
with the jumper falling away from the plane at 45 degrees for
the whole jump.

That would have really stirred things up :-) :-)

Skr


dpreguy  (D 835)

Apr 11, 2012, 7:54 PM
Post #56 of 87 (803 views)
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Re: [billvon] Why Groundspeed? [In reply to] Can't Post

Dear Mr Kallend,
Don't be so quick to insult. If you are correct, a cement block tossed out of the plane will soon achieve 80 knots speed across the ground, as your theory is that it will reach "equalibrium" in 6 to 8 seconds. Your theory is also, that a cement block tossed off a bridge will go downstream the same speed as the river, if you just give it some time. Wrong. It will never go the speed of the river, as it is affected by gravity. The only way an object can equal the speed of it's medium, whether it be water or air, is to stay on the level it enters, such as a wood chip on water.
Go ahead, drop a cement block into a river, and even if it is a mile deep, it will NEVER reach the speed of the flow of the water. Nor will a skydiver ever reach the speed of the wind. Nor will a sailboat ever reach the speed of the wind on a downwind. The heaver and more aerodynamic an object is, the less it is affected by the moving medium it is in. Pause for a moment and consider the cement block being tossed from an airplane. Agreed: the angle it goes down is not going to be vertical, because the wind will have "some" effect, but it will never go 80, as you suppose. As the angle from vertical increases, it tells-reflects the amount of effect the moving medium has on it. The closer it comes to staying at the same level, the closer it's speed will match the speed of the moving air/water.wind or whtever medium. Such as a feather tossed from the plane. It would go downward so slowly, that it woould get close to 80.

If your thoery is correct, then it would make no
difference if what leaves the plane is a feather or a cement block or something in between, like a person, because in 6 to 8 seconds they will all be going 80 across the ground ("reach equailbrium").. Kinda silly to believe that. Nothing that is going downward will ever reach 80 across the ground, even if it falls for an hour.


skr  (D 981)

Apr 11, 2012, 9:59 PM
Post #57 of 87 (792 views)
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Re: [dpreguy] Why Groundspeed? [In reply to] Can't Post

Hey, Walt, I've read this 3 times trying to get
why you're saying this.

An aerodynamic, plutonium brick will take longer
than a feather, but they both get to 80 knots.

The sideways motion is independent of the downward
motion.

Unless .. Do you mean by this:

> As the angle from vertical increases, it tells-reflects the amount of effect the moving medium has on it.

that the motion is asymptotic, that it gets closer
and closer to 80 but never actually gets there?

Skr


pchapman  (D 1014)

Apr 11, 2012, 10:24 PM
Post #58 of 87 (790 views)
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Re: [dpreguy] Why Groundspeed? [In reply to] Can't Post

In reply to:
If you are correct, a cement block tossed out of the plane will soon achieve 80 knots speed across the ground, as your theory is that it will reach "equalibrium" in 6 to 8 seconds. Your theory is also, that a cement block tossed off a bridge will go downstream the same speed as the river, if you just give it some time.

Sorry, the cement block (in a deep enough river) will end up moving horizontally as fast as the river flows. Otherwise it would continually be getting drag from the side... and thus change its horizontal speed until there is none.

Are you somehow thinking that gravity will tend to "straighten it out", making it fall more vertically? That won't matter. The only way for the object to avoid a push from the side is to move horizontally with the medium.

You are right that something really dense, with a lot of mass compared to the drag area, will take more time until the flow around it can get it moving fully with it. I guess we think of such things as "having a lot of inertia", even if technically that has nothing to do with drag but only mass.

I'll trust Kallend since he seems to know what he's doing. (Eg, 2nd order modified Euler's method in his numerical simulation. Doesn't prove anything, but you can't do that if you're clueless.)

I will grant you that dense objects will take longer than we might expect to get to equilibrium and move almost fully at the speed of the fluid around them. That applies to concrete blocks, and to some extent skydivers.

I'll also grant you that an object will only very slowly approach the speed of the fluid around it, when close to that speed. The cement block will reach 75 mph horizontally after a while, then ever more slowly approach 80. Theoretically it only approaches 80 asymptotically and thus never reaches it, but in practice it'll be pretty much at 80, so close that you'll never measure the difference.

Mr Kallend might have been hasty in mentioning 6-8 seconds -- given that his own numerical simulation out on the web suggests longer periods. For example, one can run his freefall simulation with a 115 fps airplane speed, about 80 mph. That's an equivalent problem -- there's an 80 mph horizontal wind on the object, so when (or to what degree) does the object start to move 80 mph horizontally?

It takes about 15 seconds for the 80 mph forward throw to be largely lost, but about 25 sec for the trajectory to be essentially vertical to the eye. In other words, 15-25 seconds rather then 6-8 for a skydiver to conform to the horizontal airflow around him.

The time will be significantly less for less of a horizontal relative flow -- so for all I know, 6-8 seconds might work when thinking about 20 mph wind, not the 80 mph being used in the recent posts as an example.

(I'll guess that one simplification in the freefall program is that the modelled drag area of the skydiver is the same from any angle, rather than calculating different drags for different angles, from a maximum for wind on the belly, and least for wind on the head. That'll affect the results somewhat.)

To reiterate:
Although it can certainly take longer than a few seconds for a dense object to reach the horizontal speed of the fluid flow around it, a concrete block in freefall or in a river WILL reach the horizontal speed of the flow around it.


Premier billvon  (D 16479)
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Apr 11, 2012, 11:49 PM
Post #59 of 87 (788 views)
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Re: [dpreguy] Why Groundspeed? [In reply to] Can't Post

>as your theory is that it will reach "equalibrium" in 6 to 8 seconds.

Or thereabouts, yes. Small, dense objects take longer to reach equilibrium than large, light objects. But they all reach it.

>Your theory is also, that a cement block tossed off a bridge will go
>downstream the same speed as the river, if you just give it some time.

It certainly will - until it hits the bottom, that is.

>Go ahead, drop a cement block into a river, and even if it is a mile deep,
>it will NEVER reach the speed of the flow of the water.

Incorrect. SCUBA divers understand this; a descending SCUBA diver, if he uses his buddy as a reference instead of an anchor line, can find himself far from his intended dive site, even though to him it looks like he's going straight down. That's because he is moving at exactly the same speed as the current (unless he swims, of course.) If he sees a cinderblock drop past, perhaps dropped by a bored skydiver from a bridge, to HIM it will be going straight down - and like him it will be moving with the current.

>The heaver and more aerodynamic an object is, the less it is affected by
>the moving medium it is in.

The longer it takes to reach equilibrium - correct. But they all eventually reach it. If that wasn't true you'd be doing 80kts towards the direction of jump run when you opened because you'd keep all that speed, rather than reach equilibrium with the air you are in.

>Nothing that is going downward will ever reach 80 across the ground, even
>if it falls for an hour.

It will indeed.

This is one of those arguments like the "I know skydivers go up when they open - I saw it on a video!" Yes, it looks like that, and it sounds like that could happen. But the reality simply does not support that idea.


FastRon

Apr 12, 2012, 1:21 AM
Post #60 of 87 (782 views)
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Re: [pchapman] Why Groundspeed? [In reply to] Can't Post

I don't have the experience to provide input on the winds aloft stuff, but I do know something about external ballistics. A dense object (like a bullet) will eventually (very quickly in fact) be moving at the rate and 'value' ( angle of incidence) of the moving air/wind. There is a measurable difference in the drift of objects moving at velocities above and below the speed of sound, but freefall speeds are not that high (yet).


skr  (D 981)

Apr 12, 2012, 2:49 PM
Post #61 of 87 (759 views)
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Re: [skr] Why Groundspeed? [In reply to] Can't Post

Well! That wasn't the discussion I was expecting
from that first post. I know .. Expectations ..

I thought we were going to talk about training students.

But I'm worn out, so I'm going to restate the central
point, and go have a beer or something.

When jumper 2 exits:

|<------>|         is the separation of exit points 

|<=======.======>| is the separation jumper 2 sees


UPPERS ----------- > < ----------- AIRPLANE

|<------>|
EP2 EP1
|<=======.======>|
. . J1
. .
. .
. .
. .
. .
. .
. .
. .
. .
. . OP2 OP1
. . .<------>.
. . . .
. . . .
# # # # # GROUND # # # # # # # GROUND # # # # #

The relevance of this is that almost everybody's
first intuition for leaving separation is to look
out the door, see |<=======.======>| for separation,
and expect that to still be there at opening time.

That's the ground based, mockup intuition of how
the world works.

They need to understand that they're going to spend
all but the first and last bits of the freefall in
a steady state situation where jumper 2 is in a higher,
faster moving layer, and that they're going to end up
with |<------>| at the bottom.

And I think it's possible for experienced jumpers and
USPA and the Safety and Training Committee to explain
this to new jumpers.

I know ..
"Look Martha! Hand me my camera!
There's an idealistic optimist!" :-) :-)

Skr


popsjumper  (D 999999999)

Apr 12, 2012, 4:28 PM
Post #62 of 87 (751 views)
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Re: [skr] Why Groundspeed? [In reply to] Can't Post

It STILL seems as though you are saying that J1 didn't pass through that same higher, faster moving layer.

Why is this not true in your example:
When J1 passes through, it will take him horizontally farther away from J2 and when J2 pass through it will take him horizontally closer to J1, back to the original horizontal separation at exit.

J1 will see J2 first moving horizontally away from him and then horizontally closer to him.


Premier billvon  (D 16479)
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Apr 12, 2012, 4:34 PM
Post #63 of 87 (747 views)
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Re: [popsjumper] Why Groundspeed? [In reply to] Can't Post

>J1 will see J2 first moving horizontally away from him and then
>horizontally closer to him.

Three time periods here:

J1 and J2 are in the same air. Their relationship stays the same.

J1 passes through the layer. He sees himself moving farther and farther from J2.

J2 now hits the layer. J1 sees himself stop moving farther and farther away from J2, and now maintains the new (farther) distance to J2.


popsjumper  (D 999999999)

Apr 13, 2012, 4:10 AM
Post #64 of 87 (724 views)
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Re: [billvon] Why Groundspeed? [In reply to] Can't Post

My example dealt with three layers. Yours with two.
I somewhat agree with the outcome of your two-layer scenario. You just didn't carry it on out to opening. You didn't account for the continued push that J2 is still getting when J1 opens.

And yes, J1's canopy will drift.

And it should be noted that your two-layer scenario assumes that the second layer wind is in the same direction as the first.

From J2's perspective, should it be in an opposing direction, J1 would be approaching J2 until he (J2) hits the second layer and reaches equilibrium.

Or, you could say it the other way around:
J2 would be approaching J1 until he (J2) hits the second layer and reaches equilibrium.


Now, can you say that the horizontal separation at opening would be more or less than at exit?

Don't forget that when J1 opens, J2 is still being pushed and he is being pushed for 10 seconds worth.

And consider J1s canopy drift...is it going to be the same speed and relative distance as his freefall drift?


(This post was edited by popsjumper on Apr 13, 2012, 4:36 AM)


kallend  (D 23151)

Apr 13, 2012, 11:07 AM
Post #65 of 87 (709 views)
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Re: [dpreguy] Why Groundspeed? [In reply to] Can't Post

In reply to:
Dear Mr Kallend,
Don't be so quick to insult. If you are correct, a cement block tossed out of the plane will soon achieve 80 knots speed across the ground, as your theory is that it will reach "equalibrium" in 6 to 8 seconds.

Nope, that is NOT what I said. My comment related to SKYDIVERS, not concrete blocks. A concrete block will still eventually reach an effective equilibrium but it will take a little longer.

Your comments show a distinct lack of knowledge of basic physics.


(This post was edited by kallend on Apr 13, 2012, 11:08 AM)


Premier billvon  (D 16479)
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Apr 13, 2012, 3:12 PM
Post #66 of 87 (690 views)
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Re: [popsjumper] Why Groundspeed? [In reply to] Can't Post

>I somewhat agree with the outcome of your two-layer scenario. You just
>didn't carry it on out to opening.

Well, you still need sufficient separation at opening. The two layer scenario illustrates how those two different layers can increase your separation in freefall. Is that enough increase? Depends on time between jumpers, degree of difference in the winds etc.

>From J2's perspective, should it be in an opposing direction, J1 would be
>approaching J2 until he (J2) hits the second layer and reaches equilibrium.

Yes! And that's why:

1) opposite winds are the worst case and why
2) sometimes the groundspeed thing alone does not work.

Fortunately those cases are rare.

>And consider J1s canopy drift...is it going to be the same speed
>and relative distance as his freefall drift?

Same speed? Yes. Same relative distance, given the same time? Yes. However it will have a MUCH bigger effect on his track relative to the ground. Not because his average horizontal speed has changed (it hasn't) but because his vertical speed has decreased a lot.

(Of course if he heads off in a specific direction his canopy's speed will now increase/decrease his horizontal speed. But that's true no matter what the wind.)


skr  (D 981)

Apr 13, 2012, 7:12 PM
Post #67 of 87 (673 views)
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Re: [popsjumper] Why Groundspeed? [In reply to] Can't Post

> It STILL seems as though you are saying that J1 didn't pass through that same higher, faster moving layer.

Right. This region during and right after exit
is hard to get into words.

The bottom is easy. J1 arrives at O1; a little while
later J2 arrives at O2.

Once you can control opening points you can talk about
group centers and tracking and canopy motion and so on.

And the middle is easy. Each jumper is falling straight
down in his little patch of air, and the higher patch
is moving slowly toward the lower patch.

But the exit stage is hard. It's not linear, and J1 and
J2 are in corresponding parts of their trajectory at
different times, and so on.

I find myself fantasizing about taking John's program
and modifying it to have one jumper exit, and then
some time later a similar jumper exit, and plot, or
somehow display, a history of their horizontal separation.

He's already done all the heavy lifting, selecting the
drag law, the numerical integration technique, step size,
and so on.

But I know I'm not going to do that. I'm just a couple
g's short of floating off to ... otherwhere.

But maybe there's some energetic math or physics student
out there who could be enticed by the fame and glory of
finally resolving the notorious separation debate :-) :-)

And then students could all look at that program.

And we experienced jumpers, too. I'd like to run a bunch
of different scenarios and get a clearer idea of exactly
how the distances play out.

----

But! To your question:

J1 and J2 are not interchangeable. It's not symmetric.
If we start the video running when J1 exits, then J2's
path doesn't look like J1's. It has an extra, horizontal
piece on the front end.

The freefall parts of the trajectories have the same
shape, but when you're comparing J1 and J2 moment by
moment you're comparing points on two dissimilar paths.

       UPPERS ------ >       <------- AIRPLANE 

T=J2 exit T=0 =J1 exit
_ _ _ _ _ _ _ _ _ _ =J2 keep going straight and level
/ /
| |
\ \
\ \
\ \T=J2 exit <--comparing J2's freefall
\ \ trajectory from top to
\ \ almost bottom with J1's
\ \ from here on down
\ \
\ \
\ \
\ \
\T=part way down \
\ \
| |
| |
| |
| |T=part way down
/ /
/ /
/ /
/ /
open open


# # # # # # # GROUND # # # # # # # # # # GROUND # # #

So, I can't answer any better than that until that mythical,
long lost, skydiving, physics grad student shows up and we
can run a bunch of examples.

It's a good question.

Skr


kallend  (D 23151)

Apr 14, 2012, 8:21 AM
Post #68 of 87 (641 views)
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Re: [billvon] Why Groundspeed? [In reply to] Can't Post

In reply to:
>I somewhat agree with the outcome of your two-layer scenario. You just
>didn't carry it on out to opening.

Well, you still need sufficient separation at opening. The two layer scenario illustrates how those two different layers can increase your separation in freefall. Is that enough increase? Depends on time between jumpers, degree of difference in the winds etc.

>From J2's perspective, should it be in an opposing direction, J1 would be
>approaching J2 until he (J2) hits the second layer and reaches equilibrium.

Yes! And that's why:

1) opposite winds are the worst case and why
2) sometimes the groundspeed thing alone does not work.

Fortunately those cases are rare.

The opposite wind case is not particularly rare in the mid west, particularly near the great lakes.


winsor  (D 13715)

Apr 15, 2012, 8:27 AM
Post #69 of 87 (610 views)
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Re: [JohnMitchell] Why Groundspeed? [In reply to] Can't Post

In reply to:
In reply to:

Einstein is often (incorrectly) credited with inventing the A-bomb, while those who actually did invent it are largely unknown. There's a pattern here?
But Einstein emailed Pres. Roosevelt about it, right?

At the urging of Leo Szilard et al.


kallend  (D 23151)

Apr 15, 2012, 11:48 AM
Post #70 of 87 (594 views)
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Re: [winsor] Why Groundspeed? [In reply to] Can't Post

In reply to:
In reply to:
In reply to:

Einstein is often (incorrectly) credited with inventing the A-bomb, while those who actually did invent it are largely unknown. There's a pattern here?
But Einstein emailed Pres. Roosevelt about it, right?

At the urging of Leo Szilard et al.

No, it wasn't Al, it was Wigner, Teller and SachsTongue

Einstein is reported to have said "Daran habe ich gar nicht gedacht!" when told about a nuclear chain reaction.


winsor  (D 13715)

Apr 15, 2012, 12:44 PM
Post #71 of 87 (584 views)
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Re: [kallend] Why Groundspeed? [In reply to] Can't Post

In reply to:
In reply to:
In reply to:
In reply to:

Einstein is often (incorrectly) credited with inventing the A-bomb, while those who actually did invent it are largely unknown. There's a pattern here?
But Einstein emailed Pres. Roosevelt about it, right?

At the urging of Leo Szilard et al.

No, it wasn't Al, it was Wigner, Teller and SachsTongue

Einstein is reported to have said "Daran habe ich gar nicht gedacht!" when told about a nuclear chain reaction.

Richard Rhodes' "The Making of the Atomic Bomb" had Szilard penning the letter and Al signing on with Wigner, Teller & Sachs.

As you note, Einstein took no credit for the work.


BSBD,

Winsor


JohnMitchell  (D 6462)

Apr 15, 2012, 3:22 PM
Post #72 of 87 (574 views)
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Re: [kallend] Why Groundspeed? [In reply to] Can't Post

In reply to:
[
Nope, that is NOT what I said. My comment related to SKYDIVERS, not concrete blocks. A concrete block will still eventually reach an effective equilibrium but it will take a little longer.
It's all about an object's ballistic coefficient. Everything has one, including skydivers. The concept is very familiar to accuracy shooters and ammo reloaders, me included. Smile


popsjumper  (D 999999999)

Apr 15, 2012, 5:37 PM
Post #73 of 87 (559 views)
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Re: [JohnMitchell] Why Groundspeed? [In reply to] Can't Post

In reply to:
In reply to:
[
Nope, that is NOT what I said. My comment related to SKYDIVERS, not concrete blocks. A concrete block will still eventually reach an effective equilibrium but it will take a little longer.
It's all about an object's ballistic coefficient. Everything has one, including skydivers. The concept is very familiar to accuracy shooters and ammo reloaders, me included. Smile

Even anvils reach equilibrium...ask Bigun.


(This post was edited by popsjumper on Apr 15, 2012, 5:38 PM)


JohnMitchell  (D 6462)

Apr 15, 2012, 5:43 PM
Post #74 of 87 (549 views)
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Re: [popsjumper] Why Groundspeed? [In reply to] Can't Post

I think he uses Mach numbers, doesn't he?Tongue


hub1100  (C 41319)

Apr 15, 2012, 5:58 PM
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Re: [popsjumper] Why Groundspeed? [In reply to] Can't Post

With respect to this thread:

Isn't equilibrium defined as: the condition of a system in which all competing influences are balanced, in a wide variety of contexts. (balance)

Aren't we all taking about Terminal velocity? : A free-falling object achieves its terminal velocity when the downward force of gravity (FG) equals the upward force of drag (Fd). This causes the net force on the object to be zero, resulting in an acceleration of zero.

I would agree Anvil's, bantam weights and free fliers all reach a different terminal velocity. Is this factor the ultimate part of this discussion, not a flat timed fall rate?

Thx -hubs-


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