# of members ?

[Me Loonn]

How many members there are, actually ?
How many active and how many not-so-active ?

(inspired by : 'Space Empires: IV Recent Visitors: 29' ... 3 full rows http://forum.shrapnelgames.com/images/icons/tongue.gif )

[Katchoo]

According to the main Shrapnel page there are 4097 people registered to post Messages. This is of couse spread out among all the forums (many non-SE) present here.

I'm sure one of the Moderators might have more acurate info on how many out of that 4097 are active and how many of the active ones frequent here.

[Taera]

it is very difficult to decide who are active and who are not. there are many people on vacation or temporarily away for RL reasons, and some only check in on weekends or weekdays. overall i'd say there are about 30 constantly active forumers

[DavidG]

Originally posted by Taera:
it is very difficult to decide who are active and who are not. there are many people on vacation or temporarily away for RL reasons, and some only check in on weekends or weekdays. overall i'd say there are about 30 constantly active forumers<font size="2" face="Verdana, Helvetica, sans-serif">Really? Sounds way to low to me.

[Taz-in-Space]

I did a little unofficial count and got over 60 regulars (Posts at least once/week). But maybe this is just a good week! http://forum.shrapnelgames.com/images/icons/icon10.gif

WooHoo! 10 more Posts and I'm a First Looie! http://forum.shrapnelgames.com/images/icons/icon6.gif

[ June 07, 2003, 04:38: Message edited by: Taz-in-Space ]

[Fyron]

He said constantly active, which means a number of Posts per day at least. One post a week is active, but not "constantly active". http://forum.shrapnelgames.com/images/icons/icon12.gif

[Kamog]

Well, I started counting, and I got to about 40, but it's difficult to say who is "constantly active" and who is not. A few people are obviously constantly active, but many do not post every single day but do post often, so it's hard to say.

[Taera]

for one thing you are http://forum.shrapnelgames.com/images/icons/icon10.gif

[Slynky]

Originally posted by Imperator Fyron:
He said constantly active, which means a number of Posts per day at least. One post a week is active, but not "constantly active". http://forum.shrapnelgames.com/images/icons/icon12.gif <font size="2" face="Verdana, Helvetica, sans-serif">Perhaps we shouldn't measure "constantly" active by Fyron standards... http://forum.shrapnelgames.com/images/icons/icon10.gif .

Also, I count active by whom I see when I pop in...not necessarily who Posts something. Of course, the term, "active", may infer more than just reading http://forum.shrapnelgames.com/images/icons/icon12.gif .

[Fyron]

Yes, "active" requires more than just lurking. http://forum.shrapnelgames.com/images/icons/icon12.gif

[narf poit chez BOOM]

me active.

taz hyperactive. hey, i'm a luie? when did that happen? over 600 Posts? guess i do post alot.

[ June 07, 2003, 21:37: Message edited by: narf poit chez BOOM ]

[Stone Mill]

Originally posted by Slynky:
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Imperator Fyron:
He said constantly active, which means a number of Posts per day at least. One post a week is active, but not "constantly active". http://forum.shrapnelgames.com/images/icons/icon12.gif <font size="2" face="Verdana, Helvetica, sans-serif">Perhaps we shouldn't measure "constantly" active by Fyron standards... http://forum.shrapnelgames.com/images/icons/icon10.gif .

Also, I count active by whom I see when I pop in...not necessarily who Posts something. Of course, the term, "active", may infer more than just reading http://forum.shrapnelgames.com/images/icons/icon12.gif .</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Quantifying Fyron's forum activity is undefinable by our current mathimatical tools. http://forum.shrapnelgames.com/images/icons/icon12.gif

[Fyron]

You must have failed Algebra then. http://forum.shrapnelgames.com/images/icons/icon12.gif

[Taera]

i will be taking calculus next year, perharps that'll help http://forum.shrapnelgames.com/images/icons/icon10.gif

[Fyron]

Nah, too advanced. Actually, algebra is almost too advanced. Its counting, really. http://forum.shrapnelgames.com/images/icons/icon12.gif

[Taera]

well, if you know, why dont you answer? http://forum.shrapnelgames.com/images/icons/icon10.gif

[Fyron]

The answer is 42, of course.

[Taera]

no, your wrong. as everyone else is. muwahahahahhahaha
oh

[Stone Mill]

Originally posted by Imperator Fyron:
You must have failed Algebra then. http://forum.shrapnelgames.com/images/icons/icon12.gif <font size="2" face="Verdana, Helvetica, sans-serif">Ack! You found my weakness! http://forum.shrapnelgames.com/images/icons/icon12.gif

Perhaps, but does not the word "constantly" best relate to physics, or perhaps the study of word use when applied to describing someone's behavior?

Your posting may be constant, like the force of gravity, for all we know. http://forum.shrapnelgames.com/images/icons/icon10.gif

[Taera]

constant related to what? the higher you go the less the gravity so it depends http://forum.shrapnelgames.com/images/icons/icon10.gif

[Fyron]

Gravity is constant in that the rate of change of the force due to gravity is constant. http://forum.shrapnelgames.com/images/icons/icon12.gif

[PvK]

I think you need to study anti-logic to quantify Fryon. http://forum.shrapnelgames.com/images/icons/icon12.gif

PvK

[Jack Simth]

Originally posted by Imperator Fyron:
Gravity is constant in that the rate of change of the force due to gravity is constant. http://forum.shrapnelgames.com/images/icons/icon12.gif <font size="2" face="Verdana, Helvetica, sans-serif">That only works in specialized circumstances; perhaps gravity is constant in that that G, the gravitational constant, is constant.

[David E. Gervais]

Well, I'm definately a 'Constant', and that's alot better than being a 'Variable' IHMO. http://forum.shrapnelgames.com/images/icons/icon12.gif

hmmm,.. but then again my post count is always changing, so I might be a 'variable' disguised as a 'constant'.

Come to think of it, many of the people here would best be described as 'Regulars',..

So I'm happy to announce I'm a regular guy. http://forum.shrapnelgames.com/images/icons/tongue.gif

Cheers! http://forum.shrapnelgames.com/images/icons/icon10.gif

P.S. I thought Gravity was 'relitive' not 'constant'. And if I'm not wrong it's relitive to mass, right?

[ June 08, 2003, 21:49: Message edited by: David E. Gervais ]

[Jack Simth]

Force from gravity is relative, sure:

F = G((M * m)/(d^2))

Where F is the magnitude of the force due to gravity, G is the gravitational constant (it is only sort of constant, as it depends on the units used), M and m are the two masses involved and d is the distance between the two masses. The direction is towards the other object.

Also, as Acceleration = Force/Mass, the acceleration is also relative.

[Kamog]

Originally posted by David E. Gervais:
Well, I'm definately a 'Constant', and that's alot better than being a 'Variable' IHMO. http://forum.shrapnelgames.com/images/icons/icon12.gif

hmmm,.. but then again my post count is always changing, so I might be a 'variable' disguised as a 'constant'.

Come to think of it, many of the people here would best be described as 'Regulars',..
<font size="2" face="Verdana, Helvetica, sans-serif">Yeah, a variable can be static, or it can be const, and also a variable can be a member.

[Stone Mill]

In this forum, you can be bailed out at any given time when you stumble into a scientific discussion. Gotta love it! http://forum.shrapnelgames.com/images/icons/icon7.gif

[Fyron]

Originally posted by PvK:
I think you need to study anti-logic to quantify Fryon. http://forum.shrapnelgames.com/images/icons/icon12.gif

PvK<font size="2" face="Verdana, Helvetica, sans-serif">Not sure who this Fryon is... but it is quite impossible to quantify a person. http://forum.shrapnelgames.com/images/icons/tongue.gif

[Kamog]

Is the gravitational constant really constant?

[Jack Simth]

Originally posted by Kamog:
Is the gravitational constant really constant?<font size="2" face="Verdana, Helvetica, sans-serif">To the best of my knoweledge, within a given system of units (e.g. metric: meters, seconds, kilograms; "English": feet, seconds, slugs) the gravitational constant is constant. However, we have only had the tools to measure it for a fairly short period of time, and only on Earth.

[Suicide Junkie]

And around earth, and all over the solar system, and all over the galaxy...

If the gravitational constant was different over at alpha centauri, for example, then the stars would be orbiting each other at the "wrong" speed.

PS: And, variables can also be functions in certain cases.

[ June 09, 2003, 14:15: Message edited by: Suicide Junkie ]

[Perrin]

Wait I thought that there was only Fyron. At that the rest of us are just sub personalities or figments of his imagination.

[Fyron]

No silly, we are all Puke. http://forum.shrapnelgames.com/images/icons/icon6.gif

And the gravitational constant is indeed constant everywhere. We may not have the exact value down to the nanometer (and probably never will), but that doesn't really matter. http://forum.shrapnelgames.com/images/icons/icon12.gif

[Jack Simth]

Originally posted by Suicide Junkie:
If the gravitational constant was different over at alpha centauri, for example, then the stars would be orbiting each other at the "wrong" speed.<font size="2" face="Verdana, Helvetica, sans-serif">But to tell how fast they "should" be orbiting each other, you need to know how massive they are, which is calculated from G and how fast they are orbiting each other. Using G to calculate G in such a manner is circular reasoning; it isn't reliable. It is probably the same everywhere, but we can't be certain until we send people over there to take local measurements.

[ June 09, 2003, 18:57: Message edited by: Jack Simth ]

[Suicide Junkie]

Star masses can also be found by looking at colour, age, size, etc.

And having a third body in the system to observe helps a lot.

G can be calculated from the masses, distance between, and the observed acceleration.

[ June 09, 2003, 21:14: Message edited by: Suicide Junkie ]

[Jack Simth]

Originally posted by Suicide Junkie:
Star masses can also be found by looking at colour, age, size, etc.
<font size="2" face="Verdana, Helvetica, sans-serif">Getting the mass from the color, age, size, etc. implicitly uses G, as the plasma physics that produce such results include gravitational effects from the mass of the star. Again, G is used to calculate G, and as such is circular logic and is not reliable. Mind you, G is probably constant throughout the universe - but there isn't any good way to be certain of that until we get out there and measure things up close. Until then, G is constant makes for a good working theory, but it can't be proven. Originally posted by Suicide Junkie:

And having a third body in the system to observe helps a lot.

G can be calculated from the masses, distance between, and the observed acceleration.<font size="2" face="Verdana, Helvetica, sans-serif">For that, the masses have to be known. Getting the masses uses G, although sometimes it is implicit rather than explicit. Again, G is used to obtain G, which is circular reasoning; not reliable. Mind you, G is probably constant throughout the universe - but there isn't any good way to be certain of that until we get out there and measure things up close.

Without knowing both the masses and G, some simple numerical manipulation on the gravitational formulas can tell you that the distance and acceleration alone won't help:

F = G(M*m)/(d^2)
F' = G'(M'*m')/(d'^2)
A = F/m = (G*M)/(d^2)
A' = F'/m' = (G'*M')/(d'^2)
If A = A' and d = d', then
(G'*M')/(d^2) = (G*M)/(d^2)
-> (G'*M') = (G*M)

Example: Suppose G' = 2G:
-> 2G*M' = G*M -> 2M' = M -> M' = M/2

Then M' = M/2 results in the same acceleration for the same distance. The number of bodies won't make a difference for this aspect of things.

[Atrocities]

F = G(M*m)/(d^2)
F' = G'(M'*m')/(d'^2)
A = F/m = (G*M)/(d^2)
A' = F'/m' = (G'*M')/(d'^2)
If A = A' and d = d', then
(G'*M')/(d^2) = (G*M)/(d^2)
-> (G'*M') = (G*M) <font size="2" face="Verdana, Helvetica, sans-serif">The universe made simple. http://forum.shrapnelgames.com/images/icons/icon7.gif

[mottlee]

Man did this one get DEEP! (way over my head)

[Greybeard]

Well, I consider my reading the Forum as active. I try to read it every day. However, I only post occassionally, usually because someone has already made the point I was going to make and I don't want to just say, "yea, me too!"

Greybeard http://forum.shrapnelgames.com/images/icons/icon10.gif http://forum.shrapnelgames.com/images/icons/icon10.gif http://forum.shrapnelgames.com/images/icons/icon10.gif

[Suicide Junkie]

Gravity is not the only force out there, so if gravity is off, and mass is off to compensate, you'll also have to adjust just the speed of light (E=MC^2), the electromagnetic, weak and strong nuclear forces, etc because the mass of nucleons has all changed.

Is that really what you meant to imply?
I am not a physicist, but I suspect you'd be hard pressed to find a stable universe with different constants and still have anything close to the same observations made.

[ June 10, 2003, 14:46: Message edited by: Suicide Junkie ]

[Loser]

As in any act of speculation, the speculator must consider that there may be something else happening that s/he does not understand.

Not that you should leave it alone, go ahead and poke at it. Hypothesize, experiment, publish, play with wooden blocks!

There are some things in this reality that we have only a limited ability to perceive. The studies of history, psychoanalysis (and psychology to a lesser degree), and astrophysics require some amount of assumption and intuition. Someday maybe we'll be able to clearly perceive and test with all three. Someday maybe we'll cure halitosis. And someday I might find a girl who is both fun and a suitable wife/mother-of-my-children. But for now we're just doing what we can (which isn't so bad: humans kick aft!).

[Jack Simth]

Originally posted by Suicide Junkie:
Gravity is not the only force out there, so if gravity is off, and mass is off to compensate, you'll also have to adjust just the speed of light (E=MC^2), the electromagnetic, weak and strong nuclear forces, etc because the mass of nucleons has all changed.
<font size="2" face="Verdana, Helvetica, sans-serif">Interesting - how did you jump from the mass of stars perhaps being other than what people think it is to the mass of the particles perhaps being different? Originally posted by Suicide Junkie:

Is that really what you meant to imply? <font size="2" face="Verdana, Helvetica, sans-serif">Strange - This started when I said G is constant on earth; people stated that G is constant everywhere as an absolute fact, and I replied that we can't be certain of that, as we have only been measuring for a short time on Earth; that the possibility can't be ruled out that it is different elsewhere. I have always indicated that G is probably constant elswhere each time I reply; however, there have been arguments in the past which suggest that G has changed such as the one connected to this abstract. (http://www.science-frontiers.com/sf017/sf017p10.htm) Originally posted by Suicide Junkie:

I am not a physicist, but I suspect you'd be hard pressed to find a stable universe with different constants and still have anything close to the same observations made.<font size="2" face="Verdana, Helvetica, sans-serif">Actually, all figures on such things make references to changes of particular magnitudes; less than that, and no compensation is necessary for a stable universe.

[ June 11, 2003, 03:42: Message edited by: Jack Simth ]

[Me Loonn]

Sigh
As usual, this thread faced the same fate as all the others ...