| Topic: |
Science > Physics |
| User: |
"Don1" |
| Date: |
29 Aug 2005 07:47:17 PM |
| Object: |
If an equal arm scale balances |
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
It is the policy of NIST to calibrate spring scales to measure mass,
and they too, must be recalibrated each time their location is
significantly altered.
Don
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| User: "Steve Ralph" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 10:47:47 AM |
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"Don1" <dcshead@charter.net> wrote in message
news:1125362837.387381.52620@z14g2000cwz.googlegroups.com...
If an equal arm balance scale balances in one place, it will balance
anywhere.
Wrong. Totally completely utterly wrong.
Are you this stupid naturally, or do you work at it?
sr
That's why people think they measure mass; which is constant
anywhere.
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
It is the policy of NIST to calibrate spring scales to measure mass,
and they too, must be recalibrated each time their location is
significantly altered.
Don
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| User: "Herman Trivilino" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 05:59:43 PM |
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"Steve Ralph" <steve@steveralph.f9.co.uk> wrote ...
If an equal arm balance scale balances in one place, it will balance
anywhere.
Wrong. Totally completely utterly wrong.
Are you this stupid naturally, or do you work at it?
No one knows for sure. So, let's change the subject.
He's practicing his act. He throws it out there to see what it'll get in
the way of a reaction. The stuff that he gets wrong he omits in his future
posts. Well, he tries to do that, to the best of his ability.
If he were better at it he'd be a troll.
----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
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| User: "" |
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| Title: Re: If an equal arm scale balances |
29 Aug 2005 08:55:52 PM |
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Did you know that pecans are a good source of polyunsaturated oil???
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| User: "pete" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 07:06:46 AM |
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Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
--
pete
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 08:13:38 AM |
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pete wrote:
Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
--
pete
Yes pete but it will balance with a different reading for weight: About
six times less than on Earth.
Don
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| User: "Mark Fergerson" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 10:48:22 AM |
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Don1 wrote:
pete wrote:
Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
Absolutely.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
Absolutely right.
Yes pete but it will balance with a different reading for weight: About
six times less than on Earth.
Absolutely wrong. Do you actually believe that you'd have to move
the counterbalance if the load stays the same? Why would you think that?
Mark L. Fergerson
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 04:39:06 PM |
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Mark Fergerson wrote:
Don1 wrote:
pete wrote:
Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
Absolutely.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
Absolutely right.
Yes pete but it will balance with a different reading for weight: About
six times less than on Earth.
Absolutely wrong. Do you actually believe that you'd have to move
the counterbalance if the load stays the same? Why would you think that?
It's not the difference in weight, it's the product of the weight and
its distance from the fulcrum, that must balance that on the opposite
side. On the moon a body's weight is only one sixth as much as on
Earth.
Don
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| User: "Mark Fergerson" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 07:58:35 AM |
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Don1 wrote:
Mark Fergerson wrote:
Don1 wrote:
pete wrote:
Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
Absolutely.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
Absolutely right.
Yes pete but it will balance with a different reading for weight: About
six times less than on Earth.
Absolutely wrong. Do you actually believe that you'd have to move
the counterbalance if the load stays the same? Why would you think that?
It's not the difference in weight, it's the product of the weight and
its distance from the fulcrum, that must balance that on the opposite
side.
Ah, then you _do_ know how a steelyard works.
On the moon a body's weight is only one sixth as much as on
Earth.
Correct, for both the load _and_ the counterbalance. Since the
weight of both change by exactly the same ratio but the distances
don't change at all, the balance will still balance. Why can't you
see that?
Mark L. Fergerson
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 09:39:53 AM |
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Mark Fergerson wrote:
Don1 wrote:
Mark Fergerson wrote:
Don1 wrote:
pete wrote:
Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
Absolutely.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
Absolutely right.
Yes pete but it will balance with a different reading for weight: About
six times less than on Earth.
Absolutely wrong. Do you actually believe that you'd have to move
the counterbalance if the load stays the same? Why would you think that?
It's not the difference in weight, it's the product of the weight and
its distance from the fulcrum, that must balance that on the opposite
side.
Ah, then you _do_ know how a steelyard works.
On the moon a body's weight is only one sixth as much as on
Earth.
Correct, for both the load _and_ the counterbalance. Since the
weight of both change by exactly the same ratio but the distances
don't change at all, the balance will still balance. Why can't you
see that?
I see that Mark, but the distances don't mean the same thing on a scale
calibrated for use on the moon: The distances between markings for
weight on the moon will be about six times greater than for weight on
Earth.
Mark L. Fergerson
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| User: "odin" |
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| Title: Re: If an equal arm scale balances |
01 Sep 2005 09:13:01 AM |
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Correct, for both the load _and_ the counterbalance. Since the
weight of both change by exactly the same ratio but the distances
don't change at all, the balance will still balance. Why can't you
see that?
I see that Mark, but the distances don't mean the same thing on a scale
calibrated for use on the moon: The distances between markings for
weight on the moon will be about six times greater than for weight on
Earth.
But when the weights balance, the distance you are talking about is zero.
You talk about distances between markings, but in a scale where one weight
is compared to another, there are no markings, except for the one center
point. Idiot.
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
01 Sep 2005 09:38:40 AM |
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odin wrote:
Correct, for both the load _and_ the counterbalance. Since the
weight of both change by exactly the same ratio but the distances
don't change at all, the balance will still balance. Why can't you
see that?
I see that Mark, but the distances don't mean the same thing on a scale
calibrated for use on the moon: The distances between markings for
weight on the moon will be about six times greater than for weight on
Earth.
But when the weights balance, the distance you are talking about is zero.
You talk about distances between markings, but in a scale where one weight
is compared to another, there are no markings, except for the one center
point. Idiot.
You must be thinking of an equal arm balance.
Don
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| User: "odin" |
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| Title: Re: If an equal arm scale balances |
01 Sep 2005 11:58:54 AM |
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"Don1" <dcshead@charter.net> wrote in message
news:1125585520.521117.83270@g44g2000cwa.googlegroups.com...
odin wrote:
Correct, for both the load _and_ the counterbalance. Since the
weight of both change by exactly the same ratio but the distances
don't change at all, the balance will still balance. Why can't you
see that?
I see that Mark, but the distances don't mean the same thing on a scale
calibrated for use on the moon: The distances between markings for
weight on the moon will be about six times greater than for weight on
Earth.
But when the weights balance, the distance you are talking about is zero.
You talk about distances between markings, but in a scale where one
weight
is compared to another, there are no markings, except for the one center
point. Idiot.
You must be thinking of an equal arm balance.
Don you are an idiot. The markings you are talking about on a Steelyard
Platform Scale are not relevant to the point I was making. The balance still
requires the deviation from the center point to be zero for a reading to be
made. The markings you refer to are there to establish a variable of force
times distance (a variable torque) that is compared against a fixed force x
distance (a fixed torque). When the two torques are equal, the deviation of
the balance from center is zero. The same markings will work the same way
and give the same readings on earth or on the moon. Remember that the lower
gravitational field on the moon will effect the weight of the fixed mass and
the movable mass in the proportionality. The change in g cancels out, and
the result is the same.
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| User: "Mark Fergerson" |
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| Title: Re: If an equal arm scale balances |
01 Sep 2005 09:57:14 AM |
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Don1 wrote:
Mark Fergerson wrote:
Don1 wrote:
Mark Fergerson wrote:
Don1 wrote:
pete wrote:
Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
Absolutely.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
Absolutely right.
Yes pete but it will balance with a different reading for weight: About
six times less than on Earth.
Absolutely wrong. Do you actually believe that you'd have to move
the counterbalance if the load stays the same? Why would you think that?
It's not the difference in weight, it's the product of the weight and
its distance from the fulcrum, that must balance that on the opposite
side.
Ah, then you _do_ know how a steelyard works.
On the moon a body's weight is only one sixth as much as on
Earth.
Correct, for both the load _and_ the counterbalance. Since the
weight of both change by exactly the same ratio but the distances
don't change at all, the balance will still balance. Why can't you
see that?
I see that Mark, but the distances don't mean the same thing on a scale
calibrated for use on the moon: The distances between markings for
weight on the moon will be about six times greater than for weight on
Earth.
What are you talking about? Are you trying to say that the
steelyard will give "false" readings in that it's not reading the
"true, absolute weights" of the load and counterbalance? That would
be a silly thing to say because there just ain't no such thing as
"true, absolute weight". There's only weight as caused by the
_local_ acceleration due to gravity. Gravitational acceleration is
simply less on the Moon than it is on Earth, period, and there's
nothing special about its value on Earth. A steelyard balanced here
(say with a pint jar of water as the load and a pound mass as the
counterweight) will balance anywhere there's gravity without adjustment.
A steelyard alone cannot tell you what the local acceleration due
to gravity is. You'd have to drop it and time its fall. It will
honestly tell you that the pint jar of water feels the same
acceleration due to gravity as does the pound mass, _anywhere_.
Got it now?
Mark L. Fergerson
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| User: "snapdragon31" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 05:40:24 PM |
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Don1 wrote:
Mark Fergerson wrote:
Don1 wrote:
pete wrote:
Don1 wrote:
An unequal arm platform balance scale, or steelyard, is calibrated to
balance only when the weight placed on the platform is equal to the
scale reading. They must be recalibrated each time their location is
significantly altered.
Absolutely wrong.
Absolutely.
If the heavy weight on the short arm
balances the light weight on the long arm, on earth,
then it will also balance on the moon.
Absolutely right.
Yes pete but it will balance with a different reading for weight: About
six times less than on Earth.
Absolutely wrong. Do you actually believe that you'd have to move
the counterbalance if the load stays the same? Why would you think that?
It's not the difference in weight, it's the product of the weight and
its distance from the fulcrum, that must balance that on the opposite
side. On the moon a body's weight is only one sixth as much as on
Earth.
Don
Hi Don,
Assume on the left of the balance is a man whose mass is Ml at a
distance of Dl from the fulcrum.
The weight on the right has a mass of Mr at a distance of Dr from the
fulcrum.
Let gE be the gravity on Earth.
gM be the gravity on the Moon
Total counterclockwise moment = Ml * gE * Dl
Total clockwise moment = Mr * gE * Dr
If they balance on Earth then
Ml * gE * Dl = Mr * gE * Dr
==> Ml * Dl = Mr * Dr
==> Ml * gM * Dl = Mr * gM * Dr
they also balance on the Moon.
Snapdragon
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| User: "tadchem" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 06:14:50 PM |
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"snapdragon31" <snapdragon31@gmail.com> wrote in message
news:1125441624.244712.302690@g49g2000cwa.googlegroups.com...
<snip>
Assume on the left of the balance is a man whose mass is Ml at a
distance of Dl from the fulcrum.
The weight on the right has a mass of Mr at a distance of Dr from the
fulcrum.
Oops! You just lost sHead. He is obsessed with the idea that 'weight' is a
name for a force, and goes ballistic when anyone uses the common noun in any
other sense, such as you have just done in referring to an object used for
comparison with other objects with mass in an apparatus that exploits the
effect of gravity on mass to exert a measurable force.
He cannot understand that 'weight' can be the name of an object.
He also cannot understand that it is possible to compare the mass of two
objects (or sets of objects) via their weights *without* first quantifying
the weight.
He also does not understand concepts of elementary algebra such as the
'multiplicative inverse' or 'transitive operations.'
Read a few of the several threads in which he repeatedly posts to see how
completely naive his thought processes are.
Tom Davidson
Richmond, VA
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| User: "Paul Cardinale" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 06:07:15 PM |
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Gee let's try this out:
(view with mono-spaced font)
W1__________o______W2
<---L1--->^<-L2->
On Earth L1 * W1 = L2 * W2
So L1 = L2 * W2 / W1
But on the Moon, W1' = W1 / 6 and W2' = W2 / 6
So in order to balance, L1 = L2 * W2' / W1'
Subsituting, we get L1 = L2 * (W2/6) / (W1/6)
Multiply by 6/6 and get:
L1 = L2 * W2 / W1
Now do you see why you're an idiot?
Paul Cardinale
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| User: "Paul Cardinale" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 09:08:48 AM |
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Wow. New low level of stupitude for you. I'm sure that some people
thought it was impossible, but you're actually getting stupider. Why
is it that you think that if you make up crap out of whole cloth, that
it must be true?
Paul Cardinale
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| User: "Uncle Al" |
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| Title: Re: If an equal arm scale balances |
30 Aug 2005 10:59:49 AM |
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Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 03:35:37 AM |
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In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "tadchem" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 04:46:30 AM |
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<mmeron@cars3.uchicago.edu> wrote in message
news:tZdRe.11$45.3728@news.uchicago.edu...
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al
<UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Since balances balance force, and
F = -GMm/r^2
to get a change of 0.1 ppm in F (not easy, but also not too difficult to
observe empirically) we need only change r by 0.05 ppm.
0.1 ppm from the earth's radius R (about 6400 km) is 0.64 meters.
FWIW, the tide-raising force of the moon varies from about 0.055 g (minimum)
to 0.11 g (maximum) [the sun's tide raising force is 5/11ths that of the
moon]. But then if your balance arms (the oceans) are thousands of
kilometers long, you can get some pretty sensitive detection.
Tom Davidson
Richmond, VA
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| User: "" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 04:54:39 AM |
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In article <tIydnZL-IPM44YjeRVn-gg@comcast.com>, "tadchem" <tadchemNOSPAM@comcast.net> writes:
<mmeron@cars3.uchicago.edu> wrote in message
news:tZdRe.11$45.3728@news.uchicago.edu...
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al
<UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Since balances balance force, and
F = -GMm/r^2
to get a change of 0.1 ppm in F (not easy, but also not too difficult to
observe empirically) we need only change r by 0.05 ppm.
0.1 ppm from the earth's radius R (about 6400 km) is 0.64 meters.
Aha. But, the balance baalances (no pun intended) when it is
horizontal. Meaning, to first order, the arm is normal to the local
direction of gravitational force. Now, estimate what sort of local
anisotropy you need so that balance which is level in one location, is
not level in another.
FWIW, the tide-raising force of the moon varies from about 0.055 g (minimum)
to 0.11 g (maximum) [the sun's tide raising force is 5/11ths that of the
moon]. But then if your balance arms (the oceans) are thousands of
kilometers long, you can get some pretty sensitive detection.
No argument about it. But, the balance is only sensitive to the
anisotropy, not the force itself. So, what sort of anisotropy you
need for it to be detectable over a span of, say, 1 meter.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Uncle Al" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 12:14:22 PM |
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wrote:
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Nope. Space station spinning for gravity will do it. Coriolis effect
will depend on balance orientation and placement adn there will be
severe gradients even for immodest sizes. Dumb Donny ***** is a
boring psychotic ididot.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 05:59:20 PM |
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In article <4315E56E.26A45035@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
mmeron@cars3.uchicago.edu wrote:
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Nope. Space station spinning for gravity will do it.
That *exactly* goes under "one hell of anisotropy". There are
inherent assumptions made when using a balance (same as there are
inherent assumptions made when using *any* physical instrument) and a
key assumption is that the length of the balance is small in
comparison to the length scale over which gravity changes appreciably.
Dumb Donny ***** is a boring psychotic ididot.
true, but irrelevant to the case at hand.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
31 Aug 2005 01:41:47 PM |
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Uncle Al wrote:
mmeron@cars3.uchicago.edu wrote:
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Nope. Space station spinning for gravity will do it. Coriolis effect
will depend on balance orientation and placement adn there will be
severe gradients even for immodest sizes. Dumb Donny ***** is a
boring psychotic ididot.
Are you telling us that a balance is useless in a space station? Tell
us something we don't know Unk dopy.
Don
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| User: "Mark Fergerson" |
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| Title: Re: If an equal arm scale balances |
01 Sep 2005 10:04:47 AM |
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Don1 wrote:
Uncle Al wrote:
mmeron@cars3.uchicago.edu wrote:
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Nope. Space station spinning for gravity will do it. Coriolis effect
will depend on balance orientation and placement adn there will be
severe gradients even for immodest sizes. Dumb Donny ***** is a
boring psychotic ididot.
Are you telling us that a balance is useless in a space station? Tell
us something we don't know Unk dopy.
He just said it _isn't_ useless, you just have to be careful with
orientation etc. In fact, if you do several weighings while
carefully noting how you orient and place the scale, you can
determine the axis and rate of rotation without looking out a window.
BTW Don, if the space station example is too "unEarthly" for you,
ever been on a merry-go-round?
Mark L. Fergerson
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
01 Sep 2005 02:23:10 PM |
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Mark Fergerson wrote:
Don1 wrote:
Uncle Al wrote:
mmeron@cars3.uchicago.edu wrote:
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Nope. Space station spinning for gravity will do it. Coriolis effect
will depend on balance orientation and placement adn there will be
severe gradients even for immodest sizes. Dumb Donny ***** is a
boring psychotic ididot.
Are you telling us that a balance is useless in a space station? Tell
us something we don't know Unk dopy.
He just said it _isn't_ useless, you just have to be careful with
orientation etc. In fact, if you do several weighings while
carefully noting how you orient and place the scale, you can
determine the axis and rate of rotation without looking out a window.
Who on Earth would bother, when the whole world's available? The Earth
provides a better reference frame.
BTW Don, if the space station example is too "unEarthly" for you,
ever been on a merry-go-round?
I certainly wouldn't choose it as a reference for measuring in any way,
shape or form. It's enough to cause a severe headache. Only guys like
you and Unk Dopey would even think of such a dumb thing.
Don
Mark L. Fergerson
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
01 Sep 2005 07:28:59 PM |
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Don1 wrote:
Mark Fergerson wrote:
Don1 wrote:
Uncle Al wrote:
mmeron@cars3.uchicago.edu wrote:
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Nope. Space station spinning for gravity will do it. Coriolis effect
will depend on balance orientation and placement adn there will be
severe gradients even for immodest sizes. Dumb Donny ***** is a
boring psychotic ididot.
Are you telling us that a balance is useless in a space station? Tell
us something we don't know Unk dopy.
He just said it _isn't_ useless, you just have to be careful with
orientation etc. In fact, if you do several weighings while
carefully noting how you orient and place the scale, you can
determine the axis and rate of rotation without looking out a window.
Who on Earth would bother, when the whole world's available? The Earth
provides a better reference frame.
BTW Don, if the space station example is too "unEarthly" for you,
ever been on a merry-go-round?
I certainly wouldn't choose it as a reference for measuring in any way,
shape or form. It's enough to cause a severe headache. Only guys like
you and Unk Dopey would even think of such a dumb thing.
Don
Mark L. Fergerson
Your speaking of merry-go-rounds brings to mind teeter-totters: Of
course you won't be able to get great accuracy with it, but it will
show how different weights placed on it; at various locations, will
cause differences in its balance.
Don
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| User: "odin" |
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| Title: Re: If an equal arm scale balances |
02 Sep 2005 09:24:08 AM |
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Your speaking of merry-go-rounds brings to mind teeter-totters: Of
course you won't be able to get great accuracy with it, but it will
show how different weights placed on it; at various locations, will
cause differences in its balance.
Idiot.... Why would various locations cause differences in its balance?
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| User: "Don1" |
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| Title: Re: If an equal arm scale balances |
03 Sep 2005 07:38:41 PM |
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odin wrote:
Your speaking of merry-go-rounds brings to mind teeter-totters: Of
course you won't be able to get great accuracy with it, but it will
show how different weights placed on it; at various locations, will
cause differences in its balance.
Idiot.... Why would various locations cause differences in its balance?
The heaviest kid has to sit closest to the fulcrum.
Don
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| User: "Mark Fergerson" |
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| Title: Re: If an equal arm scale balances |
02 Sep 2005 10:12:18 AM |
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Don1 wrote:
Don1 wrote:
Mark Fergerson wrote:
Don1 wrote:
Uncle Al wrote:
mmeron@cars3.uchicago.edu wrote:
In article <43148275.A9180EBF@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
Don1 wrote:
If an equal arm balance scale balances in one place, it will balance
anywhere. That's why people think they measure mass; which is constant
anywhere.
No, idiot Dumb Donny *****. Try it in an anisotropic background.
It will take one hell of anisotropy to produce a nociable difference.
For once Don said something pretty correct, you don't have to activate
automatic responses.
Nope. Space station spinning for gravity will do it. Coriolis effect
will depend on balance orientation and placement adn there will be
severe gradients even for immodest sizes. Dumb Donny ***** is a
boring psychotic ididot.
Are you telling us that a balance is useless in a space station? Tell
us something we don't know Unk dopy.
He just said it _isn't_ useless, you just have to be careful with
orientation etc. In fact, if you do several weighings while
carefully noting how you orient and place the scale, you can
determine the axis and rate of rotation without looking out a window.
Who on Earth would bother, when the whole world's available? The Earth
provides a better reference frame.
Sigh. As usual, you manage to miss the point via your usual
technique of not addressing it.
BTW Don, if the space station example is too "unEarthly" for you,
ever been on a merry-go-round?
I certainly wouldn't choose it as a reference for measuring in any way,
shape or form. It's enough to cause a severe headache. Only guys like
you and Unk Dopey would even think of such a dumb thing.
Ever been on one and tried to hand something to somebody else
also on it?
Your speaking of merry-go-rounds brings to mind teeter-totters: Of
course you won't be able to get great accuracy with it, but it will
show how different weights placed on it; at various locations, will
cause differences in its balance.
But not the same weights.
Now go answer my other post in this thread about steelyard scales.
Mark L. Fergerson
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