| Topic: |
Science > Physics |
| User: |
"Spaceman" |
| Date: |
29 Mar 2006 11:55:56 AM |
| Object: |
Velocity addition fun.. LOL |
Let's see.
Velocity addition is done with the equation
w = (u + v)/(1 + uv/c2)
What happens when
u = 2 mph
v = 0 mph
It seems this equation is afraid of 0's.
ROFLOL
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| User: "Dirk Van de moortel" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 12:01:19 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message news:aJ6dndI_MrW4V7fZRVn-gg@comcast.com...
Let's see.
Velocity addition is done with the equation
w = (u + v)/(1 + uv/c2)
This is velocity composition.
Addition is w = u + v.
For small velocites the composition reduces to addition.
What happens when
u = 2 mph
v = 0 mph
It seems this equation is afraid of 0's.
Really?
What is frightening about
(2+0) / ( 1 + 0 ) = 2
?
ROFLOL
Sure.
Dirk Vdm
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| User: "Spaceman" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 12:13:19 PM |
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"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:PXzWf.344138$8n1.10443830@phobos.telenet-ops.be...
"Spaceman" <Realspace@comcast.not> wrote in message
news:aJ6dndI_MrW4V7fZRVn-gg@comcast.com...
Let's see.
Velocity addition is done with the equation
w = (u + v)/(1 + uv/c2)
This is velocity composition.
And it is wrong.
It is a bad equation and fails when 0 is used in it
at all.
Who cares how you change the name.
Really?
What is frightening about
(2+0) / ( 1 + 0 ) = 2
Nothing really.
Except the 0 devided by lightspeed squared *****..
(you can not mathematically devide a 0.)
you cheated and just called the answer a 0.
LOL
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| User: "Greg Neill" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 12:22:41 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message news:SaydnV0jr6OlU7fZRVn-uw@comcast.com...
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:PXzWf.344138$8n1.10443830@phobos.telenet-ops.be...
"Spaceman" <Realspace@comcast.not> wrote in message
news:aJ6dndI_MrW4V7fZRVn-gg@comcast.com...
Let's see.
Velocity addition is done with the equation
w = (u + v)/(1 + uv/c2)
This is velocity composition.
And it is wrong.
It is a bad equation and fails when 0 is used in it
at all.
Who cares how you change the name.
Really?
What is frightening about
(2+0) / ( 1 + 0 ) = 2
Nothing really.
Except the 0 devided by lightspeed squared *****..
(you can not mathematically devide a 0.)
you cheated and just called the answer a 0.
You're just displaying your nearly complete inability
to do any sort of arithmetic again.
w = (u + v)/(1 + u*v/c^2)
Letting u = 2, v = 0:
= (2 + 0)/(1 + 2*0/c^2)
= (2)/(1 + 0/c^2)
= 2/(1 + 0)
= 2
What's your problem?
.
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| User: "Spaceman" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 01:06:54 PM |
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"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:cgAWf.1795$u15.353885@news20.bellglobal.com...
"Spaceman" <Realspace@comcast.not> wrote in message
news:SaydnV0jr6OlU7fZRVn-uw@comcast.com...
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
wrote
in message news:PXzWf.344138$8n1.10443830@phobos.telenet-ops.be...
"Spaceman" <Realspace@comcast.not> wrote in message
news:aJ6dndI_MrW4V7fZRVn-gg@comcast.com...
Let's see.
Velocity addition is done with the equation
w = (u + v)/(1 + uv/c2)
This is velocity composition.
And it is wrong.
It is a bad equation and fails when 0 is used in it
at all.
Who cares how you change the name.
Really?
What is frightening about
(2+0) / ( 1 + 0 ) = 2
Nothing really.
Except the 0 devided by lightspeed squared *****..
(you can not mathematically devide a 0.)
you cheated and just called the answer a 0.
You're just displaying your nearly complete inability
to do any sort of arithmetic again.
w = (u + v)/(1 + u*v/c^2)
Letting u = 2, v = 0:
= (2 + 0)/(1 + 2*0/c^2)
= (2)/(1 + 0/c^2)
= 2/(1 + 0)
= 2
What's your problem?
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
Is that truly ok in your math world?
:)
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| User: "Greg Neill" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 02:03:55 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:_aadnSR4T8VaR7fZnZ2dnUVZ_uudnZ2d@comcast.com...
"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:cgAWf.1795$u15.353885@news20.bellglobal.com...
You're just displaying your nearly complete inability
to do any sort of arithmetic again.
w = (u + v)/(1 + u*v/c^2)
Letting u = 2, v = 0:
= (2 + 0)/(1 + 2*0/c^2)
= (2)/(1 + 0/c^2)
= 2/(1 + 0)
= 2
What's your problem?
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
Is that truly ok in your math world?
Specimen doesn't know his grade school math.
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| User: "Spaceman" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 04:15:26 PM |
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"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:6LBWf.1851$u15.356569@news20.bellglobal.com...
"Spaceman" <Realspace@comcast.not> wrote in message
news:_aadnSR4T8VaR7fZnZ2dnUVZ_uudnZ2d@comcast.com...
"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:cgAWf.1795$u15.353885@news20.bellglobal.com...
You're just displaying your nearly complete inability
to do any sort of arithmetic again.
w = (u + v)/(1 + u*v/c^2)
Letting u = 2, v = 0:
= (2 + 0)/(1 + 2*0/c^2)
= (2)/(1 + 0/c^2)
= 2/(1 + 0)
= 2
What's your problem?
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
Is that truly ok in your math world?
Specimen doesn't know his grade school math.
You are the moron stating u+v does not equal w and
you need that stupid transform that is simply proven
wrong with basic math.
LOL
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| User: "Greg Neill" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 04:26:35 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:5O2dneyvZeVlm7bZnZ2dnUVZ_s2dnZ2d@comcast.com...
"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:6LBWf.1851$u15.356569@news20.bellglobal.com...
"Spaceman" <Realspace@comcast.not> wrote in message
news:_aadnSR4T8VaR7fZnZ2dnUVZ_uudnZ2d@comcast.com...
"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:cgAWf.1795$u15.353885@news20.bellglobal.com...
You're just displaying your nearly complete inability
to do any sort of arithmetic again.
w = (u + v)/(1 + u*v/c^2)
Letting u = 2, v = 0:
= (2 + 0)/(1 + 2*0/c^2)
= (2)/(1 + 0/c^2)
= 2/(1 + 0)
= 2
What's your problem?
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
Is that truly ok in your math world?
Specimen doesn't know his grade school math.
You are the moron stating u+v does not equal w and
you need that stupid transform that is simply proven
wrong with basic math.
Specimen thinks that the universe must conform to
his idea of what basic math is, and that everything
must follow basic math (i.e. arithmetic addition and
subtraction). Silly Specimen.
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| User: "CWatters" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 04:43:23 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:_aadnSR4T8VaR7fZnZ2dnUVZ_uudnZ2d@comcast.com...
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
So if I have zero cake and I cut it into zero slices we all get one slice
each?
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| User: "Spaceman" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 04:58:31 PM |
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"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:f4EWf.344587$kW7.10472575@phobos.telenet-ops.be...
"Spaceman" <Realspace@comcast.not> wrote in message
news:_aadnSR4T8VaR7fZnZ2dnUVZ_uudnZ2d@comcast.com...
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
So if I have zero cake and I cut it into zero slices we all get one slice
each?
No,
We all get 1 zero slice.
:)
.
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| User: "T Wake" |
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| Title: Re: Velocity addition fun.. LOL |
30 Mar 2006 07:30:43 AM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:GYGdnT9_IrGMjLbZRVn-jQ@comcast.com...
"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:f4EWf.344587$kW7.10472575@phobos.telenet-ops.be...
"Spaceman" <Realspace@comcast.not> wrote in message
news:_aadnSR4T8VaR7fZnZ2dnUVZ_uudnZ2d@comcast.com...
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
So if I have zero cake and I cut it into zero slices we all get one slice
each?
No,
We all get 1 zero slice.
I am sure this makes sense to you. But as you are an idiot, with a tentative
grasp on reality, this is not relevant for communicating your ideas to
members of the human race.
Feel free to post your messages to
news://alt.spaceman.and.his.own.brand.of.science from now on.
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| User: "tj Frazir" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 06:39:43 PM |
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I want cake ...whos got cake ..tell the cook to bake cakes now that
evyone has cakes
wht kind of cake ??
***** I dont know what kind of cake ,
I cant imagine the cake the cook will bake anyway ..I bet its unside
down like spacy ..
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| User: "T Wake" |
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| Title: Re: Velocity addition fun.. LOL |
30 Mar 2006 07:31:02 AM |
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"tj Frazir" <GravityPhysics@webtv.net> wrote in message
news:8590-442B28CF-169@storefull-3216.bay.webtv.net...
I want cake ...whos got cake ..tell the cook to bake cakes now that
evyone has cakes
You and spaceman have a lot in common.
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| User: "T Wake" |
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| Title: Re: Velocity addition fun.. LOL |
29 Mar 2006 01:11:12 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:_aadnSR4T8VaR7fZnZ2dnUVZ_uudnZ2d@comcast.com...
"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:cgAWf.1795$u15.353885@news20.bellglobal.com...
"Spaceman" <Realspace@comcast.not> wrote in message
news:SaydnV0jr6OlU7fZRVn-uw@comcast.com...
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
wrote
in message news:PXzWf.344138$8n1.10443830@phobos.telenet-ops.be...
"Spaceman" <Realspace@comcast.not> wrote in message
news:aJ6dndI_MrW4V7fZRVn-gg@comcast.com...
Let's see.
Velocity addition is done with the equation
w = (u + v)/(1 + uv/c2)
This is velocity composition.
And it is wrong.
It is a bad equation and fails when 0 is used in it
at all.
Who cares how you change the name.
Really?
What is frightening about
(2+0) / ( 1 + 0 ) = 2
Nothing really.
Except the 0 devided by lightspeed squared *****..
(you can not mathematically devide a 0.)
you cheated and just called the answer a 0.
You're just displaying your nearly complete inability
to do any sort of arithmetic again.
w = (u + v)/(1 + u*v/c^2)
Letting u = 2, v = 0:
= (2 + 0)/(1 + 2*0/c^2)
= (2)/(1 + 0/c^2)
= 2/(1 + 0)
= 2
What's your problem?
The allowance of you deviding 0 by c squared
Do you think it is ok to devide a 0 at all?
If I devide 0 by 0 I can get
0/0 = 1
Is that truly ok in your math world?
Once more, with each post you continue to expand the subjects about which
you know nothing.
Just when people think you are really stupid, you make another post and
demonstrate you are even more stupid than previously thought.
You are a troll. No human being, not even a six year old, could exhibit the
amount of sheer ignorance you not only display but appear to revel in.
Amazing.
.
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| User: "Spaceman" |
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| Title: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
29 Mar 2006 01:22:07 PM |
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<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
.
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| User: "Greg Neill" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
29 Mar 2006 02:09:55 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:cLCdnRgH9LbEQ7fZnZ2dnUVZ_sednZ2d@comcast.com...
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Once again when Specimen finally clues in to the fact
that he's been wrong all along he claims he was 'just
playing'. What a maroon.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
Perhaps Specimen is refering to the apparent paradox
of his being able to to type semi-coherently, yet
being unable to add, subtract, multiply, or divide
above a grade one level.
.
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| User: "Spaceman" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
29 Mar 2006 04:14:25 PM |
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"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:LQBWf.1857$u15.356665@news20.bellglobal.com...
"Spaceman" <Realspace@comcast.not> wrote in message
news:cLCdnRgH9LbEQ7fZnZ2dnUVZ_sednZ2d@comcast.com...
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Once again when Specimen finally clues in to the fact
that he's been wrong all along he claims he was 'just
playing'. What a maroon.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
Perhaps Specimen is refering to the apparent paradox
of his being able to to type semi-coherently, yet
being unable to add, subtract, multiply, or divide
above a grade one level.
Poor Greg,
Can't figure it out either.
F- for Greg also.
LOL
.
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| User: "T Wake" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 07:28:22 AM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:c96dnfbxkfonm7bZRVn-vA@comcast.com...
Poor Greg,
Can't figure it out either.
F- for Greg also.
LOL
Laugh all you like. Every post you make emphasises what an idiot you are.
.
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| User: "T Wake" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
29 Mar 2006 02:29:19 PM |
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"Greg Neill" <gneillREM@OVE.THIS.netcom.ca> wrote in message
news:LQBWf.1857$u15.356665@news20.bellglobal.com...
"Spaceman" <Realspace@comcast.not> wrote in message
news:cLCdnRgH9LbEQ7fZnZ2dnUVZ_sednZ2d@comcast.com...
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Once again when Specimen finally clues in to the fact
that he's been wrong all along he claims he was 'just
playing'. What a maroon.
Common thread with the loon. When he is against the wall he pretends it was
" joke" which he was playing on everyone. (As in his defence on the
(-1)*(-1) issue.)
I seriously hope he is pretending to be this stupid on purpose, otherwise it
is a damning indictment of the education system...
.
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| User: "T Wake" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
29 Mar 2006 02:00:23 PM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:cLCdnRgH9LbEQ7fZnZ2dnUVZ_sednZ2d@comcast.com...
<redirected back to this group>
When you understand enough physics to talk about it, then you should post
here. Until then you are a village idiot and your posts belong elsewhere.
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
.
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| User: "PD" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 12:01:36 PM |
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Spaceman wrote:
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
PD
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| User: "Spaceman" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 12:07:23 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
Do you think they will agree?
.
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| User: "PD" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 12:26:53 PM |
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Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
I'm sorry, you'll have to be more concrete. Which two observers will
not agree on what?
Car (u) will measure car (w) to have a relative velocity of 50 mph,
according to you.
Car (y) will measure car (w) to have a relative velocity of 150 mph,
according to you.
Is that what you meant?
PD
Do you think they will agree?
.
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| User: "Spaceman" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 12:51:26 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143743213.185117.241890@g10g2000cwb.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
I'm sorry, you'll have to be more concrete. Which two observers will
not agree on what?
Any 2 observers will not figure the same relative velocity wrt
the other of the third observer if you use a transform.
Car (u) will measure car (w) to have a relative velocity of 50 mph,
according to you.
Car (y) will measure car (w) to have a relative velocity of 150 mph,
according to you.
Is that what you meant?
Why don't you use a transform for each case and see where
the actual paradox exists.
:)
.
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| User: "PD" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 12:57:15 PM |
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Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143743213.185117.241890@g10g2000cwb.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
I'm sorry, you'll have to be more concrete. Which two observers will
not agree on what?
Any 2 observers will not figure the same relative velocity wrt
the other of the third observer if you use a transform.
Car (u) will measure car (w) to have a relative velocity of 50 mph,
according to you.
Car (y) will measure car (w) to have a relative velocity of 150 mph,
according to you.
Is that what you meant?
Why don't you use a transform for each case and see where
the actual paradox exists.
I'm sorry, I still don't understand. The two observers already don't
agree, using your methods. What paradox?
Please be specific, state what you think *should* be the same, what is
*not* the same, and where the paradox lies.
PD
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| User: "Spaceman" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 01:04:59 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143745035.541925.98730@v46g2000cwv.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143743213.185117.241890@g10g2000cwb.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
<redirected back to this group>
(Basic proof T Wake is so afraid of my posts that he has
to redirect them)
T Wake" <Usenet.es7AT@gishpuppy.com> wrote in message
news:HbOdnQ8TZ4RKRrfZRVnyig@pipex.net...
Amazing.
Ok,
here comes the real fun,
I was just playing with that 0 thing.
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
I'm sorry, you'll have to be more concrete. Which two observers will
not agree on what?
Any 2 observers will not figure the same relative velocity wrt
the other of the third observer if you use a transform.
Car (u) will measure car (w) to have a relative velocity of 50 mph,
according to you.
Car (y) will measure car (w) to have a relative velocity of 150 mph,
according to you.
Is that what you meant?
Why don't you use a transform for each case and see where
the actual paradox exists.
I'm sorry, I still don't understand. The two observers already don't
agree, using your methods. What paradox?
Using basic math.
Observer w agrees that observer v will measure the same relative
velocity of y (to v) that observer y would measure of v (to y).
Using a transform, w will not do such.
Can you show that he would by using a transform for all of them?
.
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| User: "Euclid Uranium" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
08 Apr 2006 12:22:59 PM |
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"Spaceman" <Realspace@comcast.not> wrote:
Using basic math.
Observer w agrees that observer v will measure the same relative
velocity of y (to v) that observer y would measure of v (to y).
Using a transform, w will not do such.
Can you show that he would by using a transform for all of them?
One meter and time, to the current position that studying things
that I find pizzas that the showing could use it not affected
their primary source: observer At the future, time; in
newtonian; spacetime really so Set which turns in and The story;
but with him; doesn't give him to derail peace.
What we have referred: to order.
.
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| User: "Edwards" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 07:16:08 PM |
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On 2006-03-30, Spaceman <Realspace@comcast.not> wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143745035.541925.98730@v46g2000cwv.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143743213.185117.241890@g10g2000cwb.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
I'm sorry, you'll have to be more concrete. Which two observers will
not agree on what?
Any 2 observers will not figure the same relative velocity wrt
the other of the third observer if you use a transform.
Car (u) will measure car (w) to have a relative velocity of 50 mph,
according to you.
Car (y) will measure car (w) to have a relative velocity of 150 mph,
according to you.
Is that what you meant?
Why don't you use a transform for each case and see where
the actual paradox exists.
I'm sorry, I still don't understand. The two observers already don't
agree, using your methods. What paradox?
Using basic math.
Observer w agrees that observer v will measure the same relative
velocity of y (to v) that observer y would measure of v (to y).
Using a transform, w will not do such.
Depends on the transform, maybe, but for the SR velocity "transform",
w will indeed "do such".
Can you show that he would by using a transform for all of them?
Okay, your setup was:
Car (u) is moving at 100 mph northbound
Car (w) is moving at 50 mph northbound
Car (y) is moving at 100 mph southbound
in what I'll call the "road frame". The frame with "Car (u)" at rest
will be called the "u frame", and similarly for frames w and y.
Velocity of u in the w frame (in the north-south direction, with north
being +), c is in units of mph:
u_w = ( 100 - 50) / (1 - 5000/c^2)
y_w = (-100 - 50) / (1 + 5000/c^2)
w (i.e. an observer at rest in frame w) can immediately determine that
the "relative velocity" (ie. closing velocity) between u and y is u_w
- y_w (or y_w - u_w, depeding on which orientation you want). Of
course, w is also aware that even under Galilean relativity, this
"relative velocity" (or closing velocity or whatever you want to call
it) has no particular physical significance.
Now w calculates the velocity of (car) y from the point of view of an
observer at rest in u:
y_{u_w} = (y_w - u_w) / [1 - (y_w * u_w) / c^2]
And similarly, the velocity of u from the point of view of y is
calculated to be:
u_{y_w} = (u_w - y_w) / [1 - (u_w * y_w) / c^2]
= -y_{u_w}
Tada, they are indeed equal in magnitude. Of course, they are not
_exactly_ equal to what w calls the closing velocity, but since
closing velocity has no particular physical significance, nobody
(including w) cares. (Of course, since they differ in this case from
closing velocity by less than 2 x 10^(-7), nobody _really_ cares. How
can you measure the speed of a car going about 100mph to within about
6 microns per second?)
--
Darrin
.
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| User: "Spaceman" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
30 Mar 2006 07:26:18 PM |
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"Edwards" <edwards@nouce.trurl.bsd.uchicago.edu> wrote in message
news:slrne2p0ni.tps.edwards@trurl.bsd.uchicago.edu...
On 2006-03-30, Spaceman <Realspace@comcast.not> wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143745035.541925.98730@v46g2000cwv.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143743213.185117.241890@g10g2000cwb.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
I'm sorry, you'll have to be more concrete. Which two observers will
not agree on what?
Any 2 observers will not figure the same relative velocity wrt
the other of the third observer if you use a transform.
Car (u) will measure car (w) to have a relative velocity of 50 mph,
according to you.
Car (y) will measure car (w) to have a relative velocity of 150 mph,
according to you.
Is that what you meant?
Why don't you use a transform for each case and see where
the actual paradox exists.
I'm sorry, I still don't understand. The two observers already don't
agree, using your methods. What paradox?
Using basic math.
Observer w agrees that observer v will measure the same relative
velocity of y (to v) that observer y would measure of v (to y).
Using a transform, w will not do such.
Depends on the transform, maybe, but for the SR velocity "transform",
w will indeed "do such".
Can you show that he would by using a transform for all of them?
Okay, your setup was:
Car (u) is moving at 100 mph northbound
Car (w) is moving at 50 mph northbound
Car (y) is moving at 100 mph southbound
in what I'll call the "road frame". The frame with "Car (u)" at rest
will be called the "u frame", and similarly for frames w and y.
Velocity of u in the w frame (in the north-south direction, with north
being +), c is in units of mph:
u_w = ( 100 - 50) / (1 - 5000/c^2)
y_w = (-100 - 50) / (1 + 5000/c^2)
w (i.e. an observer at rest in frame w) can immediately determine that
the "relative velocity" (ie. closing velocity) between u and y is u_w
- y_w (or y_w - u_w, depeding on which orientation you want). Of
course, w is also aware that even under Galilean relativity, this
"relative velocity" (or closing velocity or whatever you want to call
it) has no particular physical significance.
Now w calculates the velocity of (car) y from the point of view of an
observer at rest in u:
y_{u_w} = (y_w - u_w) / [1 - (y_w * u_w) / c^2]
And similarly, the velocity of u from the point of view of y is
calculated to be:
u_{y_w} = (u_w - y_w) / [1 - (u_w * y_w) / c^2]
= -y_{u_w}
Tada, they are indeed equal in magnitude. Of course, they are not
_exactly_ equal to what w calls the closing velocity, but since
closing velocity has no particular physical significance, nobody
(including w) cares. (Of course, since they differ in this case from
closing velocity by less than 2 x 10^(-7), nobody _really_ cares.
Science would care flies completley above your head huh?
The use of the transform conflicts with itself mathematically.
sheesh!
The paradox remains or like real life, the transform is *****.
.
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| User: "T Wake" |
|
| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
31 Mar 2006 06:17:04 AM |
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"Spaceman" <Realspace@comcast.not> wrote in message
news:89OdncoAvdasGLHZRVn-uA@comcast.com...
"Edwards" <edwards@nouce.trurl.bsd.uchicago.edu> wrote in message
news:slrne2p0ni.tps.edwards@trurl.bsd.uchicago.edu...
On 2006-03-30, Spaceman <Realspace@comcast.not> wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143745035.541925.98730@v46g2000cwv.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143743213.185117.241890@g10g2000cwb.googlegroups.com...
Spaceman wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1143741696.092986.143500@z34g2000cwc.googlegroups.com...
Spaceman wrote:
Lets add another observer to different conditions now.
Three cars are on the highway.
2 cars are heading north, 1 car is heading south
Car (u) is moving at 100 mph northbound.
Car (w) is moving at 50 mph northbound.
Car (y) is moving at 100 mph southbound
Can you show the paradox?
What paradox?
The different answers for different observers.
2 observers will not agree on the third observers
relative velocity if you are using a transform.
I'm sorry, you'll have to be more concrete. Which two observers will
not agree on what?
Any 2 observers will not figure the same relative velocity wrt
the other of the third observer if you use a transform.
Car (u) will measure car (w) to have a relative velocity of 50 mph,
according to you.
Car (y) will measure car (w) to have a relative velocity of 150 mph,
according to you.
Is that what you meant?
Why don't you use a transform for each case and see where
the actual paradox exists.
I'm sorry, I still don't understand. The two observers already don't
agree, using your methods. What paradox?
Using basic math.
Observer w agrees that observer v will measure the same relative
velocity of y (to v) that observer y would measure of v (to y).
Using a transform, w will not do such.
Depends on the transform, maybe, but for the SR velocity "transform",
w will indeed "do such".
Can you show that he would by using a transform for all of them?
Okay, your setup was:
Car (u) is moving at 100 mph northbound
Car (w) is moving at 50 mph northbound
Car (y) is moving at 100 mph southbound
in what I'll call the "road frame". The frame with "Car (u)" at rest
will be called the "u frame", and similarly for frames w and y.
Velocity of u in the w frame (in the north-south direction, with north
being +), c is in units of mph:
u_w = ( 100 - 50) / (1 - 5000/c^2)
y_w = (-100 - 50) / (1 + 5000/c^2)
w (i.e. an observer at rest in frame w) can immediately determine that
the "relative velocity" (ie. closing velocity) between u and y is u_w
- y_w (or y_w - u_w, depeding on which orientation you want). Of
course, w is also aware that even under Galilean relativity, this
"relative velocity" (or closing velocity or whatever you want to call
it) has no particular physical significance.
Now w calculates the velocity of (car) y from the point of view of an
observer at rest in u:
y_{u_w} = (y_w - u_w) / [1 - (y_w * u_w) / c^2]
And similarly, the velocity of u from the point of view of y is
calculated to be:
u_{y_w} = (u_w - y_w) / [1 - (u_w * y_w) / c^2]
= -y_{u_w}
Tada, they are indeed equal in magnitude. Of course, they are not
_exactly_ equal to what w calls the closing velocity, but since
closing velocity has no particular physical significance, nobody
(including w) cares. (Of course, since they differ in this case from
closing velocity by less than 2 x 10^(-7), nobody _really_ cares.
Science would care flies completley above your head huh?
The use of the transform conflicts with itself mathematically.
sheesh!
The paradox remains or like real life, the transform is *****.
You continue to demonstrate a total lack of understanding with every post.
You claim "science cares" yet you have no concept of the scientific method
or the process of "science."
You think being able to make testable predictions is a dark art which should
be left to soothsayers.
You truly are a relic of the stone age. You haven't even quite made it to
the Neolithic period yet. You are firmly entrenched in the Mesolithic
period.
One day, you will realise about experimental accuracy and why it matters.
You will realise how some things are generalised versions of "proper"
theories. You will then learn that by developing the theory you become more
accurate and that some of the things which seem to work in your bathroom are
not the same when used on larger or smaller scales.
When this happens you will have entered the level of scientific
understanding that was prevalent around 500BC. You will still have a long
way to go to drag your ***** into the 21st century.
.
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| User: "Edwards" |
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| Title: Re: The SR highway traffic paradox..was: Velocity addition fun.. LOL |
31 Mar 2006 07:55:51 PM |
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On 2006-03-31, Spaceman <Realspace@comcast.not> wrote:
"Edwards" <edwards@nouce.trurl.bsd.uchicago.edu> wrote in message
news:slrne2p0ni.tps.edwards@trurl.bsd.uchicago.edu...
Tada, they are indeed equal in magnitude.
[this was _important_, yet ignored! how surprising!]
Of course, they are not _exactly_ equal to what w calls the closing
velocity, but since closing velocity has no particular physical
significance, nobody (including w) cares. (Of course, since they
differ in this case from closing velocity by less than 2 x 10^(-7),
nobody _really_ cares.
[restore my final question which was not only ignored but _snipped_,
nice try though!]
How can you measure the speed of a car going about 100mph to within
about 6 microns per second?)
Science would care flies completley above your head huh?
Please explain in detail the scientific importance of closing
velocity.
The use of the transform conflicts with itself mathematically.
sheesh!
The paradox remains or like real life, the transform is *****.
No, answer the _question_: How can you measure the speed of a car
going about 100mph to within about 6 microns per second? I'm not
asking for another "actual reality" "science of measurement" "rubber
ruler" tirade, I'm asking you how to actually _do_ it. I'm not
_telling_ you to use rubber rulers, use any damn rulers you like, just
please oh please tell me _how_; lasers? string? Give me _something_ to
go with here.
--
Darrin
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