Re: Have you ever wondered.....
From: Kees Roos (croos_at_xs4all.nl)
Date: 11/19/04
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Date: Fri, 19 Nov 2004 07:57:54 +0100
"AllYou!" <idaman@conversent.net> schreef in bericht
news:BNSdnT0olI9CbwHcRVn-oQ@conversent.net...
>
> "Kees Roos" <croos@xs4all.nl> wrote in message
> news:419ced92$0$568$e4fe514c@news.xs4all.nl...
[snip]
>> >> > "Kees Roos" <croos@xs4all.nl> wrote in message
>> >> > news:419bab35$0$78738$e4fe514c@news.xs4all.nl...
>> >> [snip]
>> >> >> You talk about the same value of your 'motion' for every point on
>> >> >> the
>> >> >> earth's surface.
>> >> >> 'My' angular speed has that property also. Could you indicate how
>> >> >> the
>> >> >> two
>> >> >> relate
>> >> >> to each other, i.o.w. how you calculate your value of 'motion' from
>> >> >> a
>> >> >> given
>> >> >> value of
>> >> >> angular speed?
>> >
>> [snip]
>> I think you missed my point instead.
>> You say the case of the sphere is one example of how to quantify motion.
>> My
>> question is how
>> this one particular quantitifying of motion you mention relates to the
>> angular speed of the earth's
>> rotation. Never mind any other possible way to quantify motion, my
>> question
>> relates to this one
>> particular way.
>> Could you have a go as yet?
>
> Why you would want to so is beyond me unless you're trolling.
>
I'm ot trolling. I'm curious.
> My units of motion use the
> same labels and are equivalent to what you use for time now. One day is
> the motion of one
> full revolution of the Earth, and the units for fractions of that are also
> expressed
> similarly. Therefore, you'd express the angular speed of the Earth in the
> same way that
> you do now.
>
That's still no answer to my question of a relation between the two.
So, let me propose one, so that you can approve or disprove.
You say the earth has motion one day per revolution: m = d / r = 1
I say one day is the time interval of one revolution of the earth,
so the angular speed of the earth is one revolution per day: O = r / d = 1
Consequently your motion relates to O as : O / m = ( d / r )/( r / d ) = d^2
/ r ^2 = 1
So, the two are equivalent.
Agreed?
>> > Secondly, as I've said over and over again but
>> > which you fail to comprehend, it's my hypothysis that motion is a
>> > fundamental property of
>> > nature and should not be calculated by measured directly. Here was my
>> > exact response to
>> > your question:
>> >
>> > "With that said, motion is a fundamental phenomenon just like distance.
>> > So, I would
>> > directly measure motion using the same techniques we use to measure
>> > distance or any other
>> > fundamental phenomenon, and then I'd express quantities of speed in
>> > terms
>> > of the ratio of
>> > distance to motion."
>> >
>> > Now answer my question........what is there in this answer you don't
>> > get?
>> >
>> Never mind what I do or don't get. It is not an answer to my question.
>> My question is how two things relate. Nowhere in your answer is there any
>> mention of such a relation.
>
> What you do and don't get is germane to how extensive and detailed my
> responses must be.
> I just gave the answer to your question above,
>
My question was about the relation between two things. You did not answer
that
question at all.
> but the answer was so obvious from what
> I've explained to you already that I'm sttruggling to figure out where the
> gap in your
> understanding lies. So now you answer my question.
>
Your question is contained in your answer to my original question, but does
not relate to
that original question, so it is irrelevant.
[snip]
>> > So lets' say that when a car passes a certain position in space, I
>> > start
>> > my motion
>> > measuring device. When the car passes a second point in space, I stop
>> > my
>> > motion measuring
>> > device. I then look at this device and read the total value of that
>> > motion and express
>> > the result (e.g., 50 hours).
>> >
>> So the car has motion 50 hours?
>
> The car has a motion of 50 hours.
>
However, during your measurement the car passed 49 other points along the
road
as well, so it also has motion 1 to 49, and of course all real number in
between.
In fact it has all possible real numbers of motion simultaneously. Is that
correct?
>> What when you choose end points at twice the distance? Would that give
>> the
>> result that
>> the same car has motion 100 hours instead of motion 50 hours?
>> Mind you, I am not trying to ridicule; I try to understand what you say.
>
> If the car maintains the same speed over twice a distance, it would have a
> motion of 100
> hours.
[snip]
> And now?
>
Let's carry on.
-- Regards, Kees Roos
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