Mathematics/Philosophies ( Q & A ): 101???

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jatslo

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What? Gravity is proportional to mass, so the gravity constant is the constant of proportionality, just like -(3.14) is to a circle, and constant speed is to time, etc. <br /><br />The speed of light is proportional 186,000 miles per second, so the speed of light is the constant of proportionality, right?<br /><br />I don't know what you want me to say.<br /><br />Whoops! Here's your question.<br /><br />Your question: <font color="yellow">Force is directly proportional to acceleration. What is the constant of proportionality?</font><br /><br />You would pick force; the laws of motion break down as we leave our star, and in quantum.<br /><br />The constant of proportionality is force.<br /><br />
 
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igorsboss

Guest
<font color="yellow">What? Gravity is proportional to mass, so the gravity constant is the constant of proportionality, just like -(3.14) is to a circle, and constant speed is to time, etc. </font><br /><br />Nope. Big G is a constant of proportionality, but not betweeen force and acceleration.<br /><br /><font color="yellow">The speed of light is proportional 186,000 miles per second, so the speed of light is the constant of proportionality, right? </font><br /><br />The speed of light is a constant, but not a constant of proportionality... at least, not in the context of our current conversation.<br /><br /><font color="yellow">I don't know what you want me to say. </font><br /><br />Good! If it were simple, you wouldn't learn anything...<br /><br /><font color="yellow">Whoops! Here's your question. <br /><br />Your question: Force is directly proportional to acceleration. What is the constant of proportionality? </font><br /><br />Yes. That is the correct question.<br /><br /><font color="yellow">You would pick force; the laws of motion break down as we leave our star, and in quantum. </font><br /><br />I mean in simple, ordinary conditions. No need to get fancy here. This is just ordinary Newtonian stuff.<br /><br /><font color="yellow">The constant of proportionality is force.</font><br /><br />Nope. Please stop guessing and start thinking. Perhaps go review what a constant of proportionality is. Here is the question, restated to be as simple as possible:<br /><br />What is the constant of proportionality between force and acceleration?
 
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jatslo

Guest
Take it to the next level: as in -[ E = m( c <sub>v</sub> + c <sub>i</sub> ) <sup>2</sup> ], whereas -( c <sub>i</sub> ) may or may not be zero. However, I am more inclined to change -(c) to -(v), as in -[ E = m( v <sub>v</sub> + v <sub>i</sub> ) <sup>2</sup> ] , because I may need a direct pulse to permeate from absolute zero relative to the Earth's Sun for example. However, to do this, I need to change mass to the radius of the mass as in -[ E = m <sub>r</sub> ( v <sub>v</sub> + v <sub>i</sub> ) <sup>2</sup> ] , whereas -( m <sub>r</sub> ) is a undecided radius. I am leaning towards the gyration of radius, so that I can account for gravitational anomalies.<br /><br />I also may need an imaginary friend like -(+), as in -[ E = m <sub>r</sub> ( v <sub>v</sub> + v <sub>i</sub> ) <sup>2</sup> ] <sub>a</sub> + [ E = m <sub>r</sub> ( v <sub>v</sub> + v <sub>i</sub> ) <sup>2</sup> ] = { 0, Infinity }
 
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igorsboss

Guest
<font color="yellow">Take it to the next level</font><br /><br />No. Not yet. There is plenty of time to take it to the next level later. You must earn that ability. This is your personal rite of passage.<br /><br />This question is exceptionally simple, and exceptionally deep. It is very important that you fully understand it, or your later work will continue to be flawed.<br /><br />What is the constant of proportionality between force and acceleration?
 
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thermionic

Guest
<br />There you go, Jatso! Don't let the 'boss distract you with his mass of questions. You should get back to your analysis of free energy so I can power my flying saucer with it. So far, I haven't had any luck with the cold fusion generator.
 
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jatslo

Guest
Quickly: There are multiple answers depending how I word it. It is Mass, Acceleration, or Force, so I say Mass. With respect to acceleration it could be displacement or speed, or any. Let's call the constant of proportionality +(k), acceleration -(a), and mass -(m). What is the force?
 
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thermionic

Guest
>>Where you headed?<br /><br />Definitely off to outer space, the final frontier! There's too much mass down in the planet core. Unless of course it is hollow after all! Cheers!
 
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jatslo

Guest
You need a high-pressure vacuum, not a helium balloon, but empty high pressure. Pure high pressure is empty and opaque, and empty is lighter than helium. I have not figured out the thermals though; however, you should be able to navigate on the EM waves or push against gravity.
 
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thermionic

Guest
<br />WoW, Thanks! That's what I've been doing wrong all this time. I thought I needed a low pressure vacuum, silly me! Well, gotta go! I'm off to work on my 'saucer and prove those doubters like SteveHW wrong! Excelsior!
 
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igorsboss

Guest
<font color="yellow">There are multiple answers depending how I word it.</font><br /><br />Which one of your multiple answers is the right one?<br /><br /><font color="yellow">It is Mass, Acceleration, or Force, so I say Mass.</font><br /><br />Are you sure? Can you prove it?<br /><br /><font color="yellow">With respect to acceleration it could be displacement or speed, or any.</font><br /><br />Can you explain this further? What do you mean by this?<br /><br /><font color="yellow">What is the force?</font><br /><br />Your ma.
 
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jatslo

Guest
I am having trouble getting the math to display properly in these threads.
 
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igorsboss

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<font color="yellow">I am having trouble getting the math to display properly in these threads.</font><br /><br />For exponentiation, you can use ^ instead of superscript, as in e=mc^2. I'll know what you mean.
 
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jatslo

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<h1>‡ = Infinity</h1> <--- See I can't even get infinity to work
 
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jatslo

Guest
+[ E = m <sub>r</sub> ( v <sub>v</sub> + v <sub>i</sub> ) <sup>2</sup> ]<br /><br />Did I break it, or does the above fragment still work?<br /><br />[1] E = Energy<br />[2] m <sub>r</sub> = Radius of Gyration (Much work to do here)<br />[3] v <sub>v</sub> = Velocity of Visible Light<br />[4] v <sub>i</sub> = Velocity of Invisible Light<br />[5] v <sub>L</sub> = 1.86 Miles, or whatever
 
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h9c2

Guest
I cannot see your image yet, but the formula:<br />+[ E = m r ( v v + v i )^2 ] can never be correct if m is a radius. Your E would then be expressed as m^3/s^2.<br /><br />Math gets easier if you keep track of your units. <br /><br />to add something:<br /><br />force = (kg*m)/s^2<br />acceleration = m/s^2<br /><br /><img src="/images/icons/wink.gif" />
 
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jatslo

Guest
Thanx for responding; my fragments must be frustrating for you. +( m <sub>r</sub> ) = is an unspecified radius matrix involving Doppler gyration from a radial perspective that I have not worked out yet. What I want to do is merge velocity and speed in a geometric shape to show speed and velocity simultaneously, whereas I might want to show two shapes relative to one another, etc. I want to make sure that it could work, provided I had a repeatable matrix variable(s).<br /><br />I need to show that my mass undergoes a radial transformation, if that makes sense.
 
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h9c2

Guest
From your previous posts, I'm assuming that you want to substitute either "speed" or "velocity" for acceleration in the post above? <br />Even then, I don't know what it means.<br /><br />are we still working on the premise that E is expressed as kg*m^2/s^2 ?
 
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igorsboss

Guest
<font color="yellow">+[ E = m r ( v v + v i ) 2 ] <br /><br />Did I break it, or does the above fragment still work? <br /><br />[1] E = Energy <br />[2] m r = Radius of Gyration (Much work to do here) <br />[3] v v = Velocity of Visible Light <br />[4] v i = Velocity of Invisible Light <br />[5] v L = 1.86 Miles, or whatever</font><br /><br />Well, let's see...<br /><br />Starting with E=mr(vv+vi)^2, I'll substitute according to vL=vv+vi, giving E=mr(vL)^2. According to your definitions, your statement reads as follows:<br /><br />Energy is the same as much work to do, times whatever squared.<br /><br />Hmmm.... Well, you've written a whole sentance, so kudos for that... And, it's concise, too. But, it could be... uh... clearer.<br /><br />Congratulations, your gobblygook has progressed to jabberwocky.
 
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igorsboss

Guest
<font color="yellow">+( m r ) = is an unspecified radius matrix involving Doppler gyration from a radial perspective</font><br /><br />You might try spherical coordinates.<br /><br /><font color="yellow">What I want to do is merge velocity and speed in a geometric shape to show speed and velocity simultaneously</font><br /><br />This gives me an idea... what if you used a directed line segment as this geometric shape, with the length of the segment proportional to the speed, and the angle in the direction of travel?
 
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jatslo

Guest
Speed (c = magnitude) implies sphere, and velocity (v = magnitude and direction) implies direction, and mass (m) is where I want a repeatable matrix that works in both quantum and relativity, because I know what gravity is. I cannot use the gravity constant of proportionality in quantum, unless I substitutive and satisfy requirements.<br /><br />I will go look at quantum theory and gravity again. <br />
 
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igorsboss

Guest
<font color="yellow">I need to show that my mass undergoes a radial transformation</font><br /><br />Lately, I've been trying to hide the radial transformation that my mass has been undergoing, but then again, I haven't tried Doppler gyrations yet.
 
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jatslo

Guest
Yeah, I need to show curvature, because I figured out how to curve the masses length too; however, I need alll of this in one repeatable package. I think my keyboard needs new batteries.
 
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jatslo

Guest
Oh, and I forgot to add that I can curve mass regardless of charge or no charge and regardless weight or no weight; because mass does not require weight or charge to be massive. <img src="/images/icons/wink.gif" />
 
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