imatfaal Posted July 13, 2014 Share Posted July 13, 2014 ! Moderator Note Everyone, No more comments directed towards the person please - either talk about the argument or don't comment. Link to comment Share on other sites More sharing options...
Ophiolite Posted July 13, 2014 Share Posted July 13, 2014 You are 100% correct, but lets look at it this way: If I am 100% correct, which I lay no claim to, but you do, then it would be completely pointless to look at it any other way. Yet, you proceed to take up a page of seemingly trivial discussion that takes us nowhere. It is true that many breakthroughs in science come through the juxtaposition of observations and theories from diverse fields, or arise from the asking of apparently bizarre questions. Yet in all the instances of which I am aware a link, no matter how tentative, has been perceived by the investigator. In contrast your technique appears to be to throw ideas together randomly, or to ask questions whose semantic content is very low. Will you not attempt to bring more focus to your thinking in future? Link to comment Share on other sites More sharing options...
Bignose Posted July 13, 2014 Share Posted July 13, 2014 I can't control round off errors using online calculations sorry. As John pointed out, it's not roundoff errors. It's in error by many orders of magnitude. This is why I pointed it out. If your calculations are in error, there is zero point in reading about any kind of conclusions about them. Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted July 13, 2014 Author Share Posted July 13, 2014 As John pointed out, it's not roundoff errors. It's in error by many orders of magnitude. This is why I pointed it out. If your calculations are in error, there is zero point in reading about any kind of conclusions about them. Is it because it is that "Complex" to figure out?? Why should someone just give up?? Link to comment Share on other sites More sharing options...
Bignose Posted July 13, 2014 Share Posted July 13, 2014 Is it because it is that "Complex" to figure out?? Why should someone just give up?? it isn't that complex. I spotted the error very quickly. Once I saw the error, there is no point in continuing because it is all based on an erroneous premise. If you are building a house, and the foundation sinks or cracks, you don't just continue putting the 1st floor and the 2nd floor and roof on it. You fix the foundation. Your error is a broken foundation. Get it fixed and then see what conclusions can be drawn. Link to comment Share on other sites More sharing options...
Sensei Posted July 13, 2014 Share Posted July 13, 2014 (edited) Is it because it is that "Complex" to figure out?? Why should someone just give up?? It's not complex, but you're not taking care of anything, writing 6.626e-33 (post #40) (10x higher than should you use) but you should write 6.62607e-34 (Planck const)... etc. etc. etc. These values are not some random values made up by scientists, but we can prove these values to be true. Simply make thread "how to calculate Planck const at home" and I will show you experiment confirming it's value that we know... Edited July 13, 2014 by Sensei Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted July 13, 2014 Author Share Posted July 13, 2014 (edited) If I am 100% correct, which I lay no claim to, but you do, then it would be completely pointless to look at it any other way. Yet, you proceed to take up a page of seemingly trivial discussion that takes us nowhere. It is true that many breakthroughs in science come through the juxtaposition of observations and theories from diverse fields, or arise from the asking of apparently bizarre questions. Yet in all the instances of which I am aware a link, no matter how tentative, has been perceived by the investigator. In contrast your technique appears to be to throw ideas together randomly, or to ask questions whose semantic content is very low. Will you not attempt to bring more focus to your thinking in future? Sorry, was your question based on constitution?? I am not familiar with some terminologies many people here are accustomed too. I am a high-school drop out, never been in a laboratory, and never will be in a college institution either.. WOW! now this is what you call negative thinking right? Unless of coarse I live in a country where the citizens accept " people for whom they are." Its called "racial barriers" I am latino by the way,,, So, this is confirmation that " again " That I am not in your categories and level of education.. But is this a bad thing??? Still no'one has answer my OP question: Here it is again if you may.. If pi ratio " was" squared and = 9.8 m/s/s how would this change the whole of science? Or have they? I will re check... It's not complex, but you're not taking care of anything, writing 6.626e-33 (post #40) (10x higher than should you use) but you should write 6.62607e-34 (Planck const)... etc. etc. etc. These values are not some random values made up by scientists, but we can prove these values to be true. Simply make thread "how to calculate Planck const at home" and I will show you experiment confirming it's value that we know... Ok, I will Edited July 13, 2014 by Iwonderaboutthings Link to comment Share on other sites More sharing options...
Bignose Posted July 13, 2014 Share Posted July 13, 2014 (edited) If pi ratio " was" squared and = 9.8 m/s/s how would this change the whole of science?[/size] This has been demonstrated to be approximately true (ignoring the units, which isn't usually a good idea). It is accurate to within about 1/2%. And, the historical reason why was also shown (e.g. trying to define a second based on a pendulum -- and the equation describing the motion of a pendulum has pi in it.) The whole of science has not changed. Edited July 13, 2014 by Bignose Link to comment Share on other sites More sharing options...
Sensei Posted July 13, 2014 Share Posted July 13, 2014 (edited) If pi ratio " was" squared and = 9.8 m/s/s how would this change the whole of science? PI^2 (PI multiplied by PI) is 9.8696044010893586188344909998762 which in your vision is close to 9.8.. Accident. On Moon or other planet it won't be close each other. Edited July 13, 2014 by Sensei Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted July 13, 2014 Author Share Posted July 13, 2014 (edited) PI^2 (PI multiplied by PI) is 9.8696044010893586188344909998762 which in your vision is close to 9.8.. Accident. On Moon or other planet it won't be close each other. I don't understand " What"? what wont be close to each other??? pi and the other pi like pi (1) and pi (2) ?? Out of phase like this? Edited July 13, 2014 by Iwonderaboutthings Link to comment Share on other sites More sharing options...
Sensei Posted July 13, 2014 Share Posted July 13, 2014 PI^2 = 9.8696044010893586188344909998762 g = Earth's acceleration at sea level is 9.81 m/s These two things are not equal. Link to comment Share on other sites More sharing options...
Strange Posted July 13, 2014 Share Posted July 13, 2014 what wont be close to each other??? On the moon, for example, the value of g would be different (i.e. it would not be 9.8) therefore it would not by pi^2. Link to comment Share on other sites More sharing options...
Sensei Posted July 13, 2014 Share Posted July 13, 2014 9.8696044010893586188344909998762 / 9.81 = 1.0060758818643586767415383282239 So in percentage it's difference by 0.6% higher.. Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted July 13, 2014 Author Share Posted July 13, 2014 (edited) 1.98650352574581e+31* pi^12 is not 6.24441297605896e+43 pi^12 is 924269.18 (or so- there may be rounding errors). 1.98650352574581e+31* 924269.18 is about 1.836 e 37 So you are wrong by about three million fold. Google's calculator says that "3.14159265*3.14159265*3.14159265*3.14159265*3.14159265*3.14159265*3.14159265*3.14159265*3.14159265*3.14159265*3.14159265*3.14159265" is 924269.16885 and that pi^12 is 924269.181523 The difference is because pi isn't exactly 3.14159265 Does that really matter when dealing with a simple circumference though? I've seen pi used like this: 3.14*8=25.12 " some" circumference just that number as 3.14, the rest really doesn't matter I guess. Things like this image also don't make sense, pi ratio seems to have vertual ways of use. it must square at some point, when I say point I mean the center of what ever the circumference belongs too.. To add, imaginary units, have been said to follow a pi ratio circular path: one is real the other is imaginary making the complete calculation 1/2 true. Then which one is " real "?????? Sorry, you can't have one without the other... Can you??? On the moon, for example, the value of g would be different (i.e. it would not be 9.8) therefore it would not by pi^2. WOOOOOOOOOOOOOOOOE You just threw me back like " UM " millions of years.. I thought that gravity on the moon was UM, had no such relation to the acceleration g on earth. Are you saying that 9.8 meters per second g on earth is the same on other planets too??? Or " references other planets? Edited July 13, 2014 by Iwonderaboutthings Link to comment Share on other sites More sharing options...
Sensei Posted July 13, 2014 Share Posted July 13, 2014 Are you saying that 9.8 meters per second g on earth is the same on other planets too??? Or " references other planets? He said that acceleration is different on different planets/moons.... Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted July 13, 2014 Author Share Posted July 13, 2014 (edited) He said that acceleration is different on different planets/moons.... How do we find their acceleration, as we do on earth for g? is it only the moons????????? Our moon goes in reverse... Other moons of other planets go in reverse to? Edited July 13, 2014 by Iwonderaboutthings Link to comment Share on other sites More sharing options...
Strange Posted July 13, 2014 Share Posted July 13, 2014 Does that really matter when dealing with a simple circumference though? I've seen pi used like this: 3.14*8=25.12 " some" circumference just that number as 3.14, the rest really doesn't matter I guess. It just depends how accurate you want (or need) to be. one is real the other is imaginary making the complete calculation 1/2 true. The words "real" and "imaginary" are just names; they don't mean anything; they certainly don't mean that part of the number is more real than the other. Are you saying that 9.8 meters per second g on earth is the same on other planets too??? No, I am saying it is NOT the same. How do we find their acceleration, as we do on earth for g? We calculate it from their mass and radius (using Newton's equation for gravity). Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted July 13, 2014 Author Share Posted July 13, 2014 (edited) It just depends how accurate you want (or need) to be. The words "real" and "imaginary" are just names; they don't mean anything; they certainly don't mean that part of the number is more real than the other. No, I am saying it is NOT the same. We calculate it from their mass and radius (using Newton's equation for gravity). G [m1]*[m2]/r^2 Got it... So, this equation gives you the " acceleration of " g " But on another planet then right? Just an example here.. earth = g 9.8 m/s/s venus = g 7.5 m/s/s mars = g 12.7 m/s/s Edited July 13, 2014 by Iwonderaboutthings Link to comment Share on other sites More sharing options...
Endy0816 Posted July 13, 2014 Share Posted July 13, 2014 It provides the force. You can get back to acceleration via F=ma. Link to comment Share on other sites More sharing options...
Mordred Posted July 13, 2014 Share Posted July 13, 2014 (edited) just a side note here is the procedure to calculate a planets mass lets say you just spotted a new planet and would like its basic characteristics http://www4.wittenberg.edu/sgmoa/supplemental/FindExtrasolarPlanetMass.pdf http://www.sfu.ca/colloquium/PDC_Top/astrobiology/discovering-exoplanets/calculating-exoplanet-properties.html Edited July 13, 2014 by Mordred 1 Link to comment Share on other sites More sharing options...
Iwonderaboutthings Posted July 13, 2014 Author Share Posted July 13, 2014 (edited) It provides the force. You can get back to acceleration via F=ma. When you say " provides the force" I am concluding that this force, is no difference that the one here on earth as 9.8 m/s/s for other planets.. That they have an acceleration much like g here on earth, but that their values vary,, Again: The force of acceleration is the same as planet earth's as 9.8 m /s /s but only has different values on other worlds. Is it wrong to question: Who or what calculated 9.8 m /s /s on earth in the first place.. In other words if you can calculate the force of acceleration on other worlds, and have different values. Where did 9.8 m /s/s come from?? just a side note here is the procedure to calculate a planets mass lets say you just spotted a new planet and would like its basic characteristics http://www4.wittenberg.edu/sgmoa/supplemental/FindExtrasolarPlanetMass.pdf http://www.sfu.ca/colloquium/PDC_Top/astrobiology/discovering-exoplanets/calculating-exoplanet-properties.html Something to that nature thanks.. I think I have just realized, that the acceleration of g here on earth, is no difference than on other planets, accept that the "values" vary...I hope I am interpreting this correctly... So that F=ma, would be. F= "mass of the planet" * [ the acceleration of the planet] My example for earth then: F=ma F= "earth's mass" * [ 9.8 m/s/s]??????? Was it Issac Newton that found 9.8 m/s/s ??? And we just go by that??? You know online information even in books are not so clear.. Edited July 13, 2014 by Iwonderaboutthings Link to comment Share on other sites More sharing options...
Strange Posted July 13, 2014 Share Posted July 13, 2014 When you say " provides the force" He means that the equation you quoted gives you force, not acceleration: F = G * m1 * m2 / r^2 The acceleration is given by dividing the force by the mass: g = F / m1 = G m2 / r^2 where m2 is the mass of the planet and r is it's radius. That they have an acceleration much like g here on earth, but that their values vary,, In other words if you can calculate the force of acceleration on other worlds, and have different values. Correct. Where did 9.8 m /s/s come from? Initially, it was measured (using techniques similar to Sensei's example in post 34). But now we know how to calculate it as well. Link to comment Share on other sites More sharing options...
Fuzzwood Posted July 13, 2014 Share Posted July 13, 2014 The quality of the force might be the same, but not its magnitude. Hence you find different values for the acceleration constant on different orbital bodies. Link to comment Share on other sites More sharing options...
Strange Posted July 13, 2014 Share Posted July 13, 2014 The acceleration is given by dividing the force by the mass: g = F / m1 = G m2 / r^2 where m2 is the mass of the planet and r is it's radius. By the way, this explains why a canonball falls as fast as a pea (ignoring air resistance): the force on the canonball is greater, but it takes exactly that much more force to accelerate it the same as the pea. 1 Link to comment Share on other sites More sharing options...
Endy0816 Posted July 13, 2014 Share Posted July 13, 2014 (edited) What Strange said F = G [m1]*[m2]/r^2 assume [m1] is your object so via F=[m1]a we can state [m1]a = G [m1]*[m2]/r^2 and after dividing by [m1] on both sides... a = G*[m2]/r^2 'g' causes way too much confusion for a lowly acceleration. Edited July 13, 2014 by Endy0816 1 Link to comment Share on other sites More sharing options...
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