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How to use formula x-y =m c squared

  1. Mar 2, 2009 #1
    Not sure if I wrote it right ,but I am intrested in knowing how to use this formula to find an angle ,relative to a pool table.
     
  2. jcsd
  3. Mar 2, 2009 #2

    HallsofIvy

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    I have no idea where you got that so it is impossible to say how it should be used. What do "x", "y", "m" and "c" represent? I presume that "c" is NOT the speed of light since nothing on a pool table will be moving at near light speed!
     
  4. Mar 2, 2009 #3
    the formula to find any angle ,I am not 100 percent up to speed on this kind of thing. This is why I joined the forum.Pool is a game of angles,geomotry etc.I have another post on this site of simular content based on a statment made about pool ,it atated that there are only 3 anle for any given shot on the table,I believe this to be true .The guy who wrote the article ,his statement was made from a gemetrical point of view.The three angles are 45,30,15.I have worked figuing out how this is possible for 2 months.I use this thepry in my pool game but have not masrered it yet nor can i fully explain it .So i came here to look for answres.The formulawas originated by Einstine I belive.I will try to word it another way.XoverY= MCsquared.I heard that this is the formula used to determine an angle from given points or a given point.So if thats not the formula to use ,what would be the correct one and how would you use it?
    Thanks for responding hope you can help me out.
     
  5. Mar 2, 2009 #4
    pardon me for not previewing my post before submitting,SPELLING lol sorry
     
  6. Mar 2, 2009 #5
    one other way I presume this ,form a triagle from 2 given points,to one side point,the side is the side of a rectangle. a second point given,form a triangle from the given points off of the first triangle, meaning,draw a line from given point 2 to the side point of the first triangle formed,then draw a line from given point 2 to aprox.1/2 the distance of the side of the first triangle that was formed,the triangle now formed is = to the opposite side of the top triangle thus meaning the formula was x/y, point x divided by Y = mc squared. sound right? all points are squared
     
  7. Mar 2, 2009 #6
    Would it be Pythagoras' Theorem you want mate?

    Which is a2 + b2 = c2

    You've got c is the distance from the white ball to the ball you want to hit, d is the distance from the white ball to the pocket, and e is the distance between the pocket and the ball you want to hit, and these distances form a triangle between white ball, pocket, and ball to be hit.

    Then you drop a line from the ball you want to hit to the line from the white ball to the pocket, these lines should meet at a distance b from the white ball and be perpendicular, and should split the original triangle into two right angled triangles, one of which has sides a,b,c where a is the distance between the ball you want to hit and the point b.

    Apologies if I've misunderstood the question.

    EDIT.

    Sorry I'm getting this mixed up, see red text for corrections. I'll upload a picture to show what I mean if you want. Actually the formula d2=c2+e2-2ce cos [tex]\theta[/tex] might be better. Sorry for the mistakes.
     
    Last edited: Mar 2, 2009
  8. Mar 2, 2009 #7
    yes i think thats what im lookinf g foe diagam yes plz
     
  9. Mar 2, 2009 #8
    in the pool game how would you use this formula quickly if understood ,the origanall statement was made only three angle for any given shot 15,30,45,begrees. he said to start on the side that does not face the pocket, that is the part iam confused on
     
  10. Mar 2, 2009 #9
    the angles I think he is saying is meant to be triangles. that sound right?
     
  11. Mar 2, 2009 #10
    in cobonation they add up to 90 degrees
    ?
     
  12. Mar 2, 2009 #11
    split in 1/2 =45 degrees thecenter of the pocket
    something like that
     
  13. Mar 2, 2009 #12
    Have you got the original paper/source where you read this from, or a diagram maybe?

    See the attachment for a pic of what I was describing. Though now it doesn't quite sound like what you were looking for.
     

    Attached Files:

    Last edited: Mar 2, 2009
  14. Mar 2, 2009 #13
    Do you mean something like this:
     

    Attached Files:

  15. Mar 3, 2009 #14
    I did not read this anywhere I determind this by looking at the pool table over and over for a while this is the only soulition I could come up with.then manualy proving to my self bye drawing it out on papper then actually cutting out the triangle wnd plscingit where it needed to be.Mind you I have no math or geomotry backround at all to speak of.
     
  16. Mar 3, 2009 #15
    so far I am not able to open file ,it says pending aproval,any sugestions on how to open or can you E-mailt the attachments lineup@comcast.net Thanks
     
    Last edited: Mar 3, 2009
  17. Mar 3, 2009 #16
    I'll have to send you them later today mate. I think I better understand now what you're describing. I'll attach the pic here in case my attachments get approved before I get a chance to email you.
     

    Attached Files:

    Last edited: Mar 3, 2009
  18. Mar 3, 2009 #17
    Yes ,this is what I am looking for,but a little more imformation on how to explain that.Things like ,is there a formula to use,what would that be?
    ,how is it quickly and easy to use ?and a easy way to prove to someone else ,maybe without all the techinal stuff that most people would not understand,I believe to put it in (laymens terms)in other words, words that most people would understand.I am looking for both ,the techinical side to it and the easy way to understand it.Do you understand what I am saying?
    Backto the techinal side ,the three angle are considered to be tri-angles correct?
    In the pool game what side of the triangle would be considered not to be facing the pocket?
    Alot of questions ,to many?
     
  19. Mar 3, 2009 #18
    I have an article that I am trying to figure out may it send you this articl via E-Mail?
     
  20. Mar 3, 2009 #19
    Or go to google and type in Hal Houle point and piviot ,you will find the article there. this is what lead me to this forum after 2 months of me working on it wih no help.
     
  21. Mar 4, 2009 #20
    Ah okay I found the article. You want to know how to identify which is a 45 degree, 30 degree, or 15 degree shot?

    I think you first need to find where the 45, 30, 15 degree lines are, which I tried to draw on the attached picture from the info in the article. Then if the cue ball and the ball you want to pot are lined up on one of these lines that is the angle of the shot.

    Note in the picture I only drew the lines (left side of pic) for when the cue ball is on the bottom left of the picture and you're shooting into the two left corner pockets or top centre, and on the other side for when the cue ball is on the bottom right and shooting into the right corner pockets or top centre. If the cue ball is somewhere else or you're shooting into another pocket then the lines would be different.

    I could be completely wrong though, so you should check with someone who knows more about pool.
     

    Attached Files:

    Last edited: Mar 4, 2009
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