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Need some math help

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  • Need some math help

    I need to find 2 points on this sketch and can't seem to get past a major brain fart....I have been staring at it and .....

    Anybody have a cad program and care to help me out ?
    Need the 2 points shown as +, at the top where the 30 deg line would intersect to the zero line if it ran all the way up and to the right where it hits the flat on the side.. 0, 0, is the center of the sketch.
    It is a .778 dia. .66 across all flats


  • #2
    Can't happen except in one case. That is if the vertical flat is of zero length and the angle is 90 degrees.

    From the X at the center to the lower right + sign is the hypotenuse of a right triangle.

    The formula c squared = a squared plus bsquared. C is the hypotenuse, a is the distance from X to the right flat. b is the vertical height of the vertical flat.

    If the width really is .66 and the angle really is 60 degrees, then the length of the vertial flat will be .33 and the vertical heigth will be .165 from X to the lower right +.

    However, the distance from X to the 60 degree angle flat will be .381 instead of .33. The height to the upper + would be .762.

    Since c squared equals a squared, then b must equal zero.
    Last edited by Jpfalt; 12-27-2012, 08:03 PM.


    • #3
      Consider a triangle from the center to the mid point of the flat and out to the 30 degree corner. That would be a right angle triangle (90,60,30 degrees) with a height of .33.



      • #4
        I get (.33, .136) and (0, .707)

        That assumes the shape is an octagon, and 60 degrees was intended.
        Last edited by Tony Ennis; 12-27-2012, 08:43 PM.


        • #5
          What's shown in the sketch is an octagon, but 60 degree sides would make a hexagon. With a hexagon the x on the right side would be on the horizontal centerline.


          • #6
            I looked at this again and realized the angle from X to lower right + could be 15 degrees from horizontal.

            The height of lower right + would be .088"

            The height of the upper + would be .660


            • #7
              Hi Uncle O,

              *not sure* what you're aiming at! I'm guessing it's an 8-sided part, and the 30* angle is *not* tangent to the next flat around?

              If the + on the right is on the corner of the flats a bit of jommetry should find that, like so:

              the angle between the axis and the corner is half of 360/8, so 22.5*

              so... the X coordinate is half the A/F distance 0.330"

              Tan of 22.5* is 0.4142 = opposite (the Y coordinate) / adjacent (half the A/F) = Y/0.33

              multiply both sides by 0,33, Y coordinate = 0.4142 x 0.330 = 0.137 (to 3 sig' fig's)

              So the right-hand + is at 0.330, 0.137

              To get the top + position, draw a right-angled triangle with a 30* angle at the top, the opposite side from the edge of the flats we just worked out horizontally to meet the vertical line - this will be 0.330" long (half the A/F dimension). The hypotenuse goes from our new + point to the edge of the flat and the adjacent from the intersection of the vertical line through the axis to the 30* angle.

              tan 30* is 0.5773 = opposite over adjacent, so...

              0.5773 = 0.330 (the half-A/F dimension) / adjacent side

              multiply both sides by (adjacent) gives 0.5773 x adjacent = 0.330 (opposite)

              hmmm, not there yet!

              divide both sides by tan 30* (0.5773) gives adjacent = 0.330 (opposite) / 0.5773

              We we now know the adjacent side is 0.330 / 0.5773 = 0.5716"

              to get the position of the upper +, we have to add the Y-coordinate of the first (right-hand) +, so:

              x = 0 (as it's on the vertical from the axis), Y = 0.5716 + 0.137 = 0.7086"

              If I can attach it, diagram illustrates....


              Hope this helps, rather than confuses!

              Dave H. (the other one)

              P.S. - just noticed for the first time that tan 22.5 is root(2) - 1... I'll remember that
              Last edited by Hopefuldave; 12-27-2012, 09:12 PM.
              Rules are for the obedience of fools, and the guidance of wise men.

              Holbrook Model C Number 13 lathe, Testa 2U universal mill, bikes and tools


              • #8
                Your diagram is indeterminate. Here's a labeled version of your drawing that may help you sort it out though:


                • #9
                  Originally posted by Machine View Post
                  Your diagram is indeterminate. Here's a labeled version of your drawing that may help you sort it out though:
                  Thanks for everybodies attempts to help, I have made these before, it's been a quite awhile tho.....
                  I need to find d as shown above, and from there I will be good. In the a.m. I will put a protractor to the corner of that flat ( on the print ) and take it from there. I will post results later.
                  Thanks again.


                  • #10
                    I get d = .136. See my post above.


                    • #11
                      I stick by my calc that d=.088.

                      To check it out I laid it out in Solidworks and got the same answer.

                      I hereby declare victory and will now run like hell.


                      • #12
                        Edit: Scratch that
                        Last edited by Jaakko Fagerlund; 12-28-2012, 05:41 AM.
                        Amount of experience is in direct proportion to the value of broken equipment.


                        • #13
                          I have to agree with Machine. Your drawing is too rough to really know what you are asking. It could be a regular octagon, meaning 8 equal sides and 8 equal angles or it could be just a plain octagon, meaning it has 8 sides but they are not necessairly equal and the angles may not be equal. Then again, the top and bottom seem to be highly curved so it could be a figure with six flat sides and two curved ones. Or perhaps something else.

                          And is your 60 degree line congruent with one of the sides or is it at a different angle and only touches it at the corner marked with a cross.

                          In any case, you do not give enough details/dimensions to figure the location of the points. I would suggest a better sketch and more dimensions.
                          Paul A.
                          SE Texas

                          Make it fit.
                          You can't win and there IS a penalty for trying!


                          • #14
                            Edited totally

                            The piece is round with six flats on it. Dimensions as shown, the points are here:

                            Highest point is 0.66, the rightmost point is 0.088 above Y axis
                            Last edited by Jaakko Fagerlund; 12-28-2012, 05:47 AM.
                            Amount of experience is in direct proportion to the value of broken equipment.


                            • #15
                              The CAD solution is good, but here's a trig solution:-

                              AB = 0.33
                              AC = AB*cos(30) = 0.33*0.8660254 = 0.2857884
                              BC = AB*sin(30) = 0.33*0.5 = 0.165

                              DB = AC
                              ED = DB/tan(30) = 0.495

                              DA = BC
                              EA = ED+DA = 0.495 + 0.165 = 0.66

                              FG = 0.33
                              EF = FG/tan(30) = 0.5715768

                              GH = FA
                              GH = EA-EF = 0.66 - 0.5715768 = 0.0884232


                              Last edited by Barrington; 12-28-2012, 08:34 AM.