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TrojanPoem

  • one year ago

Engineering Drawing.

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  1. TrojanPoem
    • one year ago
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  2. TrojanPoem
    • one year ago
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    @BAdhi

  3. BAdhi
    • one year ago
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    you can find R_1 by looking at the triangle created by ( aligning R_1 and 120 radius) and (R_1 and 20 radius on the lower right) and 160

  4. BAdhi
    • one year ago
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  5. TrojanPoem
    • one year ago
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    But we have 2 unknown sides

  6. TrojanPoem
    • one year ago
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    Oh, 20 + R and 120 + R

  7. TrojanPoem
    • one year ago
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    (100)^2 + ( 20 + R)^2 = (120 + R)^2

  8. TrojanPoem
    • one year ago
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    How about R3

  9. BAdhi
    • one year ago
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    two sides and an angle ;)

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  10. TrojanPoem
    • one year ago
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    (30 + R3) * 0.5 = R3 15 + 0.5R3 = R3 0.5 R3 = 15 R3 = 30 ?

  11. BAdhi
    • one year ago
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    I think so

  12. BAdhi
    • one year ago
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    Find x by looking at the complete 120 line (i hope you can find it easily) ;)

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  13. TrojanPoem
    • one year ago
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    I am dumb.

  14. TrojanPoem
    • one year ago
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    x= 20 ?

  15. BAdhi
    • one year ago
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    sum of these three will equal to????

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  16. TrojanPoem
    • one year ago
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    30 + 10 + x = 60 x = 20 ? what am I mistaken in ?

  17. Michele_Laino
    • one year ago
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    for \(R_3\) I got this result: |dw:1444412934196:dw| we have: \(L=80 \tan(30) \) so we can write this proportion: \[\frac{{80 - x}}{x} = \frac{{80}}{L}\] from which I get: \[x = {R_3} = \frac{{80}}{{\sqrt 3 + 1}}\]

  18. Michele_Laino
    • one year ago
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    for \(R_2\) I got this: |dw:1444413551298:dw| so, if we apply the theorem of Pitagora, we have to solve this equation: \[{\left( {30 + x} \right)^2} = {50^2} + {\left( {40 - x} \right)^2}\]

  19. TrojanPoem
    • one year ago
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    @Michele_Laino , can't we use some drawing operations to find the center point ?

  20. TrojanPoem
    • one year ago
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    With more than 3 unknown centers , It would be difficult to think. Estimated time is 13 min for each figure.

  21. Michele_Laino
    • one year ago
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    from my experience, using AutoCAD, all of your arcs will made automatically, we have to use the "grip" at each points, showed by AutoCAD

  22. Michele_Laino
    • one year ago
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    I recommend you to use a realease at least AutoCAD 2000, not AutoCAD LT 2000

  23. TrojanPoem
    • one year ago
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    My professor said " You all have to master hand drawing before moving to AutoCAD" Yeah, my problem is the unknown center points it uses concepts like |dw:1444413932573:dw| two tangents to the same circle from the same point has the same length and a line from the mid of both will direct to the center

  24. Michele_Laino
    • one year ago
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    you have to draw the axes of both segments, and the intersection point of such axes is the center of your circle

  25. TrojanPoem
    • one year ago
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    That's it , if we know the radius of the circle :/ but it's unknown

  26. Michele_Laino
    • one year ago
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    |dw:1444414133266:dw|

  27. TrojanPoem
    • one year ago
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    The radius is perpendicular to the tangent line xD Nice

  28. TrojanPoem
    • one year ago
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    @Michele_Laino , can't we find R_3 with some drawing operation like what you did ?

  29. Michele_Laino
    • one year ago
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    I'm thinking...

  30. Michele_Laino
    • one year ago
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    here is my method: |dw:1444415217388:dw| try to draw a circle which is tangent to both lines #1 and #2 contemporarily

  31. TrojanPoem
    • one year ago
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    |dw:1444415579441:dw| maybe we can say that the two lines are tangents to the circle so the center of the angle of the intersection point of both lines will cut the the 30 degrees line on the circle mid point

  32. TrojanPoem
    • one year ago
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    How about R_1 ?

  33. Michele_Laino
    • one year ago
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    I'm thinking...

  34. TrojanPoem
    • one year ago
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    go http://openstudy.com/updates/5617f969e4b086e96bfdedad as it started to lag here

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