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TrojanPoem
 one year ago
Engineering Drawing.
TrojanPoem
 one year ago
Engineering Drawing.

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BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1you 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

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But we have 2 unknown sides

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Oh, 20 + R and 120 + R

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(100)^2 + ( 20 + R)^2 = (120 + R)^2

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1two sides and an angle ;)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(30 + R3) * 0.5 = R3 15 + 0.5R3 = R3 0.5 R3 = 15 R3 = 30 ?

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1Find x by looking at the complete 120 line (i hope you can find it easily) ;)

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1sum of these three will equal to????

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.030 + 10 + x = 60 x = 20 ? what am I mistaken in ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for \(R_3\) 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: \[x = {R_3} = \frac{{80}}{{\sqrt 3 + 1}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for \(R_2\) I got this: dw:1444413551298:dw so, if we apply the theorem of Pitagora, we have to solve this equation: \[{\left( {30 + x} \right)^2} = {50^2} + {\left( {40  x} \right)^2}\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino , can't we use some drawing operations to find the center point ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0With more than 3 unknown centers , It would be difficult to think. Estimated time is 13 min for each figure.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0from my experience, using AutoCAD, all of your arcs will made automatically, we have to use the "grip" at each points, showed by AutoCAD

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I recommend you to use a realease at least AutoCAD 2000, not AutoCAD LT 2000

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0My 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0you have to draw the axes of both segments, and the intersection point of such axes is the center of your circle

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0That's it , if we know the radius of the circle :/ but it's unknown

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444414133266:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0The radius is perpendicular to the tangent line xD Nice

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino , can't we find R_3 with some drawing operation like what you did ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here is my method: dw:1444415217388:dw try to draw a circle which is tangent to both lines #1 and #2 contemporarily

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0dw: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

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0go http://openstudy.com/updates/5617f969e4b086e96bfdedad as it started to lag here
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