Engineering Drawing.

- TrojanPoem

Engineering Drawing.

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- TrojanPoem

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- TrojanPoem

@BAdhi

- BAdhi

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

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- BAdhi

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- TrojanPoem

But we have 2 unknown sides

- TrojanPoem

Oh, 20 + R and 120 + R

- TrojanPoem

(100)^2 + ( 20 + R)^2 = (120 + R)^2

- TrojanPoem

How about R3

- BAdhi

two sides and an angle ;)

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- TrojanPoem

(30 + R3) * 0.5 = R3
15 + 0.5R3 = R3
0.5 R3 = 15
R3 = 30 ?

- BAdhi

I think so

- BAdhi

Find x by looking at the complete 120 line (i hope you can find it easily) ;)

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- TrojanPoem

I am dumb.

- TrojanPoem

x= 20 ?

- BAdhi

sum of these three will equal to????

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- TrojanPoem

30 + 10 + x = 60
x = 20 ? what am I mistaken in ?

- Michele_Laino

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:
\[x = {R_3} = \frac{{80}}{{\sqrt 3 + 1}}\]

- Michele_Laino

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

- TrojanPoem

@Michele_Laino , can't we use some drawing operations to find the center point ?

- TrojanPoem

With more than 3 unknown centers , It would be difficult to think. Estimated time is 13 min for each figure.

- Michele_Laino

from 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

I recommend you to use a realease at least AutoCAD 2000, not AutoCAD LT 2000

- TrojanPoem

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

- Michele_Laino

you have to draw the axes of both segments, and the intersection point of such axes is the center of your circle

- TrojanPoem

That's it , if we know the radius of the circle :/ but it's unknown

- Michele_Laino

|dw:1444414133266:dw|

- TrojanPoem

The radius is perpendicular to the tangent line xD
Nice

- TrojanPoem

@Michele_Laino , can't we find R_3 with some drawing operation like what you did ?

- Michele_Laino

I'm thinking...

- Michele_Laino

here is my method:
|dw:1444415217388:dw|
try to draw a circle which is tangent to both lines #1 and #2 contemporarily

- TrojanPoem

|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

- TrojanPoem

How about R_1 ?

- Michele_Laino

I'm thinking...

- TrojanPoem

go
http://openstudy.com/updates/5617f969e4b086e96bfdedad
as it started to lag here

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