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TrojanPoem
 one year ago
Engineering drawing.
TrojanPoem
 one year ago
Engineering drawing.

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! let me think...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I know the position of this point: dw:1444416115829:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0You know it, on the drawn figure, but if you are drawing it yourself, you don't :/

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I have this part solved, but I can't grasp it , surely, you will.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that the equation below can solve your proble, please note trhat \(R_1=x\) dw:1444416517572:dw so we can write this: \[{\left( {120 + x} \right)^2} = {160^2} + {\left( {x + 20} \right)^2}\] namely, I have applied the theorem of Pitagora

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got: \(x=580/11\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and, using my formula above, I got \(R_2=160/7\)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, You got the right radius but in Engineering drawing, I can't write steps of getting them so , I'd be considered cheater

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I see, nevertheless when you go in laboratory, in order to manufacture your object, you have to provide to the worker the measures of the radiuses

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Lab ? is this used in physics too ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry, I meant in mechanical workshop

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I think redraw this figures is just to prepare us to draw object parts
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