MHB Trig Assistance for CNC Machining/Engineering

[megani]

Trig help please! CNC Machining/Engineering

View attachment 4834

 

[Amer]

I redraw the picture. you mean by DIA ( diameter ) right ?. you can notice that the two triangles the one with the big circle and the one with small circle are similar.

$dR = \frac{39}{\sin y}, cr = \frac{23.5}{\sin y}, ch = \frac{86}{\tan y}$ now

$dR + cr + ch = 126 $ hence

$126 = \frac{39}{\sin y} + \frac{23.5}{\sin y} + \frac{86}{\tan y} $

$126 \sin y = 62.5 + 86 \cos y $
View attachment 4837

 

[megani]

Yes DIA means the diameter...

 

[megani]

I redraw the picture. you mean by DIA ( diameter ) right ?. you can notice that the two triangles the one with the big circle and the one with small circle are similar

Sorry your picture is not working now.

 

[Amer]

I edited my post

 

[megani]

I redraw the picture. you mean by DIA ( diameter ) right ?. you can notice that the two triangles the one with the big circle and the one with small circle are similar.

$dR = \frac{39}{\sin y}, cr = \frac{23.5}{\sin y}, ch = \frac{86}{\tan y}$ now

$dR + cr + ch = 126 $ hence

$126 = \frac{39}{\sin y} + \frac{23.5}{\sin y} + \frac{86}{\tan y} $

$126 \sin y = 62.5 + 86 \cos y $

Just not too sure with the distance from S - H?

 

[Amer]

Just not too sure with the distance from S - H?

you mean ch ? it is typo I meant $dh$. ch is perpendicular to SR so we have a right triangle $\tan y = \frac{86}{hd} $
$hd + dR + cr = 126$

 

[megani]

Yes sorry HD...Initially I though 126-39-23.5 but there's a little gap between the circle, do you see?

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View attachment 4838

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Actually...gap next to big circle too! :-\

 

[Amer]

yea I see it but it dose not affect the solution. "c" is the intersection between the tangent and the horizontal line "rs". and "ra " is perpendicular to the tangent it is a theorem the angle $acr = y$ I used sine definition to come up with $\sin y = \frac{23.5}{rc} \rightarrow rc = \frac{23.5}{\sin y}$

 

[Jameson]

Hi megani,

Welcome to MHB! This appears to be part of a graded assignment. If so, are you allowed to receive outside help? I ask only to confirm that we aren't taking part in unallowed assistance.

 

[megani]

Hi megani,

Welcome to MHB! This appears to be part of a graded assignment. If so, are you allowed to receive outside help? I ask only to confirm that we aren't taking part in unallowed assistance.

Hi. No. I'm actually a learning support teacher (main focus numeracy) and my colleague and I are developing a revision workbook for some Dip. Engineering students and this was one of their resources.