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Credits go to Edgar Reynaldo, Niunio, Peter Hull, and Polybios for helping out!
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[MATH] distance in X between two parallel lines
Ariesnl
Member #2,902
November 2002
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Hey everyone, I ran into a math problem, I hope someone can help me out

When the distance of two parallel lines is given and the angle of those lines with the Y axis is also known I want to know the distance in x

Anyone ? thnx

Perhaps one day we will find that the human factor is more complicated than space and time (Jean luc Picard)
Current project: [Star Trek Project ] Join if you want ;-)

Niunio
Member #1,975
March 2002
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What about pitagoras? I mean, trace a line perpendicular to both paralels. Then trace a line with the given angle from one of the new vertex. This should give to you a triangle and you know all angles (the rect one, the one given and its supplementary) and one of the cathetus, so you should be able to calculate the hypotenuse, can't you?

I'm not able to doodle now but I hope you can picture it. ;)

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Current projects: Allegro.pas | MinGRo

Edgar Reynaldo
Member #8,592
May 2007
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Peter Hull
Member #1,136
March 2001

<math>x=d/{cos \theta}</math>

Polybios
Member #12,293
October 2010

Distracting... I think it's x = d / cos(phi).

Edit: beaten.

Edit2: We need a sketch:
{"name":"610966","src":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/e\/c\/ec67707f7a1439cd8c62ba6517fbbef9.png","w":437,"h":357,"tn":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/e\/c\/ec67707f7a1439cd8c62ba6517fbbef9"}610966

So:
<math>\varphi' = 90^\circ - \varphi</math>
and
<math>\frac{d}{x} = \sin(\varphi') = \sin(90^\circ - \varphi) = \cos(\varphi) \Rightarrow x = \frac{d}{\cos(\varphi)}</math>

Ariesnl
Member #2,902
November 2002
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Thanks a lot !

Perhaps one day we will find that the human factor is more complicated than space and time (Jean luc Picard)
Current project: [Star Trek Project ] Join if you want ;-)

Niunio
Member #1,975
March 2002
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Polybios said:

Edit2: We need a sketch:

Just that. Thank you. :)

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Current projects: Allegro.pas | MinGRo

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