X from a curved triangle

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I'm having a hard time figuring out what to do here and it feels as though it involves converting the function into polar coordinates but I'm not a hundred percent sure.

Essentially, I have a triangle in which two sides are straight and one side is a curved function. Knowing only the length of one side, the angle between them, and the function, I need to either be able to find the other straight side or x. I do not need help with solving it but more so just guidance on how I would go about solving it. I've included a mspaint attempt at drawing it.

Thanks in advance
 

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  • #2
Redbelly98
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What is your application? Or is this a homework or self-study problem?
 
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The application is towards image processing. I am a coop student and am deriving equations for a project so I cannot go into very much detail.
I've tried using polar coordinate system but my equation blows up when I approach 90 degrees, arc length seems like a reasonable approach but I am not fully certain on what to do.
 
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Redbelly98
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Okay, we are careful around here about helping people with homework. So it's good to know this is not a homework problem :smile:

.

attachment.php?attachmentid=31641&d=1296076080.jpg

I guess you've tried some obvious ways to do this, but let me just suggest the first thing that comes to my mind. There may be reasons why it won't work; if so I'll leave it to you to provide more info.

May I presume line L is in a known orientation (perhaps vertical)? That means the slop of the other straight side -- let's call it M -- is known.

From the slope of M and the coordinates of the common vertex of L and M, you can write an equation for M in y=mx+b form. Then you need to solve for the intersection of the line with f(x); there are standard techniques for doing this numerically.

Once you know the common vertex of L and M, as well as the intersection of M with f(x), you can get the length of M using the Distance Formula from geometry.

The application is towards image processing. I am a coop student and am deriving equations for a project so I cannot go into very much detail.
I've tried using polar coordinate system but my equation blows up when I approach 90 degrees, arc length seems like a reasonable approach but I am not fully certain on what to do.
Hmmm, when θ is 90°, the line is simply y=whatever (the y-coordinate of the vertex common to L and M. Not sure why that would be a problem.
 
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Okay, we are careful around here about helping people with homework. So it's good to know this is not a homework problem :smile:
.

attachment.php?attachmentid=31641&d=1296076080.jpg

I guess you've tried some obvious ways to do this, but let me just suggest the first thing that comes to my mind. There may be reasons why it won't work; if so I'll leave it to you to provide more info.

May I presume line L is in a known orientation (perhaps vertical)? That means the slop of the other straight side -- let's call it M -- is known.

From the slope of M and the coordinates of the common vertex of L and M, you can write an equation for M in y=mx+b form. Then you need to solve for the intersection of the line with f(x); there are standard techniques for doing this numerically.

Once you know the common vertex of L and M, as well as the intersection of M with f(x), you can get the length of M using the Distance Formula from geometry.


Hmmm, when θ is 90°, the line is simply y=whatever (the y-coordinate of the vertex common to L and M. Not sure why that would be a problem.
Its understandable for your skepticism, and so thank you for your time. I guess I could be a little less vague. Your correct the orientation is vertical and also the function is an exponential function, we can say exp(x).

I just finished trying your method, it works to a point. When I equate the linear function with the known exponential function, it requires a 'Lambert W" (also known as "Product Log") function. I will look into these functions, but at the same time, are there any other suggestions you could make?
 
  • #6
Redbelly98
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Okay, glad I was pretty much on target with things.

You have to solve an equation that looks something like
Aekx = mx + b​
or
g(x) = Aekx - (mx + b) = 0​

You can use the Newton-Raphson method to do this numerically; it is described here:
You'll need an initial trial value for x. I suggest using a very large value for x to guarantee convergence to the desired solution. If theta is close to 90 degrees, or greater, than a small trial value of x will give problems. If you have some idea of the largest x could possibly be in practice, that would be a good initial value.

Hope that helps.
 
  • #7
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Okay, glad I was pretty much on target with things.

You have to solve an equation that looks something like
Aekx = mx + b​
or
g(x) = Aekx - (mx + b) = 0​

You can use the Newton-Raphson method to do this numerically; it is described here:
You'll need an initial trial value for x. I suggest using a very large value for x to guarantee convergence to the desired solution. If theta is close to 90 degrees, or greater, than a small trial value of x will give problems. If you have some idea of the largest x could possibly be in practice, that would be a good initial value.

Hope that helps.
Yes, it does help. I remember that sort of process from class when told to approximate the value of [tex]\sqrt{}2[/tex] with out a calculator. Never even crossed my mind to use it outside of that. Thank you very much for your help
 

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