How to get x-y coord. from x-z coord. and y-z coord.?

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To derive x-y coordinates from x-z and y-z coordinates in Matlab, two primary approaches can be utilized. The first method involves selecting a specific z value to plot the corresponding (x, y) data points, which is always reliable. The second method requires picking a value of x, determining the corresponding z, and then finding y, but this is only valid if the functions are well-behaved and the z value is unique. The discussion emphasizes the importance of understanding parametric plots and suggests that Matlab has functions to facilitate this process. Overall, these methods provide a structured way to visualize the relationship between the coordinates without the z component.
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I have data from tiltmeter. It has 3 components (time, x, and y) and stations. I use Matlab for calculating the data (.csv) until i get some plots. I have two plots, time-x and time-y. Each plot has a trendline. So that means i have two trendlines in two plots. Assume that time is z and x is still x and y is still y (cartesian coordinate). Then i have plot x-z and y-z. I'm confused about how i can get x-y coordinate by removing that z component and calculate it in Matlab? What's the theory i can read about that problem? Thankyou.
 
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This is called parametric plot, Matlab should have functions for it.
There are two approaches to plot this:
- pick a value of z, plot the data point (x,y).
- pick a value of x, find the corresponding z, find the corresponding y, plot (x,y)

The first one always works, the second one only if the functions are sufficiently well-behaved (in particular, the found z has to be unique).
 
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mfb said:
This is called parametric plot, Matlab should have functions for it.
There are two approaches to plot this:
- pick a value of z, plot the data point (x,y).
- pick a value of x, find the corresponding z, find the corresponding y, plot (x,y)

The first one always works, the second one only if the functions are sufficiently well-behaved (in particular, the found z has to be unique).

Thankyou so much!
 
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