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Intersection point of two lines from the data set.

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Arsal15
Arsal15 2016-1-28
评论： Star Strider 2016-1-28
Hi, How points on a line can be calculated if you are given the lower(x1,y1) and the upper(x2,y2) points and a given speed from (x1,y1)to(x2,y2).
if you can guide. ?

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回答（2 个）

Star Strider
Star Strider 2016-1-28
For a linear plot like that, the easiest way would be to use the polyfit and polyval functions:
p = polyfit([x1; x2], [y1; y2], 1); % Fit Line To Known Points With Linear Approximation
x3 = ... SOME POINT OR VECTOR OF POINTS ...; % Desired ‘x’ Values
y3 = polyval(p, x3); % Calculate ‘y’ Values
If ‘x’ is time and ‘y’ is position, the speed will be given by ‘p(1)’.
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Arsal15
Arsal15 2016-1-28
Sorry to ask again. I am confused in your variables?
Can you take these variables and guide me? As I have taken below. Now if I have another time instant like 5.5 which is in between [4.8133 and 5.9180] then what will be x and y coordinates of that point at instant 5.5? For sure that point will be on that line ?
if true
x_position = [x(1) x(2)] = [0.0988 0.0442];
y_position = [y(1) y(2)] = [8.2928 8.2941];
time_at_positions = [t(1) t(2)] = [4.8133 5.9180];
speed_x_direction = [vx(1) vx(2)] = [-0.0132 -0.1592];
speed_y_direction = [vy(1) vy(2)] = [-0.0003 0.0389];
end
I will be thankful if you can guide me . As I am stuck on this point.
Star Strider
Star Strider 2016-1-28
I would use the interp1 function:
x = [0.0988 0.0442];
y = [8.2928 8.2941];
t = [4.8133 5.9180];
tq = 5.5;
xy = interp1(t', [x; y]', tq)
xy =
64.8597e-003 8.2936e+000
I’m still not certain I’m understanding the problem, but this seems at least to be working toward a solution.

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Image Analyst
Image Analyst 2016-1-28
See attached polyfit demo.
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Arsal15
Arsal15 2016-1-28
Can you please also look into my comment above and give your suggestion.
I will be grateful for your kind reply.

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