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Chart, plotting a series at an angle

  1. #1
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    Chart, plotting a series at an angle

    Hello everyone,

    I'm having 2 problems with making a chart

    Problem 1:

    I'm trying to find a way how to plot series at an angle
    This i how that graph should look

    Screenshot_3.jpg

    Only data i have to begin with is "C" ,(x(вертикално опетрећење)=0, y(чврсточа на смицање)=17) starting point of the line
    and angle, (in this case the angle is 18) which determines at what angle with the line continue to x value 400

    Problem 2:

    I need to find the y value (чврсточа на смицање) when the plotted line series intersects major gridlines at 150, 250 and 350

    Is there a way to do it by some math or adding some series to act like gridlines, would be much better then too manually look on the chart and write the new values
    (need to automate that process, another chart need those values)

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    Re: Chart, plotting a series at an angle

    This tutorial talks about the relationship between slope and angle of a line: https://sciencing.com/convert-angle-...e-8128171.html
    and this one: https://www.onlinemath4all.com/angle...of-a-line.html

    One key thing to remember is that Excel, like other programming languages, does trig in radians, so you will need to convert your angle to radians when it is given in degrees (use RADIANS() function).

    Once you have the slope m, then you can use your slope-intercept form for the equation of a line (y=mx+b) to calculate y at any value for x. With a table of x,y values, you can use an Excel scatter chart to visualize the line.
    Quote Originally Posted by shg
    Mathematics is the native language of the natural world. Just trying to become literate.

  3. #3
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    Re: Chart, plotting a series at an angle

    Thanks for your answer, i've found a solution that works perfectly for my case so i might as well share it here if anyone needs its

    "This is more of a trigonometry question than a charting question. You basically have a right triangle. At the x=150, height would be

    tan(angle) = height/length
    tan(18) * length = height

    However, you have a y intercept of 17, so we'll need to account for that and add 17 to your height

    tan(18) * length + 17 = height
    tan(18) * 150 + 17 = height
    65.74 = height

    Visually inspecting the chart, that looks to be correct.

    In XL, the trig functions expect a radian input, so you need to use the RADIANS function to convert from degrees to radians. The XL formula then would be

    =TAN(RADIANS(18)*150+17

    Once you have that, you could make this dynamic by replacing the 150 with a cell reference

    =TAN(RADIANS(18)*A2+17

    and then in col A put whatever x-values you want to calculate for. This should then be able to solve both Problem 1 & 2."

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