Viewing 15 posts - 1 through 15 (of 15 total)
  • Maths – bearings, coordinates & fixed point distance calcs
  • steve_b77
    Free Member

    I’ve got a maths problem that I can’t figure out and I wonder if one of you delightful people on here can help me out.

    Here’s the scenario:

    Point A is a design location of a foundation with a given set of coordinates and a bearing and a fixed distance to a point.

    Easting: 345687.837
    Northing: 391242.185
    Bearing: 146.034
    Distance to fixed point: 2.500m

    Point B is the location the foundation was installed at:

    Easting: 345688.011
    Northing: 391241.965
    Bearing: unkown

    Is it possible to calculate the new distance from the fixed point using the above data, if so how?

    kevj
    Free Member

    Yes, there are a number of ways. I have (somewhere) an excel spreadsheet that does this if you need to convert from local datum to national or vice versa, but could be used in this situation.

    Another way would be to cad out the co-ordinates, and measure the vector between the points.

    mogrim
    Full Member

    Yes it’s possible.

    Plot it on a piece of paper, you’ll soon see how.

    kevj
    Free Member

    It’s 280mm out

    steve_b77
    Free Member

    Kevj,

    Could you send me the spread sheet?

    kevj
    Free Member

    I’ll look on my laptop at work tomorrow (I don’t use this any more in my current job) but should have a copy somewhere in my saved files.

    richmars
    Full Member

    The figures you give are just metres based on the national OS grid. A bit of maths (one minus the other, square, add, square root), gives 280.49mm
    between the two points.

    steve_b77
    Free Member

    Maybe I should’ve explained it a little better.

    The fixed point is a straight line, the points are perpendicular to it.

    I know the distance from the straight line to point A, assuming that point B is also perpendicular to the line what is the distance to the straight line?

    Not the distance between the two points – I can figure that out

    richmars
    Full Member

    Still don’t understand.
    What fixed point is the straight line? You need two points for a line.
    A point can’t be perpendicular to anything. (I guess you mean a line from the point to the line.)
    Suggest you draw it in CAD. I would if I was still at work.

    kevj
    Free Member

    Sorry Steve, I thought you were asking for the distance between co-ordinate A and B.

    Do you want to know the distance between co-ordinate B and the end of the straight line from A? Assuming everything is perpendicular.

    poly
    Free Member

    OK, so i think I might be able to decipher what you are asking – a picture (even a sketch) is always helpful…

    https://www.dropbox.com/s/rm57uiayx8w05r0/Slide1.jpg

    If I follow correctly you want the distance BF in this diag?

    Its relatively easy to do. You need to dust off your trigonometry. Without being rude though, if you need to ask on a bike forum how to do this – are you the right person to be working this out? How will you know if I tell you the right answer and you build the thing in the wrong place?

    steve_b77
    Free Member

    My fault for the poor explanation.

    Say the straight line is a straight wall, point A is where the foundation is meant to be positioned at those coords, 2.5m from the wall, at right angles to it.

    Point B is where it ended up, but still at right angles to the wall, how far from the line of the wall is point B?

    That’s a much better explanation (I thinks)

    poly
    Free Member

    My fault for the poor explanation.

    Say the straight line is a straight wall, point A is where the foundation is meant to be positioned at those coords, 2.5m from the wall, at right angles to it.

    Point B is where it ended up, but still at right angles to the wall, how far from the line of the wall is point B?

    That’s a much better explanation (I thinks) Where does the 146 deg come in?

    Do you mean you want BX in this diag? https://www.dropbox.com/s/pqlyfvayi2d7umx/Slide1.jpg

    steve_b77
    Free Member

    It’s the original bearing in of the centre line of the foundation

    richmars
    Full Member

    That’s a much better explanation (I thinks)

    I don’t think so!
    Again, a point can’t be perpendicular to a line.

    Forget about point A being 2.5m from the wall, just move it so it’s on point A. Now whats the difference in the positions of these points?
    I think that’s what you’re after, and I think it’s in the figures of A and B.
    Edit, move the wall so it remains parallel to its original position.

Viewing 15 posts - 1 through 15 (of 15 total)

The topic ‘Maths – bearings, coordinates & fixed point distance calcs’ is closed to new replies.