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Triangulated radius

  1. Oct 3, 2012 #1
    i have a problem working out the correct arch under a queenslander house 3 metres in the air
    The span etween the pillers is 3280mm
    So the cord is 3280mm and from the center of the cord to the radius is 450mm
    I have worked the radius out to be 5238mm
    Since the house is only 3000mm of the ground the center of the arc is aprox 2000mm below ground so i wont be able to pivot a string on the ground to scribe the arc on the pickets so i can cut them to the correct lenght.
    I need a formula caculating the base of a triangle and scribing an arc with the apex of the triangle
    A nail at the base of each piller with a piece of string around both nails up to the apex touching the middle picket. With a pencil placed at the apex in the string scribing an arc of a circle with a radius of 5238mm. what would the size of the base of the triangle have to be?
    As i have about 10 arches to do all of varrying size
  2. jcsd
  3. Oct 3, 2012 #2
    Hello bretus and welcome to Physics Forums.

    You have left out one vital piece of information

    Your first information describes a circular arc.

    Your proposed method will generate an elliptical arc.

    So which do you want?
  4. Oct 3, 2012 #3
    Hi Studiot
    OK the shappe of the curve is not important but it must be repeatable on different cords
    So if i run the cord from the center of first picket to the center of the last picket the cord lenght will be 3132mm and from center of cord to top of ellipse will be 450mm
    So i need a formula to calculate the base so i can scribe the ellipse given that i only have 3000mm in hight to work with
  5. Oct 3, 2012 #4
    Well I am not quite sure what you are trying to achieve, or what you mean by the same curve.

    You have provided two different centre measurements (3280 & 3132).

    So does this mean that the centres vary?

    The rise to span for a circular arc will, in general, be greater than for an ellipse of a similar span.

    Obviously if the spans vary then each one will have to be worked out separately as they cannot have the same rise and radius.

    Can you not use a template? Say one made of hardboard. It would be easier to mark and cut on a bench than in situ. You could easily access the centres then.

    For setting out a circular curve I usually make use of the formula

    Deviation or offset of a circular curve froma straight line = Y2/2R
  6. Oct 3, 2012 #5
    The 3280mm cord was for a circle the 3132mm cord is for an elipse
    the 450mm rise is fixed regardless of curve or span
    No it is not possable to use a template would be to difficult
    I only went for the triangular method to give me a means of laying out the curve in place over different spans
    As the first and last pickets are 1150mm long and the middle picket is 700mm long hence the 450mm rise with 29 total pickets in the 3280 span(piller to piller) 3132mm being from center first picket to center last picket
    The first, center and last picket lenghts are fixed on all spans
    Hope this makes it clearer
  7. Oct 3, 2012 #6
    Does this help?

    You can measure along the straight at the apex or between the springings as shown.

    Attached Files:

  8. Oct 3, 2012 #7
    Thats the way i first tried and is in the beginning of this post

    It gave me a 5238mm radius whichn is not possible to measure because house is only 3000mm

    of the ground. Which means i can not scribe the line because the maximum radius i can scribe

    is 3000mm -700mm lenght of middle picket which is 2300mm maximum radius.

    Thats why I opted for base triangle scribing elipse with apex

    Because adjusting the lenght and hieght of the base(from the ground) i can scribe an elipse

    maintaining a 450 rise with different spans.

    this method will involve clamping a piece of timber at a specific hieght between the pillers

    and fitting 2 screwws a set distance apart in that timber

    With the equation to work this out i can physicaly scribe the lines for the elipses on all the


    Given that all spans start and end with a 1150mm picket with a 700mm picket in the center

    with the remaining picket needing to be scribed and cut to lengh

    and the spans will very from 1500mm to 3280mm
  9. Oct 4, 2012 #8
    I'm sorry you didn't understand my diagram.

    Measuring offsets from the straight (tangent) is totally different using a radius (centre) point.

    You only need the radius to perform calculations according to my formula.

    Then you measure offsets from a straight line at distances calculated by my formula.

    It is a very easy formula and allows ready construction of a table of distances and offsets.
  10. Oct 4, 2012 #9
    Yes i know what you mean but that solution does not enable me to physically scrib a line on the end of string to the pickets.

    And will involve about 800 seperate calculations for all the spans.

    The largest single point radius i can scribe is 2300mm which means that some of the pickets wont even be reached.

    And the lenght of the first and last and center pickets cannot be changed which means the 450mm rise will be the same for all the different spans
    Last edited: Oct 4, 2012
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