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    Using laser distance meters in bright sun

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    • B Offline
      bjornkn
      last edited by

      @gaieus said:

      That's true however the majority of buildings are (fortunately) rectangular/vertical (more or less). If not, we get out our secondary school geometry books again and do some calculations. πŸ˜„
      Wonder how those formulas look? πŸ˜‰

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      • GaieusG Offline
        Gaieus
        last edited by

        The book are in the attic ATM. I should get them down one day...
        πŸ˜„

        Gai...

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        • R Offline
          Roger
          last edited by

          SOH-CAH-TOA:

          Sine = Opposite Γ· Hypotenuse
          Cosine = Adjacent Γ· Hypotenuse
          Tangent = Opposite Γ· Adjacent 
          

          or

          Some Old Hippy Caught Another Hippy Trippin' On Acid

          http://www.azcreative.com

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          • R Offline
            Roger
            last edited by

            @gaieus said:

            I also have a Bosch meter and it has a function to measure the shortest or longest route you are "sweeping through". Here are two, quick sketches what it does:

            DOH! Thanks, I don't remember seeing this in the instruction sheet. But then the whole thing looked so simple that I tossed the instruction without reading. Thanks Gaieus.

            http://www.azcreative.com

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            • GaieusG Offline
              Gaieus
              last edited by

              Well, devices of even the same brand can be very different so there may not be this function on yours. Yet of course looking it up would be best.

              Gai...

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              • B Offline
                bjornkn
                last edited by

                @Roger :
                Afaik sin/cos/tan only works on right-angled triangles, and so do the Pythagoras rule a^2+b^2=c^2 - or here : a=sqr(c^2-b^2)
                That's why I was so interested in the formulas in Gaieus' books. Too bad they are in the attic πŸ˜‰

                BTW, did you get that iPhone theodolite to work with any precision?

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                • R Offline
                  Roger
                  last edited by

                  Bjorn, any triangle can be divided into two right triangles, each of which can be solved by pythagoras or sohcahtoa.

                  http://www.azcreative.com

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                  • B Offline
                    bjornkn
                    last edited by

                    @roger said:

                    Bjorn, any triangle can be divided into two right triangles, each of which can be solved by pythagoras or sohcahtoa.

                    Yes, as long as you know some angles?
                    Or if you measure from 2 positions?

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                    • TIGT Offline
                      TIG Moderator
                      last edited by

                      While we are on 'formulas'...

                      If you know two sides of any triangle and the angle between them then you can find the length of the other side opposite that angle using the Cosine Rule
                      a^2 = b^2 + c^2 - 2bc*cosA
                      can be rearranged to give the side a thus...
                      a = sqrt(b^2 + c^2 - 2bc*cosA)
                      Notes:
                      Side 'a' is opposite angle 'A'.
                      There are obviously the other two versions of the equation for the two other sides, but you can give them the reference you want so you always find 'a' without needing to have these in memory!

                      The Sine Rule works for any two angles and one related side, from which you can find the another side
                      a / sinA = b / sinB = c / sinC = D
                      which can be rearranged to give the side a thus...
                      a = b * sinA / sinB
                      Notes:
                      Side 'a' is opposite angle 'A' and so on.
                      The ratios are all equal to D, which is the diameter of the circle that passes through all of the three vertices of the triangle - the radius of that circle is D/2.
                      To find the area of any triangle where you know two sides and the angle between them use
                      Area = a * b * sinC / 2

                      πŸ€“

                      TIG

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                      • R Offline
                        Roger
                        last edited by

                        Folding stick
                        My folding meter stick with distance measure on top

                        Long stick
                        Stick unfolded - the distance (outer edge to outer edge) of the two yellow reflective tapes is 3ft.

                        Target bar
                        Laser target bar hung magnetically on wall corner. Reflective tape is 1 ft long and entire black bar is 22 inches.

                        Neodymium magnets
                        This is the back of the target bar. The circles are neodymium super magnets affixed with double sided tape. There are also holes in the bar that can be used with push pins, nails, strings, picture hooks, and clamps.

                        Stick with compass
                        This is the stick with compass

                        Distance meter mounted on stick
                        Compass remove and distance meter in place.

                        http://www.azcreative.com

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                        • B Offline
                          bjornkn
                          last edited by

                          @tig said:

                          While we are on 'formulas'...

                          If you know two sides of any triangle and the angle between them then you can find the length of the other side opposite that angle using the Cosine Rule

                          Exactly!
                          But you need to know one angle (which may be 90 degrees) and 2 sides to calculate the length of the third, right?
                          With only 2 sides and no angle you can't know the length of the third side?
                          But if the wall is vertical (90 degrees) you can calculate without using sin/cos, but just the a^2+b^2=c^2?

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                          • B Offline
                            bjornkn
                            last edited by

                            Is there a built-in level in the Bosch laser then?

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                            • R Offline
                              Roger
                              last edited by

                              @bjornkn said:

                              Is there a built-in level in the Bosch laser then?

                              I am designing and building an accessory to attach a level and a small monocular and the whole device will clip to my meter/yard/walking stick.

                              http://www.azcreative.com

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                              • R Offline
                                Roger
                                last edited by

                                @bjornkn said:

                                @roger said:

                                Bjorn, any triangle can be divided into two right triangles, each of which can be solved by pythagoras or sohcahtoa.

                                Yes, as long as you know some angles?
                                Or if you measure from 2 positions?

                                Two angles with the enclosed side or two sides and the enclosed angle. And if you know one angle to begin with and divide the triangle into two right triangles you will know your first angle, your second angle will be the right triangle (90 degrees). Then if you add known angle to 90 degrees and subtract the result from 180 degrees you know the third angle.

                                http://www.azcreative.com

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                                • R Offline
                                  Roger
                                  last edited by

                                  @bjornkn said:

                                  Is there a built-in level in the Bosch laser then?

                                  You did not fully understand the point of Gaieus' diagram and comments. As you sweep the meter up and down the distance reading will change. It starts out long reaches its shortest point and then gets long again. The angle where the distance is shortest is a perpendicular to the wall. When the distance is the shortest, you have a 90 degree angle between the wall and the laser line.

                                  http://www.azcreative.com

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                                  • TIGT Offline
                                    TIG Moderator
                                    last edited by

                                    Also don't forget the 'Similar Triangle' method to get the height of something too tall to measure directly - this uses 'bright-sunlight' because it has to have a shadow - no gadgets required!Capture.PNG

                                    TIG

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                                    • B Offline
                                      bjornkn
                                      last edited by

                                      @roger said:

                                      You did not fully understand the point of Gaieus' diagram and comments. As you sweep the meter up and down the distance reading will change. It starts out long reaches its shortest point and then gets long again. The angle where the distance is shortest is a perpendicular to the wall. When the distance is the shortest, you have a 90 degree angle between the wall and the laser line.

                                      Yes, I understood that. Nice feature πŸ˜„ I was just curious if the Bosch lasers come with a level built-in, as I might be interested in buying one. Could be useful for measuring/calculating a longer reference, for use with tgi3D, than what is easily measured with a 5m tape measure. The longer the reference, the better the precision.
                                      My Android phone has a built-in level (but the back side is slightly curved...) and I know that some laser devices are self-levelling, thus I wouldn't be surprised if the Bosch had a level too. And the iPhone probably have an app too?

                                      Re. angles and measuring distances:
                                      My question was how to calculate a height if the walls are not exactly vertical, based on only distance measurements from one position.
                                      Gaieus' method will work for vertical walls, and the method with pointing upwards should work too, if there is something there to stop the beam, like a roof, or maybe a board laid on top of a fortress wall etc?
                                      Like TIG explained you need the length of the 2 sides + the angle between them to calculate the third side. As long as the triangle does not have a 90 degree angle (like with a non-vertical wall) you cannot use only distances, but need one angle as well.

                                      @Gaieus:
                                      What is the model number of your Bosch?

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                                      • B Offline
                                        bjornkn
                                        last edited by

                                        @tig said:

                                        Also don't forget the 'Similar Triangle' method to get the height of something too tall to measure directly - this uses 'bright-sunlight' because it has to have a shadow - no gadgets required!

                                        Nice, but it does require a vertical wall and a flat ground, which we rarely have here in Norway πŸ˜‰
                                        Like this photo of our local fortress shows..FestningaC_resize.jpg
                                        No flat ground, no vertical walls, and no right angles anywhere at all actually.

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                                        • TIGT Offline
                                          TIG Moderator
                                          last edited by

                                          There's no sun [therefore no shadows] in Norway for half of the year anyway πŸ˜†
                                          How about this make a 'quadrant'
                                          [I know professional 'levels' do it but some duct-tape and bodge never hurt anyone!]
                                          Get an old plastic protractor hold it so the flat edge is at the top and tape a drinking-straw onto the back of it so that it is at 45 degrees and aligned with the central cross-hairs.
                                          Now get a length of string [~50cm] and tape that to the front of it over the cross hairs, hanging down over the curved part. Add a small weight to the end of the string so that it acts like a plumb-bob.
                                          Now if you hold it so that the string aligns with the vertical 90 degree division line on the protractor the drinking-straw 'sight' is at 45 degrees to the horizontal.
                                          You can hold this device but it's much easier to duct-tape it onto a 'staff' about 1.5m long...
                                          You need to walk away from the building to a distance that's an approximation of the height plus a bit.
                                          Place the quadrant on/in the ground with the drinking-straw sight looking at the building and check it's level [string in central on 90 degree marker]. Look through the 'sight' and if you are too near you won't see the top if the building, and you must move away say 1m, if you are too far away you'll see sky and need to move nearer by say 1m. Move back and forth until the required top of the building is dead-center on the 'sight'. Double check for levelness and that the staff is firmly planted. Look along the top [level] edge of the protractor and that will strike the wall in the same horizontal plane. Mark it [an assistant is useful for this but you can set up a ranging-rod against the wall face and estimate up/down along it]. Now run your tape measure from the cross-hairs of the quadrant to the mark on the wall. This gives you the height of the wall above the wall mark. Measure from the wall mark to a fixed level like the floor to give a datum. If the wall is vertical that's it Capture.PNG But if the wall is battered back the true height of the wall above the wall mark is the measured height plus the amount the wall sets back - this is more tricky to establish... [we'll assume an even batter!] you can use the quadrant to estimate this - place the quadrant on the wall base and set it vertical, sight through to a point on the wall and mark it, measure the height of the quadrant above the wall base [h], and the setback of the wall from the center of the vertical quadrant [s]. The ratio is s/h = S/H, where H is the measured-height not actual-height - so the additional height you need to add onto the measured-height is H*s/h...
                                          If the wall is on a hill so that its base will always be above the horizontal plane through your quadrant's first station-point [at the wall's height horizontally away from the wall] then it's more tricky... Instead of a mark on the wall you mark the rising ground and measure horizontally to that, move the quadrant onto that mark and strike a horizontal plane again, if it again hits the ground mark that and measure that and move onto that mark [we are accumulating additional lengths to total up to the total wall height], at some point as we advance towards the wall the mark will be on the wall itself so mark it and measure, now the total distance from adding all of these horizontal distances is the height of the wall above the first station-point - to make adjustments to measure from the wall-base take off the [staff_height * intermediate_stations_number] and add on the distance from the wall-base up to the last wall-mark.

                                          Alternatively by some decent surveying equipment πŸ˜†

                                          TIG

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                                          • GaieusG Offline
                                            Gaieus
                                            last edited by

                                            @tig said:

                                            Also don't forget the 'Similar Triangle' method to get the height of something too tall to measure directly - this uses 'bright-sunlight' because it has to have a shadow - no gadgets required![attachment=1:1akttagz]<!-- ia1 -->Capture.PNG<!-- ia1 -->[/attachment:1akttagz]

                                            Yeah, that's how a Greek (forget which one) measured the height of the pyramids som 2-2.5 millennia ago. And the pyramids do not have vertical walls either - the point is to take the measurement when the sun is low enough for the pyramid to cast shadow (it will be the point which casts the longest shadow anyway)

                                            @bjornkn said:

                                            I was just curious if the Bosch lasers come with a level built-in, as I might be interested in buying one.

                                            Mine does not. It's a PLR 30 (rather simple one) and it only measures to some 30 metres altogether.

                                            But where it measures from is its "bottom", not from the middle as Roger mounted it (don't know Roger if that affects your accuracy)

                                            [attachment=0:1akttagz]<!-- ia0 -->stick_mount_sm.jpg<!-- ia0 -->[/attachment:1akttagz]


                                            stick_mount_sm.jpg

                                            Gai...

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