Computation logic of Partway in TikZ

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3















PGF Manual pg. 66 describes about Partway Calculations.



enter image description here



In this example, ($ (A)!.5!(B) $), .5 refers to 50% right? A point that is 50% away from point A. In that case, why do we require point B in the calculation. We can always use the first point for relative positioning. The comparison is made to xcolor package, which is not right (e.g orange!50!black. Here, the color would be 50% of orange and 50% of black. The second color is mandatory.) This comparison is not right and I would like to understand the purpose of providing the second point in the calculation.



Secondly, the computation of D is also not convincing:



enter image description here



After computing X, we can simply say, place point D above X at a distance of 6pts. It should also be equivalent to:



($ (X) ! sin(60)*2 ! 90:(B) $) right?



Please clarify and help me understand this tricky concept.










share|improve this question


























    3















    PGF Manual pg. 66 describes about Partway Calculations.



    enter image description here



    In this example, ($ (A)!.5!(B) $), .5 refers to 50% right? A point that is 50% away from point A. In that case, why do we require point B in the calculation. We can always use the first point for relative positioning. The comparison is made to xcolor package, which is not right (e.g orange!50!black. Here, the color would be 50% of orange and 50% of black. The second color is mandatory.) This comparison is not right and I would like to understand the purpose of providing the second point in the calculation.



    Secondly, the computation of D is also not convincing:



    enter image description here



    After computing X, we can simply say, place point D above X at a distance of 6pts. It should also be equivalent to:



    ($ (X) ! sin(60)*2 ! 90:(B) $) right?



    Please clarify and help me understand this tricky concept.










    share|improve this question
























      3












      3








      3








      PGF Manual pg. 66 describes about Partway Calculations.



      enter image description here



      In this example, ($ (A)!.5!(B) $), .5 refers to 50% right? A point that is 50% away from point A. In that case, why do we require point B in the calculation. We can always use the first point for relative positioning. The comparison is made to xcolor package, which is not right (e.g orange!50!black. Here, the color would be 50% of orange and 50% of black. The second color is mandatory.) This comparison is not right and I would like to understand the purpose of providing the second point in the calculation.



      Secondly, the computation of D is also not convincing:



      enter image description here



      After computing X, we can simply say, place point D above X at a distance of 6pts. It should also be equivalent to:



      ($ (X) ! sin(60)*2 ! 90:(B) $) right?



      Please clarify and help me understand this tricky concept.










      share|improve this question














      PGF Manual pg. 66 describes about Partway Calculations.



      enter image description here



      In this example, ($ (A)!.5!(B) $), .5 refers to 50% right? A point that is 50% away from point A. In that case, why do we require point B in the calculation. We can always use the first point for relative positioning. The comparison is made to xcolor package, which is not right (e.g orange!50!black. Here, the color would be 50% of orange and 50% of black. The second color is mandatory.) This comparison is not right and I would like to understand the purpose of providing the second point in the calculation.



      Secondly, the computation of D is also not convincing:



      enter image description here



      After computing X, we can simply say, place point D above X at a distance of 6pts. It should also be equivalent to:



      ($ (X) ! sin(60)*2 ! 90:(B) $) right?



      Please clarify and help me understand this tricky concept.







      tikz-pgf






      share|improve this question













      share|improve this question











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      asked Mar 9 at 13:12









      subham sonisubham soni

      4,91683187




      4,91683187




















          3 Answers
          3






          active

          oldest

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          7














          Why do we need (B) at the (X) calculation?



          Look at this code



          documentclass[tikz]standalone
          usetikzlibrarycalc
          begindocument
          begintikzpicture
          coordinate [label=left:$A$] (A) at (0,0);
          coordinate [label=right:$B$] (B) at (1.25,0.25);
          draw (A) -- (B);
          node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ; % (1)
          coordinate[label=above:$C$] (C) at (1,2);
          draw (A) -- (C);
          node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ; % (2)
          endtikzpicture
          enddocument


          enter image description here



          If (B) were unnecessary, (1) and (2) must have the same effect, i.e. (X) and (Y) must be the same. In fact, they aren't. Therefore (B) is necessary.



          For more information, (B) stands for the ending point of the segment on which (X) is marked.



          For the second code



          Look at this



          documentclass[tikz]standalone
          usetikzlibrarycalc
          begindocument
          begintikzpicture
          coordinate [label=left:$A$] (A) at (0,0);
          coordinate [label=right:$B$] (B) at (1.25,0.25);
          draw (A) -- (B);
          node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ;
          node [fill=red,inner sep=1pt,label=above:$D$] (D) at ($(X)!sin(60)*2!90:(B)$) ;
          coordinate[label=right:$C$] (C) at (1,-2);
          draw (A) -- (C);
          node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ;
          node [fill=red,inner sep=1pt,label=below:$E$] (E) at ($(Y)!sin(60)*2!90:(C)$) ;
          endtikzpicture
          enddocument


          enter image description here



          Again the same reason as above: (B) must be crucial otherwise (D) and (E) must be the same.






          share|improve this answer























          • Joulev we can use above and other anchors right Instead of the sin calculation. That would make it simpler. That is my actual query for the second question

            – subham soni
            Mar 9 at 14:00






          • 1





            @subhamsoni Of course using sin is not the only way to draw that. As the section in the manual is about Partway Calculations, the author uses sin in the example. In my case I would not use sin.

            – JouleV
            Mar 9 at 14:02


















          7














          The purpose of the TikZ manual tutorials is to introduce the most useful concepts and library to get started with TikZ. And not to give the best way to build the figures.



          First question: In that case, why do we require point B in the calculation?



          Because in plane geometry, there is an infinity of points located at the same distance from a given point: these are the points of a circle.



          By example, in the figure below, points M and N are located 2 cm from the centre of the circle. It is therefore necessary to indicate a second point B.



          screenshot



          Second question: Please clarify and help me understand this tricky concept



          This passage illustrates the use of the calc library. For this example, he uses geometric relationships in the equilateral triangle to place point D.



          Personally, I would have built this triangle using polar coordinates, then rotating.
          The aim here is not to make it as simple as possible, but to present TikZ's main concepts and libraries.



          screenshot



          documentclass[tikz,border=5mm]standalone
          usetikzlibrarycalc
          begindocument

          begintikzpicture
          draw[rotate=10] (0,0)coordinate(A)--(2,0)coordinate(B)--(60:2)coordinate(D)--cycle;
          node[left] at (A)A;
          node[right] at (B)B;
          node [above] at (D) D;
          endtikzpicture
          enddocument


          Translated with www.DeepL.com/Translator






          share|improve this answer
































            6















            In this example, ($ (A)!.5!(B) $), .5 refers to 50% right?




            Correct. It's simply another notation.




            A point that is 50% away from point A. In that case, why do we require point B in the calculation.




            Wrong. If you take the path from A to B, then the 50% denote that you move only 50% along that path (starting at point A) maintaining the direction. Or even easier: Take the vector (b-a), scale it with 0.5 and add it to a. Then you have your new coordinate.



            Without the second part the direction would be unspecified. That is, you would not know whether you should move upwards or sidewards or downwards. In turn you could not uniquely identify the destination which would not be what you want when drawing a picture.






            share|improve this answer

























            • thanks. Can you please clarify the second point.

              – subham soni
              Mar 9 at 13:27











            • @subhamsoni Without knowing the direction a distance says nothing. Should (a)!.5 point upwards, sidewards, …? That wouldn't be clear.

              – TeXnician
              Mar 9 at 15:01











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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7














            Why do we need (B) at the (X) calculation?



            Look at this code



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ; % (1)
            coordinate[label=above:$C$] (C) at (1,2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ; % (2)
            endtikzpicture
            enddocument


            enter image description here



            If (B) were unnecessary, (1) and (2) must have the same effect, i.e. (X) and (Y) must be the same. In fact, they aren't. Therefore (B) is necessary.



            For more information, (B) stands for the ending point of the segment on which (X) is marked.



            For the second code



            Look at this



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ;
            node [fill=red,inner sep=1pt,label=above:$D$] (D) at ($(X)!sin(60)*2!90:(B)$) ;
            coordinate[label=right:$C$] (C) at (1,-2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ;
            node [fill=red,inner sep=1pt,label=below:$E$] (E) at ($(Y)!sin(60)*2!90:(C)$) ;
            endtikzpicture
            enddocument


            enter image description here



            Again the same reason as above: (B) must be crucial otherwise (D) and (E) must be the same.






            share|improve this answer























            • Joulev we can use above and other anchors right Instead of the sin calculation. That would make it simpler. That is my actual query for the second question

              – subham soni
              Mar 9 at 14:00






            • 1





              @subhamsoni Of course using sin is not the only way to draw that. As the section in the manual is about Partway Calculations, the author uses sin in the example. In my case I would not use sin.

              – JouleV
              Mar 9 at 14:02















            7














            Why do we need (B) at the (X) calculation?



            Look at this code



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ; % (1)
            coordinate[label=above:$C$] (C) at (1,2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ; % (2)
            endtikzpicture
            enddocument


            enter image description here



            If (B) were unnecessary, (1) and (2) must have the same effect, i.e. (X) and (Y) must be the same. In fact, they aren't. Therefore (B) is necessary.



            For more information, (B) stands for the ending point of the segment on which (X) is marked.



            For the second code



            Look at this



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ;
            node [fill=red,inner sep=1pt,label=above:$D$] (D) at ($(X)!sin(60)*2!90:(B)$) ;
            coordinate[label=right:$C$] (C) at (1,-2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ;
            node [fill=red,inner sep=1pt,label=below:$E$] (E) at ($(Y)!sin(60)*2!90:(C)$) ;
            endtikzpicture
            enddocument


            enter image description here



            Again the same reason as above: (B) must be crucial otherwise (D) and (E) must be the same.






            share|improve this answer























            • Joulev we can use above and other anchors right Instead of the sin calculation. That would make it simpler. That is my actual query for the second question

              – subham soni
              Mar 9 at 14:00






            • 1





              @subhamsoni Of course using sin is not the only way to draw that. As the section in the manual is about Partway Calculations, the author uses sin in the example. In my case I would not use sin.

              – JouleV
              Mar 9 at 14:02













            7












            7








            7







            Why do we need (B) at the (X) calculation?



            Look at this code



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ; % (1)
            coordinate[label=above:$C$] (C) at (1,2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ; % (2)
            endtikzpicture
            enddocument


            enter image description here



            If (B) were unnecessary, (1) and (2) must have the same effect, i.e. (X) and (Y) must be the same. In fact, they aren't. Therefore (B) is necessary.



            For more information, (B) stands for the ending point of the segment on which (X) is marked.



            For the second code



            Look at this



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ;
            node [fill=red,inner sep=1pt,label=above:$D$] (D) at ($(X)!sin(60)*2!90:(B)$) ;
            coordinate[label=right:$C$] (C) at (1,-2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ;
            node [fill=red,inner sep=1pt,label=below:$E$] (E) at ($(Y)!sin(60)*2!90:(C)$) ;
            endtikzpicture
            enddocument


            enter image description here



            Again the same reason as above: (B) must be crucial otherwise (D) and (E) must be the same.






            share|improve this answer













            Why do we need (B) at the (X) calculation?



            Look at this code



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ; % (1)
            coordinate[label=above:$C$] (C) at (1,2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ; % (2)
            endtikzpicture
            enddocument


            enter image description here



            If (B) were unnecessary, (1) and (2) must have the same effect, i.e. (X) and (Y) must be the same. In fact, they aren't. Therefore (B) is necessary.



            For more information, (B) stands for the ending point of the segment on which (X) is marked.



            For the second code



            Look at this



            documentclass[tikz]standalone
            usetikzlibrarycalc
            begindocument
            begintikzpicture
            coordinate [label=left:$A$] (A) at (0,0);
            coordinate [label=right:$B$] (B) at (1.25,0.25);
            draw (A) -- (B);
            node [fill=red,inner sep=1pt,label=below:$X$] (X) at ($(A)!.5!(B)$) ;
            node [fill=red,inner sep=1pt,label=above:$D$] (D) at ($(X)!sin(60)*2!90:(B)$) ;
            coordinate[label=right:$C$] (C) at (1,-2);
            draw (A) -- (C);
            node [fill=red,inner sep=1pt,label=left:$Y$] (Y) at ($(A)!.5!(C)$) ;
            node [fill=red,inner sep=1pt,label=below:$E$] (E) at ($(Y)!sin(60)*2!90:(C)$) ;
            endtikzpicture
            enddocument


            enter image description here



            Again the same reason as above: (B) must be crucial otherwise (D) and (E) must be the same.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 9 at 13:35









            JouleVJouleV

            11.2k22560




            11.2k22560












            • Joulev we can use above and other anchors right Instead of the sin calculation. That would make it simpler. That is my actual query for the second question

              – subham soni
              Mar 9 at 14:00






            • 1





              @subhamsoni Of course using sin is not the only way to draw that. As the section in the manual is about Partway Calculations, the author uses sin in the example. In my case I would not use sin.

              – JouleV
              Mar 9 at 14:02

















            • Joulev we can use above and other anchors right Instead of the sin calculation. That would make it simpler. That is my actual query for the second question

              – subham soni
              Mar 9 at 14:00






            • 1





              @subhamsoni Of course using sin is not the only way to draw that. As the section in the manual is about Partway Calculations, the author uses sin in the example. In my case I would not use sin.

              – JouleV
              Mar 9 at 14:02
















            Joulev we can use above and other anchors right Instead of the sin calculation. That would make it simpler. That is my actual query for the second question

            – subham soni
            Mar 9 at 14:00





            Joulev we can use above and other anchors right Instead of the sin calculation. That would make it simpler. That is my actual query for the second question

            – subham soni
            Mar 9 at 14:00




            1




            1





            @subhamsoni Of course using sin is not the only way to draw that. As the section in the manual is about Partway Calculations, the author uses sin in the example. In my case I would not use sin.

            – JouleV
            Mar 9 at 14:02





            @subhamsoni Of course using sin is not the only way to draw that. As the section in the manual is about Partway Calculations, the author uses sin in the example. In my case I would not use sin.

            – JouleV
            Mar 9 at 14:02











            7














            The purpose of the TikZ manual tutorials is to introduce the most useful concepts and library to get started with TikZ. And not to give the best way to build the figures.



            First question: In that case, why do we require point B in the calculation?



            Because in plane geometry, there is an infinity of points located at the same distance from a given point: these are the points of a circle.



            By example, in the figure below, points M and N are located 2 cm from the centre of the circle. It is therefore necessary to indicate a second point B.



            screenshot



            Second question: Please clarify and help me understand this tricky concept



            This passage illustrates the use of the calc library. For this example, he uses geometric relationships in the equilateral triangle to place point D.



            Personally, I would have built this triangle using polar coordinates, then rotating.
            The aim here is not to make it as simple as possible, but to present TikZ's main concepts and libraries.



            screenshot



            documentclass[tikz,border=5mm]standalone
            usetikzlibrarycalc
            begindocument

            begintikzpicture
            draw[rotate=10] (0,0)coordinate(A)--(2,0)coordinate(B)--(60:2)coordinate(D)--cycle;
            node[left] at (A)A;
            node[right] at (B)B;
            node [above] at (D) D;
            endtikzpicture
            enddocument


            Translated with www.DeepL.com/Translator






            share|improve this answer





























              7














              The purpose of the TikZ manual tutorials is to introduce the most useful concepts and library to get started with TikZ. And not to give the best way to build the figures.



              First question: In that case, why do we require point B in the calculation?



              Because in plane geometry, there is an infinity of points located at the same distance from a given point: these are the points of a circle.



              By example, in the figure below, points M and N are located 2 cm from the centre of the circle. It is therefore necessary to indicate a second point B.



              screenshot



              Second question: Please clarify and help me understand this tricky concept



              This passage illustrates the use of the calc library. For this example, he uses geometric relationships in the equilateral triangle to place point D.



              Personally, I would have built this triangle using polar coordinates, then rotating.
              The aim here is not to make it as simple as possible, but to present TikZ's main concepts and libraries.



              screenshot



              documentclass[tikz,border=5mm]standalone
              usetikzlibrarycalc
              begindocument

              begintikzpicture
              draw[rotate=10] (0,0)coordinate(A)--(2,0)coordinate(B)--(60:2)coordinate(D)--cycle;
              node[left] at (A)A;
              node[right] at (B)B;
              node [above] at (D) D;
              endtikzpicture
              enddocument


              Translated with www.DeepL.com/Translator






              share|improve this answer



























                7












                7








                7







                The purpose of the TikZ manual tutorials is to introduce the most useful concepts and library to get started with TikZ. And not to give the best way to build the figures.



                First question: In that case, why do we require point B in the calculation?



                Because in plane geometry, there is an infinity of points located at the same distance from a given point: these are the points of a circle.



                By example, in the figure below, points M and N are located 2 cm from the centre of the circle. It is therefore necessary to indicate a second point B.



                screenshot



                Second question: Please clarify and help me understand this tricky concept



                This passage illustrates the use of the calc library. For this example, he uses geometric relationships in the equilateral triangle to place point D.



                Personally, I would have built this triangle using polar coordinates, then rotating.
                The aim here is not to make it as simple as possible, but to present TikZ's main concepts and libraries.



                screenshot



                documentclass[tikz,border=5mm]standalone
                usetikzlibrarycalc
                begindocument

                begintikzpicture
                draw[rotate=10] (0,0)coordinate(A)--(2,0)coordinate(B)--(60:2)coordinate(D)--cycle;
                node[left] at (A)A;
                node[right] at (B)B;
                node [above] at (D) D;
                endtikzpicture
                enddocument


                Translated with www.DeepL.com/Translator






                share|improve this answer















                The purpose of the TikZ manual tutorials is to introduce the most useful concepts and library to get started with TikZ. And not to give the best way to build the figures.



                First question: In that case, why do we require point B in the calculation?



                Because in plane geometry, there is an infinity of points located at the same distance from a given point: these are the points of a circle.



                By example, in the figure below, points M and N are located 2 cm from the centre of the circle. It is therefore necessary to indicate a second point B.



                screenshot



                Second question: Please clarify and help me understand this tricky concept



                This passage illustrates the use of the calc library. For this example, he uses geometric relationships in the equilateral triangle to place point D.



                Personally, I would have built this triangle using polar coordinates, then rotating.
                The aim here is not to make it as simple as possible, but to present TikZ's main concepts and libraries.



                screenshot



                documentclass[tikz,border=5mm]standalone
                usetikzlibrarycalc
                begindocument

                begintikzpicture
                draw[rotate=10] (0,0)coordinate(A)--(2,0)coordinate(B)--(60:2)coordinate(D)--cycle;
                node[left] at (A)A;
                node[right] at (B)B;
                node [above] at (D) D;
                endtikzpicture
                enddocument


                Translated with www.DeepL.com/Translator







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Mar 10 at 6:38

























                answered Mar 9 at 17:16









                AndréCAndréC

                10.5k11548




                10.5k11548





















                    6















                    In this example, ($ (A)!.5!(B) $), .5 refers to 50% right?




                    Correct. It's simply another notation.




                    A point that is 50% away from point A. In that case, why do we require point B in the calculation.




                    Wrong. If you take the path from A to B, then the 50% denote that you move only 50% along that path (starting at point A) maintaining the direction. Or even easier: Take the vector (b-a), scale it with 0.5 and add it to a. Then you have your new coordinate.



                    Without the second part the direction would be unspecified. That is, you would not know whether you should move upwards or sidewards or downwards. In turn you could not uniquely identify the destination which would not be what you want when drawing a picture.






                    share|improve this answer

























                    • thanks. Can you please clarify the second point.

                      – subham soni
                      Mar 9 at 13:27











                    • @subhamsoni Without knowing the direction a distance says nothing. Should (a)!.5 point upwards, sidewards, …? That wouldn't be clear.

                      – TeXnician
                      Mar 9 at 15:01















                    6















                    In this example, ($ (A)!.5!(B) $), .5 refers to 50% right?




                    Correct. It's simply another notation.




                    A point that is 50% away from point A. In that case, why do we require point B in the calculation.




                    Wrong. If you take the path from A to B, then the 50% denote that you move only 50% along that path (starting at point A) maintaining the direction. Or even easier: Take the vector (b-a), scale it with 0.5 and add it to a. Then you have your new coordinate.



                    Without the second part the direction would be unspecified. That is, you would not know whether you should move upwards or sidewards or downwards. In turn you could not uniquely identify the destination which would not be what you want when drawing a picture.






                    share|improve this answer

























                    • thanks. Can you please clarify the second point.

                      – subham soni
                      Mar 9 at 13:27











                    • @subhamsoni Without knowing the direction a distance says nothing. Should (a)!.5 point upwards, sidewards, …? That wouldn't be clear.

                      – TeXnician
                      Mar 9 at 15:01













                    6












                    6








                    6








                    In this example, ($ (A)!.5!(B) $), .5 refers to 50% right?




                    Correct. It's simply another notation.




                    A point that is 50% away from point A. In that case, why do we require point B in the calculation.




                    Wrong. If you take the path from A to B, then the 50% denote that you move only 50% along that path (starting at point A) maintaining the direction. Or even easier: Take the vector (b-a), scale it with 0.5 and add it to a. Then you have your new coordinate.



                    Without the second part the direction would be unspecified. That is, you would not know whether you should move upwards or sidewards or downwards. In turn you could not uniquely identify the destination which would not be what you want when drawing a picture.






                    share|improve this answer
















                    In this example, ($ (A)!.5!(B) $), .5 refers to 50% right?




                    Correct. It's simply another notation.




                    A point that is 50% away from point A. In that case, why do we require point B in the calculation.




                    Wrong. If you take the path from A to B, then the 50% denote that you move only 50% along that path (starting at point A) maintaining the direction. Or even easier: Take the vector (b-a), scale it with 0.5 and add it to a. Then you have your new coordinate.



                    Without the second part the direction would be unspecified. That is, you would not know whether you should move upwards or sidewards or downwards. In turn you could not uniquely identify the destination which would not be what you want when drawing a picture.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Mar 10 at 10:22

























                    answered Mar 9 at 13:22









                    TeXnicianTeXnician

                    25.9k63491




                    25.9k63491












                    • thanks. Can you please clarify the second point.

                      – subham soni
                      Mar 9 at 13:27











                    • @subhamsoni Without knowing the direction a distance says nothing. Should (a)!.5 point upwards, sidewards, …? That wouldn't be clear.

                      – TeXnician
                      Mar 9 at 15:01

















                    • thanks. Can you please clarify the second point.

                      – subham soni
                      Mar 9 at 13:27











                    • @subhamsoni Without knowing the direction a distance says nothing. Should (a)!.5 point upwards, sidewards, …? That wouldn't be clear.

                      – TeXnician
                      Mar 9 at 15:01
















                    thanks. Can you please clarify the second point.

                    – subham soni
                    Mar 9 at 13:27





                    thanks. Can you please clarify the second point.

                    – subham soni
                    Mar 9 at 13:27













                    @subhamsoni Without knowing the direction a distance says nothing. Should (a)!.5 point upwards, sidewards, …? That wouldn't be clear.

                    – TeXnician
                    Mar 9 at 15:01





                    @subhamsoni Without knowing the direction a distance says nothing. Should (a)!.5 point upwards, sidewards, …? That wouldn't be clear.

                    – TeXnician
                    Mar 9 at 15:01

















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