Finding a curve that lies on the arbitrary sphere given $alpha(0)$ and $alpha'(0)$.












0












$begingroup$



Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






differential-geometry







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




    I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






    differential-geometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




      I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






      differential-geometry







      share|cite|improve this question











      $endgroup$





      Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




      I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






      differential-geometry




      differential-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 '18 at 18:24









      Yadati Kiran

      1,751619




      1,751619










      asked Nov 25 '18 at 18:13









      Beryl1934Beryl1934

      1




      1






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
          $$ gamma: s mapsto pcos s + Zsin s $$
          parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
            $endgroup$
            – Beryl1934
            Nov 25 '18 at 18:58












          • $begingroup$
            Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
            $endgroup$
            – MisterRiemann
            Nov 25 '18 at 19:14





















          0












          $begingroup$

          Let us place here the conditions for future referencing :



          $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



          There is an obvious candidate




          • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


          • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



          $$V(t)=begin{pmatrix}1&0& 0\
          0&cos(a)&-sin(a)\
          0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
          sin(kt)\
          0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
          sin(a)sin(kt)end{pmatrix} tag{2}$$



          We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



          The first condition is clearly fullfilled.



          It remains to comply to 2nd condition (b).



          As :



          $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



          Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



          $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



          Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



          Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



          Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



          Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






          share|cite|improve this answer











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013177%2ffinding-a-curve-that-lies-on-the-arbitrary-sphere-given-alpha0-and-alpha%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14


















            0












            $begingroup$

            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14
















            0












            0








            0





            $begingroup$

            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






            share|cite|improve this answer









            $endgroup$



            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 '18 at 18:20









            MisterRiemannMisterRiemann

            5,8451624




            5,8451624












            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14




















            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14


















            $begingroup$
            That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
            $endgroup$
            – Beryl1934
            Nov 25 '18 at 18:58






            $begingroup$
            That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
            $endgroup$
            – Beryl1934
            Nov 25 '18 at 18:58














            $begingroup$
            Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
            $endgroup$
            – MisterRiemann
            Nov 25 '18 at 19:14






            $begingroup$
            Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
            $endgroup$
            – MisterRiemann
            Nov 25 '18 at 19:14













            0












            $begingroup$

            Let us place here the conditions for future referencing :



            $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



            There is an obvious candidate




            • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


            • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



            $$V(t)=begin{pmatrix}1&0& 0\
            0&cos(a)&-sin(a)\
            0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
            sin(kt)\
            0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
            sin(a)sin(kt)end{pmatrix} tag{2}$$



            We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



            The first condition is clearly fullfilled.



            It remains to comply to 2nd condition (b).



            As :



            $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



            Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



            $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



            Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



            Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



            Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



            Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Let us place here the conditions for future referencing :



              $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



              There is an obvious candidate




              • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


              • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



              $$V(t)=begin{pmatrix}1&0& 0\
              0&cos(a)&-sin(a)\
              0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
              sin(kt)\
              0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
              sin(a)sin(kt)end{pmatrix} tag{2}$$



              We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



              The first condition is clearly fullfilled.



              It remains to comply to 2nd condition (b).



              As :



              $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



              Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



              $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



              Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



              Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



              Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



              Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Let us place here the conditions for future referencing :



                $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



                There is an obvious candidate




                • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


                • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



                $$V(t)=begin{pmatrix}1&0& 0\
                0&cos(a)&-sin(a)\
                0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
                sin(kt)\
                0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
                sin(a)sin(kt)end{pmatrix} tag{2}$$



                We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



                The first condition is clearly fullfilled.



                It remains to comply to 2nd condition (b).



                As :



                $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



                Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



                $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



                Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



                Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



                Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



                Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






                share|cite|improve this answer











                $endgroup$



                Let us place here the conditions for future referencing :



                $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



                There is an obvious candidate




                • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


                • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



                $$V(t)=begin{pmatrix}1&0& 0\
                0&cos(a)&-sin(a)\
                0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
                sin(kt)\
                0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
                sin(a)sin(kt)end{pmatrix} tag{2}$$



                We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



                The first condition is clearly fullfilled.



                It remains to comply to 2nd condition (b).



                As :



                $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



                Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



                $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



                Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



                Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



                Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



                Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 26 '18 at 9:08

























                answered Nov 26 '18 at 2:03









                Jean MarieJean Marie

                28.9k42049




                28.9k42049






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013177%2ffinding-a-curve-that-lies-on-the-arbitrary-sphere-given-alpha0-and-alpha%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                    Image
                    2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
                    Read more

                    ComboBox Display Member on multiple fields

                    Image
                    2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
                    Read more

                    Is it possible to collect Nectar points via Trainline?

                    Image
                    2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
                    Read more