Extra distance travelled along a sine wave path












3












$begingroup$


Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).



If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?



What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?






geometry curves







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).



    If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?



    What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?






    geometry curves







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).



      If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?



      What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?






      geometry curves







      share|cite|improve this question









      $endgroup$




      Let's say I have a blue line which is 10 metres long. I then draw a single cycle of a sine curve along this line, in red which has a maximum distance of 0.1 meter from the line (so an amplitude of 0.1m).



      If I walk along the sine curve, I will walk further than 10 metres. But how do I calculate this new distance travelled?



      What about if I increase the frequency so that there are five full cycles along my 10 metre blue line? How does the frequency affect it, in other words?






      geometry curves




      geometry curves






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 3 '17 at 16:33









      Max WilliamsMax Williams

      1162




      1162






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          HINT



          This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.



          Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.



          If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.



          $$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$



          Solve for $ (a,alpha, e)$.



          The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.



          EDIT1:



          Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:



          $$ y_1= A sin dfrac{2 pi x}{lambda}$$



          $$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
              $endgroup$
              – Max Williams
              Nov 3 '17 at 16:40










            • $begingroup$
              Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
              $endgroup$
              – imranfat
              Nov 3 '17 at 16:43












            • $begingroup$
              I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
              $endgroup$
              – Max Williams
              Nov 3 '17 at 16:47












            • $begingroup$
              My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
              $endgroup$
              – imranfat
              Nov 3 '17 at 16:49












            • $begingroup$
              Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
              $endgroup$
              – Max Williams
              Nov 3 '17 at 16:53











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            HINT



            This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.



            Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.



            If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.



            $$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$



            Solve for $ (a,alpha, e)$.



            The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.



            EDIT1:



            Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:



            $$ y_1= A sin dfrac{2 pi x}{lambda}$$



            $$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              HINT



              This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.



              Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.



              If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.



              $$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$



              Solve for $ (a,alpha, e)$.



              The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.



              EDIT1:



              Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:



              $$ y_1= A sin dfrac{2 pi x}{lambda}$$



              $$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                HINT



                This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.



                Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.



                If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.



                $$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$



                Solve for $ (a,alpha, e)$.



                The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.



                EDIT1:



                Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:



                $$ y_1= A sin dfrac{2 pi x}{lambda}$$



                $$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$






                share|cite|improve this answer











                $endgroup$



                HINT



                This answer may not help directly for higher frequencies/waves but stating it nevertheless.., the following requires some imagination.



                Sinusoidal waves are development of slantly cut cylinder intersections seen as surface development.



                If a cylinder radius $a$ is cut at an angle $alpha$ to radial plane the ellipse axes are $ (a, a sec alpha) $ then the ellipse has an eccentricity $ e= sin alpha$.



                $$ 2 pi a = 10 m, quad a tan alpha = dfrac{.01}{2} ; $$



                Solve for $ (a,alpha, e)$.



                The perimeter is $ 4 a , E(e)$ which a standard result using Elliptic integrals; it can be found by integration or google it for a start.



                EDIT1:



                Useful to remember the waveform equations of same amplitude $A$ for single and multiple frequencies $n$:



                $$ y_1= A sin dfrac{2 pi x}{lambda}$$



                $$ y_n= A sin dfrac{2 pi xcdot n}{lambda}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 7 '17 at 7:09

























                answered Nov 3 '17 at 17:14









                NarasimhamNarasimham

                20.8k52158




                20.8k52158























                    1












                    $begingroup$

                    You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:40










                    • $begingroup$
                      Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:43












                    • $begingroup$
                      I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:47












                    • $begingroup$
                      My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:49












                    • $begingroup$
                      Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:53
















                    1












                    $begingroup$

                    You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:40










                    • $begingroup$
                      Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:43












                    • $begingroup$
                      I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:47












                    • $begingroup$
                      My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:49












                    • $begingroup$
                      Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:53














                    1












                    1








                    1





                    $begingroup$

                    You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example






                    share|cite|improve this answer









                    $endgroup$



                    You need the arc length formula for this: $intsqrt{1+f'(x)^2}dx$ on interval $[0,10]$. What you need to do is establish the formula of your curve, which is not hard to do. After all, you know the period (then use $b=frac{2pi}{period}$) and the amplitude as you stated. Anyway, the real problem is that the arc length formula is not going to come out "nice", in the sense that I greatly expect that you are going to deal with an integral that cannot be evaluated in terms of elementary functions. You are then left with an approximation using an TI for example







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 3 '17 at 16:38









                    imranfatimranfat

                    8,03141532




                    8,03141532












                    • $begingroup$
                      Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:40










                    • $begingroup$
                      Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:43












                    • $begingroup$
                      I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:47












                    • $begingroup$
                      My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:49












                    • $begingroup$
                      Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:53


















                    • $begingroup$
                      Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:40










                    • $begingroup$
                      Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:43












                    • $begingroup$
                      I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:47












                    • $begingroup$
                      My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
                      $endgroup$
                      – imranfat
                      Nov 3 '17 at 16:49












                    • $begingroup$
                      Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
                      $endgroup$
                      – Max Williams
                      Nov 3 '17 at 16:53
















                    $begingroup$
                    Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
                    $endgroup$
                    – Max Williams
                    Nov 3 '17 at 16:40




                    $begingroup$
                    Thanks - is there a non-mathematicians' version? If it approximates the distance then that's fine.
                    $endgroup$
                    – Max Williams
                    Nov 3 '17 at 16:40












                    $begingroup$
                    Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
                    $endgroup$
                    – imranfat
                    Nov 3 '17 at 16:43






                    $begingroup$
                    Of course there is a less "stringent" mathematics version. How much less mathematically you want to go?. You can "avoid" calculus in the following manner: Use increments of $0.1$ for $x$ and calculate the corresponding $y's$ for each increasing $x$. Then use Pythagorean theorem $100$ times $(10/0.1)$ and add up distances. That would be a good approximation. It's a little work but not difficult. You still need to set up your sine formula, also not difficult. (This answers the first part of your question)
                    $endgroup$
                    – imranfat
                    Nov 3 '17 at 16:43














                    $begingroup$
                    I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
                    $endgroup$
                    – Max Williams
                    Nov 3 '17 at 16:47






                    $begingroup$
                    I was hoping for a formula which doesn't require calculus (by which i mean, for example, integrating functions - sorry if that's not calculus, i'm not a mathematician), and into which I can just plug some numbers and carry out only simple operations. Imagine I'm trying to do it in a computer program where I can only do basic operations, raising to a power, sin, cos and tan and their converses.
                    $endgroup$
                    – Max Williams
                    Nov 3 '17 at 16:47














                    $begingroup$
                    My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
                    $endgroup$
                    – imranfat
                    Nov 3 '17 at 16:49






                    $begingroup$
                    My second approach in my previous comment does not require calculus. The Pythagorean theorem is what you need (remember that famous $a^2+b^2=c^2$ on a right triangle from high school?). I think that is as basic as you can go. To calculate a distance between two points, the formula translates to $c=sqrt{(0.1)^2+(y_2-y_1)^2}$. You need to do that 100 times as you have 100 increments. Then add up. Having something like Excel available would be tremendously helpful
                    $endgroup$
                    – imranfat
                    Nov 3 '17 at 16:49














                    $begingroup$
                    Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
                    $endgroup$
                    – Max Williams
                    Nov 3 '17 at 16:53




                    $begingroup$
                    Ok, thanks. I guess I was too optimistic for a simple one-hit equation then!
                    $endgroup$
                    – Max Williams
                    Nov 3 '17 at 16:53


















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                    Mario Kart Wii

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                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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                    The Binding of Isaac: Rebirth/Afterbirth

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                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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                    What does “Dominus providebit” mean?

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                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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