float/double Math.Round in C#












12















float ff = (float)31.15;



double dd = 31.15;



var frst = Math.Round(ff, 1, MidpointRounding.AwayFromZero);



var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


frst: 31.1



drst: 31.2



Can someone explain why?






c# rounding







share|improve this question




















  • 2





    Because (float)31.15 is not equal to (double)31.15. Floating-point-arithmetic allmost allways yields to rounding-erros. In paticular rounding a double works different from rounding a float.

    – HimBromBeere
    2 hours ago








  • 1





    Rounding errors are unavoidable with floating point values. They are as unavoidable as death and taxes : youtube.com/watch?v=PZRI1IfStY0

    – Christopher
    1 hour ago






  • 1





    Possible duplicate of Difference between decimal, float and double in .NET?

    – Jan S.
    1 hour ago






  • 3





    @i486 Well, not always, there is a reason why float exists.

    – SeM
    1 hour ago






  • 2





    @i486 Again, that doesn't mean, that you should never use float.

    – SeM
    56 mins ago
















12















float ff = (float)31.15;



double dd = 31.15;



var frst = Math.Round(ff, 1, MidpointRounding.AwayFromZero);



var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


frst: 31.1



drst: 31.2



Can someone explain why?






c# rounding







share|improve this question




















  • 2





    Because (float)31.15 is not equal to (double)31.15. Floating-point-arithmetic allmost allways yields to rounding-erros. In paticular rounding a double works different from rounding a float.

    – HimBromBeere
    2 hours ago








  • 1





    Rounding errors are unavoidable with floating point values. They are as unavoidable as death and taxes : youtube.com/watch?v=PZRI1IfStY0

    – Christopher
    1 hour ago






  • 1





    Possible duplicate of Difference between decimal, float and double in .NET?

    – Jan S.
    1 hour ago






  • 3





    @i486 Well, not always, there is a reason why float exists.

    – SeM
    1 hour ago






  • 2





    @i486 Again, that doesn't mean, that you should never use float.

    – SeM
    56 mins ago














12












12








12


2






float ff = (float)31.15;



double dd = 31.15;



var frst = Math.Round(ff, 1, MidpointRounding.AwayFromZero);



var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


frst: 31.1



drst: 31.2



Can someone explain why?






c# rounding







share|improve this question
















float ff = (float)31.15;



double dd = 31.15;



var frst = Math.Round(ff, 1, MidpointRounding.AwayFromZero);



var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


frst: 31.1



drst: 31.2



Can someone explain why?






c# rounding




c# rounding






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









Dmitry Bychenko

107k1092133




107k1092133










asked 2 hours ago









YaoqingYaoqing

902




902








  • 2





    Because (float)31.15 is not equal to (double)31.15. Floating-point-arithmetic allmost allways yields to rounding-erros. In paticular rounding a double works different from rounding a float.

    – HimBromBeere
    2 hours ago








  • 1





    Rounding errors are unavoidable with floating point values. They are as unavoidable as death and taxes : youtube.com/watch?v=PZRI1IfStY0

    – Christopher
    1 hour ago






  • 1





    Possible duplicate of Difference between decimal, float and double in .NET?

    – Jan S.
    1 hour ago






  • 3





    @i486 Well, not always, there is a reason why float exists.

    – SeM
    1 hour ago






  • 2





    @i486 Again, that doesn't mean, that you should never use float.

    – SeM
    56 mins ago














  • 2





    Because (float)31.15 is not equal to (double)31.15. Floating-point-arithmetic allmost allways yields to rounding-erros. In paticular rounding a double works different from rounding a float.

    – HimBromBeere
    2 hours ago








  • 1





    Rounding errors are unavoidable with floating point values. They are as unavoidable as death and taxes : youtube.com/watch?v=PZRI1IfStY0

    – Christopher
    1 hour ago






  • 1





    Possible duplicate of Difference between decimal, float and double in .NET?

    – Jan S.
    1 hour ago






  • 3





    @i486 Well, not always, there is a reason why float exists.

    – SeM
    1 hour ago






  • 2





    @i486 Again, that doesn't mean, that you should never use float.

    – SeM
    56 mins ago








2




2





Because (float)31.15 is not equal to (double)31.15. Floating-point-arithmetic allmost allways yields to rounding-erros. In paticular rounding a double works different from rounding a float.

– HimBromBeere
2 hours ago







Because (float)31.15 is not equal to (double)31.15. Floating-point-arithmetic allmost allways yields to rounding-erros. In paticular rounding a double works different from rounding a float.

– HimBromBeere
2 hours ago






1




1





Rounding errors are unavoidable with floating point values. They are as unavoidable as death and taxes : youtube.com/watch?v=PZRI1IfStY0

– Christopher
1 hour ago





Rounding errors are unavoidable with floating point values. They are as unavoidable as death and taxes : youtube.com/watch?v=PZRI1IfStY0

– Christopher
1 hour ago




1




1





Possible duplicate of Difference between decimal, float and double in .NET?

– Jan S.
1 hour ago





Possible duplicate of Difference between decimal, float and double in .NET?

– Jan S.
1 hour ago




3




3





@i486 Well, not always, there is a reason why float exists.

– SeM
1 hour ago





@i486 Well, not always, there is a reason why float exists.

– SeM
1 hour ago




2




2





@i486 Again, that doesn't mean, that you should never use float.

– SeM
56 mins ago





@i486 Again, that doesn't mean, that you should never use float.

– SeM
56 mins ago












3 Answers
3






active

oldest

votes


















17














Well, Math.Round wants double, not float, that's why



Math.Round(ff, 1, MidpointRounding.AwayFromZero);


equals to 



Math.Round((double)ff, 1, MidpointRounding.AwayFromZero);


and if we inspect (double)ff value



Console.Write(((double)ff).ToString("R"));


we'll see round up errors in action



31.149999618530273


Finally, Math.Round(31.149999618530273, 1, MidpointRounding.AwayFromZero) == 31.1 as expected






share|improve this answer
























  • I'm curious why (double)ff is exactly 31.14999961853027 and not 31.150000000000000

    – Gonzalo Lorieto
    1 hour ago






  • 4





    @Gonzalo Lorieto: double is binary and 0.15 == 15 / 100 = 3 / 20 is a periodic fraction in binary representation.

    – Dmitry Bychenko
    1 hour ago






  • 1





    @GonzaloLorieto stackoverflow.com/questions/618535/…

    – Camilo Terevinto
    1 hour ago






  • 1





    @GonzaloLorieto You can read it here -> Floating-point arithmetic

    – SeM
    1 hour ago



















7














In floating point, all numbers are represented internally as fractions where the denominator is a power of 2.



(This is a similar way to how decimals are actually fractions with power-of-10 denominators. So 31.15 is just a way of writing the fraction 3115/100)



In floating point, 31.15 must be represented internally as a binary number. The closest binary fraction is: 1111.1001001100110011001100110011001100110011001100110011001100...repeating



The 1100 recurs (repeats forever), and so the number will be truncated depending on whether it is stored in a double or a float. In a float it is truncated to 24 digits, and in a double to 53.



Exact:  1111.100100110011001100110011001100110011001100110011001100110011001100...forever

Float:  1111.10010011001100110011

Double: 1111.1001001100110011001100110011001100110011001100110


Therefore you can see that the double that this number converts to, is actually slightly larger than the float it converts to. So it is clear that it won't necessarily round to the same number, since it is not the same number to begin with.






share|improve this answer

































    0














    It is already explained why we do have that rounding problem but it is worth mentioning that with this small trick you can get away from it:



    float ff = (float)31.15;
    

    
double dd = 31.15;
    

    
var frst = Math.Round(
    
    double.Parse(ff.ToString(CultureInfo.InvariantCulture), CultureInfo.InvariantCulture),
    
    1, 
    
    MidpointRounding.AwayFromZero);
    

    
var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


    Then the result should be the same:



    3.2
    
3.2


    so while (double)ff introduce rounding problem, double.Parse(ff.ToString()) doesn't because the conversion from float to double is avoided.






    share|improve this answer

























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      17














      Well, Math.Round wants double, not float, that's why



      Math.Round(ff, 1, MidpointRounding.AwayFromZero);


      equals to 



      Math.Round((double)ff, 1, MidpointRounding.AwayFromZero);


      and if we inspect (double)ff value



      Console.Write(((double)ff).ToString("R"));


      we'll see round up errors in action



      31.149999618530273


      Finally, Math.Round(31.149999618530273, 1, MidpointRounding.AwayFromZero) == 31.1 as expected






      share|improve this answer
























      • I'm curious why (double)ff is exactly 31.14999961853027 and not 31.150000000000000

        – Gonzalo Lorieto
        1 hour ago






      • 4





        @Gonzalo Lorieto: double is binary and 0.15 == 15 / 100 = 3 / 20 is a periodic fraction in binary representation.

        – Dmitry Bychenko
        1 hour ago






      • 1





        @GonzaloLorieto stackoverflow.com/questions/618535/…

        – Camilo Terevinto
        1 hour ago






      • 1





        @GonzaloLorieto You can read it here -> Floating-point arithmetic

        – SeM
        1 hour ago
















      17














      Well, Math.Round wants double, not float, that's why



      Math.Round(ff, 1, MidpointRounding.AwayFromZero);


      equals to 



      Math.Round((double)ff, 1, MidpointRounding.AwayFromZero);


      and if we inspect (double)ff value



      Console.Write(((double)ff).ToString("R"));


      we'll see round up errors in action



      31.149999618530273


      Finally, Math.Round(31.149999618530273, 1, MidpointRounding.AwayFromZero) == 31.1 as expected






      share|improve this answer
























      • I'm curious why (double)ff is exactly 31.14999961853027 and not 31.150000000000000

        – Gonzalo Lorieto
        1 hour ago






      • 4





        @Gonzalo Lorieto: double is binary and 0.15 == 15 / 100 = 3 / 20 is a periodic fraction in binary representation.

        – Dmitry Bychenko
        1 hour ago






      • 1





        @GonzaloLorieto stackoverflow.com/questions/618535/…

        – Camilo Terevinto
        1 hour ago






      • 1





        @GonzaloLorieto You can read it here -> Floating-point arithmetic

        – SeM
        1 hour ago














      17












      17








      17







      Well, Math.Round wants double, not float, that's why



      Math.Round(ff, 1, MidpointRounding.AwayFromZero);


      equals to 



      Math.Round((double)ff, 1, MidpointRounding.AwayFromZero);


      and if we inspect (double)ff value



      Console.Write(((double)ff).ToString("R"));


      we'll see round up errors in action



      31.149999618530273


      Finally, Math.Round(31.149999618530273, 1, MidpointRounding.AwayFromZero) == 31.1 as expected






      share|improve this answer













      Well, Math.Round wants double, not float, that's why



      Math.Round(ff, 1, MidpointRounding.AwayFromZero);


      equals to 



      Math.Round((double)ff, 1, MidpointRounding.AwayFromZero);


      and if we inspect (double)ff value



      Console.Write(((double)ff).ToString("R"));


      we'll see round up errors in action



      31.149999618530273


      Finally, Math.Round(31.149999618530273, 1, MidpointRounding.AwayFromZero) == 31.1 as expected







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 1 hour ago









      Dmitry BychenkoDmitry Bychenko

      107k1092133




      107k1092133













      • I'm curious why (double)ff is exactly 31.14999961853027 and not 31.150000000000000

        – Gonzalo Lorieto
        1 hour ago






      • 4





        @Gonzalo Lorieto: double is binary and 0.15 == 15 / 100 = 3 / 20 is a periodic fraction in binary representation.

        – Dmitry Bychenko
        1 hour ago






      • 1





        @GonzaloLorieto stackoverflow.com/questions/618535/…

        – Camilo Terevinto
        1 hour ago






      • 1





        @GonzaloLorieto You can read it here -> Floating-point arithmetic

        – SeM
        1 hour ago



















      • I'm curious why (double)ff is exactly 31.14999961853027 and not 31.150000000000000

        – Gonzalo Lorieto
        1 hour ago






      • 4





        @Gonzalo Lorieto: double is binary and 0.15 == 15 / 100 = 3 / 20 is a periodic fraction in binary representation.

        – Dmitry Bychenko
        1 hour ago






      • 1





        @GonzaloLorieto stackoverflow.com/questions/618535/…

        – Camilo Terevinto
        1 hour ago






      • 1





        @GonzaloLorieto You can read it here -> Floating-point arithmetic

        – SeM
        1 hour ago

















      I'm curious why (double)ff is exactly 31.14999961853027 and not 31.150000000000000

      – Gonzalo Lorieto
      1 hour ago





      I'm curious why (double)ff is exactly 31.14999961853027 and not 31.150000000000000

      – Gonzalo Lorieto
      1 hour ago




      4




      4





      @Gonzalo Lorieto: double is binary and 0.15 == 15 / 100 = 3 / 20 is a periodic fraction in binary representation.

      – Dmitry Bychenko
      1 hour ago





      @Gonzalo Lorieto: double is binary and 0.15 == 15 / 100 = 3 / 20 is a periodic fraction in binary representation.

      – Dmitry Bychenko
      1 hour ago




      1




      1





      @GonzaloLorieto stackoverflow.com/questions/618535/…

      – Camilo Terevinto
      1 hour ago





      @GonzaloLorieto stackoverflow.com/questions/618535/…

      – Camilo Terevinto
      1 hour ago




      1




      1





      @GonzaloLorieto You can read it here -> Floating-point arithmetic

      – SeM
      1 hour ago





      @GonzaloLorieto You can read it here -> Floating-point arithmetic

      – SeM
      1 hour ago













      7














      In floating point, all numbers are represented internally as fractions where the denominator is a power of 2.



      (This is a similar way to how decimals are actually fractions with power-of-10 denominators. So 31.15 is just a way of writing the fraction 3115/100)



      In floating point, 31.15 must be represented internally as a binary number. The closest binary fraction is: 1111.1001001100110011001100110011001100110011001100110011001100...repeating



      The 1100 recurs (repeats forever), and so the number will be truncated depending on whether it is stored in a double or a float. In a float it is truncated to 24 digits, and in a double to 53.



      Exact:  1111.100100110011001100110011001100110011001100110011001100110011001100...forever
      
Float:  1111.10010011001100110011
      
Double: 1111.1001001100110011001100110011001100110011001100110


      Therefore you can see that the double that this number converts to, is actually slightly larger than the float it converts to. So it is clear that it won't necessarily round to the same number, since it is not the same number to begin with.






      share|improve this answer






























        7














        In floating point, all numbers are represented internally as fractions where the denominator is a power of 2.



        (This is a similar way to how decimals are actually fractions with power-of-10 denominators. So 31.15 is just a way of writing the fraction 3115/100)



        In floating point, 31.15 must be represented internally as a binary number. The closest binary fraction is: 1111.1001001100110011001100110011001100110011001100110011001100...repeating



        The 1100 recurs (repeats forever), and so the number will be truncated depending on whether it is stored in a double or a float. In a float it is truncated to 24 digits, and in a double to 53.



        Exact:  1111.100100110011001100110011001100110011001100110011001100110011001100...forever
        
Float:  1111.10010011001100110011
        
Double: 1111.1001001100110011001100110011001100110011001100110


        Therefore you can see that the double that this number converts to, is actually slightly larger than the float it converts to. So it is clear that it won't necessarily round to the same number, since it is not the same number to begin with.






        share|improve this answer




























          7












          7








          7







          In floating point, all numbers are represented internally as fractions where the denominator is a power of 2.



          (This is a similar way to how decimals are actually fractions with power-of-10 denominators. So 31.15 is just a way of writing the fraction 3115/100)



          In floating point, 31.15 must be represented internally as a binary number. The closest binary fraction is: 1111.1001001100110011001100110011001100110011001100110011001100...repeating



          The 1100 recurs (repeats forever), and so the number will be truncated depending on whether it is stored in a double or a float. In a float it is truncated to 24 digits, and in a double to 53.



          Exact:  1111.100100110011001100110011001100110011001100110011001100110011001100...forever
          
Float:  1111.10010011001100110011
          
Double: 1111.1001001100110011001100110011001100110011001100110


          Therefore you can see that the double that this number converts to, is actually slightly larger than the float it converts to. So it is clear that it won't necessarily round to the same number, since it is not the same number to begin with.






          share|improve this answer















          In floating point, all numbers are represented internally as fractions where the denominator is a power of 2.



          (This is a similar way to how decimals are actually fractions with power-of-10 denominators. So 31.15 is just a way of writing the fraction 3115/100)



          In floating point, 31.15 must be represented internally as a binary number. The closest binary fraction is: 1111.1001001100110011001100110011001100110011001100110011001100...repeating



          The 1100 recurs (repeats forever), and so the number will be truncated depending on whether it is stored in a double or a float. In a float it is truncated to 24 digits, and in a double to 53.



          Exact:  1111.100100110011001100110011001100110011001100110011001100110011001100...forever
          
Float:  1111.10010011001100110011
          
Double: 1111.1001001100110011001100110011001100110011001100110


          Therefore you can see that the double that this number converts to, is actually slightly larger than the float it converts to. So it is clear that it won't necessarily round to the same number, since it is not the same number to begin with.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          BenBen

          28.7k55487




          28.7k55487























              0














              It is already explained why we do have that rounding problem but it is worth mentioning that with this small trick you can get away from it:



              float ff = (float)31.15;
              

              
double dd = 31.15;
              

              
var frst = Math.Round(
              
    double.Parse(ff.ToString(CultureInfo.InvariantCulture), CultureInfo.InvariantCulture),
              
    1, 
              
    MidpointRounding.AwayFromZero);
              

              
var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


              Then the result should be the same:



              3.2
              
3.2


              so while (double)ff introduce rounding problem, double.Parse(ff.ToString()) doesn't because the conversion from float to double is avoided.






              share|improve this answer






























                0














                It is already explained why we do have that rounding problem but it is worth mentioning that with this small trick you can get away from it:



                float ff = (float)31.15;
                

                
double dd = 31.15;
                

                
var frst = Math.Round(
                
    double.Parse(ff.ToString(CultureInfo.InvariantCulture), CultureInfo.InvariantCulture),
                
    1, 
                
    MidpointRounding.AwayFromZero);
                

                
var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


                Then the result should be the same:



                3.2
                
3.2


                so while (double)ff introduce rounding problem, double.Parse(ff.ToString()) doesn't because the conversion from float to double is avoided.






                share|improve this answer




























                  0












                  0








                  0







                  It is already explained why we do have that rounding problem but it is worth mentioning that with this small trick you can get away from it:



                  float ff = (float)31.15;
                  

                  
double dd = 31.15;
                  

                  
var frst = Math.Round(
                  
    double.Parse(ff.ToString(CultureInfo.InvariantCulture), CultureInfo.InvariantCulture),
                  
    1, 
                  
    MidpointRounding.AwayFromZero);
                  

                  
var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


                  Then the result should be the same:



                  3.2
                  
3.2


                  so while (double)ff introduce rounding problem, double.Parse(ff.ToString()) doesn't because the conversion from float to double is avoided.






                  share|improve this answer















                  It is already explained why we do have that rounding problem but it is worth mentioning that with this small trick you can get away from it:



                  float ff = (float)31.15;
                  

                  
double dd = 31.15;
                  

                  
var frst = Math.Round(
                  
    double.Parse(ff.ToString(CultureInfo.InvariantCulture), CultureInfo.InvariantCulture),
                  
    1, 
                  
    MidpointRounding.AwayFromZero);
                  

                  
var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);


                  Then the result should be the same:



                  3.2
                  
3.2


                  so while (double)ff introduce rounding problem, double.Parse(ff.ToString()) doesn't because the conversion from float to double is avoided.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 1 min ago

























                  answered 1 hour ago









                  JohnnyJohnny

                  1,704615




                  1,704615






























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                      Statuo de Libereco

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                      Koordinatoj: 40°41′21″N 74°02′40″W   /   40.68917°N , 74.04444°U  / 40.68917; -74.04444  ( Statuo de Libereco ) Statuo de Libereco angle: Statue of Liberty statuo Statuo de Libereco de antaŭe Oficiala nomo: Liberty Enlightening the World Lando  Usono Ŝtato Novjorkio Regiono Novjorko Insulo Insulo de Libereco Situo Statuo de Libereco  - koordinatoj 40°41′21″N 74°02′40″W   /   40.68917°N , 74.04444°U  / 40.68917; -74.04444  ( Statuo de Libereco ) Alteco de statuo 46 m  - alteco kun soklo 93 m Pezo 205 000 kg (451 948 lb) Materialoj bronzo, ŝtalo Proponinto Gustave Eiffel  - skulptisto Frédéric Auguste Bartholdi  - proponinto de soklo Richard Morris Hunt  - kapo laŭ modelo de Isabella Eugenie Boyer Transdonita 4-an de julio 1886 Por publiko alirebla por publiko Horzono OT (UTC-5)  - somera tempo OT (UTC-4) Loko
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                      Tanganjiko

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                      Lago Tanganjiko Situo centra Afriko Ĉefaj fontoj Kalambo, Lufuko, Malagarasi kaj Ruzizi Ĉefaj elfluoj Lukugo Setlejoj Bujumbura, Kalemie Datenoj Koordinatoj 6°  S , 30°  O -6.3827777777778 29.615 782 Koordinatoj: 6°  S , 30°  O DEC Supermara alteco 782 m Surfaca areo 32 893 km² f5 Maksimuma longo 673 km f6 Maksimuma larĝo 72 km f7 Akva volumeno 18.900 km³ dep1 f8 Maksimuma profundo 1470 m f10 Averaĝa profundo 570 m f11 v   •   d   •   r Mapo de la lago Tanganjiko estas lago en orienta Afriko, inter Burundo, Kongo Kinŝasa, Tanzanio kaj Zambio. Ĝi apartenas al la Granda Rifto kaj estas unu el la Afrikaj Grandaj Lagoj. Tanzanio (46%) kaj DRK (40%) havas la majoritaton de la lago. Ĝi havas longon de 673 km, areon de ĉ. 32.900 km² kaj maksimuman profundon taksatan je 1470 metroj; pro tiu ĉi granda profundo ankaŭ ĝia volumeno estas sufiĉe granda, 18.900 km³. Oni konsi
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                      Liste der Baudenkmäler in Enneberg

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                      Die Liste der Baudenkmäler in Enneberg (ladinisch Mareo , italienisch Marebbe ) enthält die 36 als Baudenkmäler ausgewiesenen Objekte auf dem Gebiet der Gemeinde Enneberg in Südtirol. Basis ist das im Internet einsehbare offizielle Verzeichnis der Baudenkmäler in Südtirol. Dabei kann es sich beispielsweise um Sakralbauten, Wohnhäuser, Bauernhöfe und Adelsansitze handeln. Die Reihenfolge in dieser Liste orientiert sich an der Bezeichnung, alternativ ist sie auch nach der Adresse oder dem Datum der Unterschutzstellung sortierbar. Liste | Foto   Bezeichnung Standort Eintragung Beschreibung Metadaten ja Altes Gasthaus Janesch ID: 14567 46° 42′ 54″ N, 11° 52′ 42″ O 76882557800 0 8. Sep. 1986 (BLR-LAB 4974) Ländliches Wohnhaus/Bauernhof KG:  Welschellen Bauparzelle: 391 ja Asch (Prack) ID: 14540 46° 43′ 12″ N, 11° 54′ 35″ O 75496485300 20. Nov. 1950 (MD) Ansitz KG:  Enneberg Bauparzelle: 38 ja Biccia mit Stade
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