Munurin millum rættingarnar hjá "Hitastigar"

| −273.,15 °C

| −459.,67 °F

| [K] = ([°F] + 459.,67) · 5/9

| [°De] = (373.,15 − [K]) · 3/2

| [K] = 373.,15 − [°De] · 2/3

|align="center"|<math>K = {}^\circ\!C - 273.,15 = \frac{5({}^\circ\!F+459.,67)}{9} = \frac{5^\circ\!Ra}{9} = \frac{5^\circ\!R\acute{e}}{4} + 273.,15</math>

|align="center"|<math>K = \frac{100^\circ\!N}{33} + 273.,15 = \frac{40({}^\circ\!R\ddot o - 7.,5)}{21} + 273.,15 = 373.,15 - \frac{2^\circ\!De}{3}</math>

|align="center"|<math>{}^\circ\!C = K - 273.,15 = \frac{5({}^\circ\!F-32)}{9} = \frac{5(Ra-491.,67)}{9}</math>

|align="center"|<math>{}^\circ\!C = \frac{5^\circ\!R\acute{e}}{4} = \frac{100^\circ\!N}{33} = \frac{40({}^\circ\!R\ddot o - 7.,5)}{21} = \frac{2^\circ\!De}{3} - 100</math>

|align="center"|<math>{}^\circ\!F = \frac{9K}{5} - 459.,67 = \frac{9{}^\circ\!C}{5}+32 = {}^\circ\!Ra - 459.,67 = \frac{45^\circ\!R\acute{e}/4-273.,15}{5}</math>

|align="center"|<math>{}^\circ\!F = \frac{60^\circ\!N}{11}+32 = \frac{24({}^\circ\!R\ddot o-7.,5)}{7} + 32 = 212 - \frac{6De}{5}</math>

|align="center"|<math>{}^\circ\!R = \frac{9K}{5} = \frac{9({}^\circ\!C + 273.,15)}{5} = {}^\circ\!F + 459.,67 = \frac{9(5^\circ\!R\acute{e} / 5 - 273.,15)}{5}</math>

|align="center"|<math>{}^\circ\!R = \frac{60^\circ\!N}{11} + 491.,67 = \frac{24({}^\circ\!R\ddot o - 7.,5)}{7} + 491.,67 = 671.,67 - \frac{6^\circ\!De}{5}</math>

|align="center"|<math>{}^\circ\!De = \frac{3(373.,15 - K)}{2} = \frac{3(100 - {}^\circ\!C)}{2} = \frac{5(212 - {}^\circ\!F)}{6}</math>

|align="center"|<math>{}^\circ\!De = \frac{5(671.,67 - {}^\circ\!Ra)}{6} = \frac{15(80 - ^\circ\!R\acute{e})}{8} = \frac{50(33 - {}^\circ\!N)}{11} = \frac{20(60 - {}^\circ\!R\ddot o\!)}{7}</math>

|align="center"|<math>{}^\circ\!N = \frac{33(K - 273.,15)}{100} = \frac{33^\circ\!C}{100} = \frac{11({}^\circ\!F - 32)}{60}</math>

|align="center"|<math>{}^\circ\!N = \frac{11({}^\circ\!Ra - 491.,67)}{60} = \frac{33^\circ\!R\acute{e}}{80} = \frac{22(R\ddot o - 7.,5)}{35} = 33 - \frac{11 {}^\circ\!De}{50}</math>

|align="center"|<math>{}^\circ\!R\acute{e} = \frac{4(K - 273.,15)}{5} = \frac{4{}^\circ\!C}{5} = \frac{4({}^\circ\!F - 32)}{9}</math>

|align="center"|<math>{}^\circ\!R\acute{e} = \frac{4({}^\circ\!Ra - 491.,67)}{9} = \frac{80^\circ\!N}{33} = \frac{32({}^\circ\!R\ddot o - 7.5)}{21} = 80 - \frac{8^\circ\!De}{15}</math>

|align="center"|<math>{}^\circ\!R\ddot o = \frac{21(K - 273.,15)}{40} + 7.,5 = \frac{21^\circ\!C}{40} + 7.,5 = \frac{7({}^\circ\!F - 32)}{24} + 7.,5</math>

|align="center"|<math>{}^\circ\!R\ddot o = \frac{7({}^\circ\!Ra - 491.,67)}{24} + 7.,5 = \frac{21^\circ\! R\acute{e}}{32} + 7.,5 = \frac{35^\circ\!N}{22} + 7.,5 = 60 - \frac{7^\circ\!De}{20}</math>

* [http://atm.ucdavis.edu/tools/temperature_conversion.php Hitaumrokning á netinum]