subtract   32   from   the   Fahrenheit   and   divide   the remainder by 1.8. Examples: 1. F = l.8C + 32 Given:   24°C. Find: °F 24 × 1.8 = 43.2 43.2 + 32 = 75.2 or 75°F. 2. C = F 32 1.8 - Given:   96°F. Find: °C. 96 – 32 = 64 64 + 1.8 = 35.5 or 36°C To change a Celsius reading to an absolute value, add   the   Celsius   reading   to   273°   algebraically.   For example, to find the absolute value of -35°C, you would add minus 35° to 273°K algebraically. That is, you take 273° and combine 35° so you use the minus (-) function to arrive at 238°K. To  change  a  Fahrenheit  reading  to  an  absolute value,   first   convert   the   Fahrenheit   reading   to   its equivalent Celsius value.   Add this value algebraically to   273°.   Consequently,   50°F   is   equivalent   to   2830 absolute,  arrived  at  by  converting  50°F  to  10°C  and then adding the Celsius value algebraically to 273°. VERTICAL DISTRIBUTION Earth’s atmosphere is divided into layers or zones according  to  various  distinguishing  features.  (See  fig. 1-9).  The temperatures shown here are generally based on  the  latest  “U.S.  Extension  to  the  ICAO  Standard 1-15 AURORA TEMPERATURE (KINETIC) PRESSURE (MB) NOCTILUCENT CLOUDS 0 500 1000 1500 2000 600 500 400 300 200 100 700 2500 10 10 10 10 10 TROPOPAUSE NACREOUS CLOUDS MT EVEREST 180 190 200 210 220 230 240 250 260 270 280 290 300  DEGREES K DEGREES C DEGREES F 100 200 300 500 700 850 1000 50 50 25 10 100 150 200 10 -1 1 250 10^-2 -3 300 GEOMETRIC HEIGHT F REGION E REGION D REGION 10 KILOMETERS 100 90 60 80 70 40 50 10 20 30 STRATOSPHERE STRATOPAUSE MESOPAUSE THERMOSHPERE MESOSPHERE TROPOSPHERE GEOMETRIC HEIGHT KILOMETERS (400) MILES (300) (200) (100) -4 -5 -6 -7 -8 (MILES) (60) PRESSURE (MB) TEMPERATURE (10) (20) (30) (40) (50) THOUS OF FEET -90 -140 -120 -100 -80 -70 -80 -60 -50 -60 -40 -40 -30 -20 -20 0 -10 20 0 32 40 10 60 20 80 30 O Z O N O S P H E R E E  REGION D  REGION DEG.  K I O N O S P H E R E F REGION 2 F REGION 1 AGf0109 Figure 1-9.—Earth's Atmosphere.