1. Estimation

2. Rounding  The simplest estimation technique is to round.  This works very well on formulas where all the values can be reduced to one significant figure.

3. Order of Magnitude Rounding  Rounding to a power of ten is the crudest form of rounding.  Order of magnitude estimates are easy to compare since they are all only powers of ten.  For comparison to work, the units need to be the same (meters and meters, not km).

4. Order of Magnitude  My Height  Lecture Hall  Faraday West  NIU Campus (EW)  DeKalb Co (EW)  Illinois (EW)  USA (north-south)  5’9” = 1.75 m = 2 x 10 0 m  8 m = 0.8 x 10 1 m  80 m = 0.8 x 10 2 m  2000 m = 2 x 10 3 m = 2 km  28,800 m = 3 x 10 4 m = 30 km  150 km = 2 x 10 5 m = 200 km  1900 km = 2 x 10 6 m = 2 Mm These lengths differ by about one order of magnitude. mapquestmapquest uses about two steps per order of magnitude.

5. Using Geometry  Geometrical shapes can often be used to approximate real shapes. Geometric formulas Geometric relationships  Appropriate shapes can simplify the problem. 2-dimensional (triangle, circle) 3-dimensional (box, sphere). h2h2 h1h1 s1s1 s2s2

6. How Big?  Assume the density of a rock is three times that of water. How many centimeters across is a one metric ton (1000 kg) rock? The rock has a density of 3 g/cm 3The rock has a density of 3 g/cm 3 The volume is 10 6 g / (3 g/cm 3 ) = 3.3 x 10 5 cm 3The volume is 10 6 g / (3 g/cm 3 ) = 3.3 x 10 5 cm 3 Estimate that the rock is a sphere, V = (4/3)  r 3Estimate that the rock is a sphere, V = (4/3)  r 3 d = 2r = 2 (3V/4  ) 1/3d = 2r = 2 (3V/4  ) 1/3 d = 85.7 cm  90 cmd = 85.7 cm  90 cm next