Math with fixed number of mantissa digits Example 63 digit mantissa numbers x x10 1 Note: Objective is to multiply these two numbers together x x The multiplication with no constraints x x 10 3 Round off to 3 mantissa digits Example shows the philosophy of how it is done. Alternate approach Truncate least significant digit that is outside the “digit space” as soon as it shows up in the calculation error is created after every arithmetic step error is multiplied if previous value is put back into another multiplication. Bottom Line When computers calculate bad STUFF happens so be skeptical of your numerical method results! (when there isn’t enough digits for intermediate steps) Computer does these calculations in the base 2 number system so it is actually done as outlined above but with 1’s and 0’s.

Lagrangian Control Volume < g > One set of symbol options to indicate a vector < linear velocity through control volume > = y x y z = (,, ) ) ( m [ m(t) ] = (t) = x < F acceleration of gravity < a acceleration of control volume Mass flow rate < g Velocity in x direction x Balance statement: I = 1 n < F n = resultant force Gravitational force on the object The reference space moves and the stuff in that space moves because the space is moving. Other perspective has the stuff moving through the non-moving control volume. < < a Example 7 n < F n = resultant force Develop the mass transport model for material referenced to the following Lagrangian control volume = < a + (ii) The actual force that is pulling the control volume up < g [ m(t) ] < F = < a + 2 < F 1 < m m [ m(t) ] For the y direction: z [m(t)] I = 1 ( +0, +0, ) = +x +z +y < This vector equation contains three scalar equations. (one for each of the direction components of the vector) ( 0) + (0) = 0 (i) For the x direction: ( 0) + (0) = 0 For the x direction:(iii) ( -0, -0, -[m(t)] ) g z + z m a z ( 0, 0, +[m(t)] ) = (-[m(t)] ) z m - g z + g z z m + a z = a z [m(t)] z m -[m(t)] g z + = dtd( ) v (t) z [m(t)] z m - g z + = dt d v (t) z )( Note: t m t=0 + m m(t) = m t m 0 + < g

Lagrangian Control Volume [ m(t) ] Balance statement: < a Example 7 n < F n = resultant force Develop the mass transport model for material referenced to the following Lagrangian control volume = < g < a + (ii) The actual force that is pulling the control volume up < g [ m(t) ] < F = < a + 2 < F 1 < m m [ m(t) ] For the y direction: I = 1 ( +0, +0, ) = +x +z +y This vector equation contains three scalar equations. (one for each of the direction components of the vector) ( 0) + (0) = 0 (i) For the x direction: ( 0) + (0) = 0 For the x direction:(iii) ( -0, -0, -[m(t)] ) g z + z m a z ( 0, 0, +[m(t)] ) = (-[m(t)] ) - g z g z z m + a z = a z [m(t)] z m -[m(t)] g z + = [m(t)] z m d( ) v (t) z [m(t)] z m - g z + = dt d v (t) z ( Note: t m t=0 + m(t) = m t m 0 + [m(t)] z m - g z + = dtd( ) v (t) z )( [m(t)] z m + dt ) z z m momentum acceleration force mass flow rate scalar m the magnitude of z m linear velocity

Numerical methods for engineers includes units mks (iii) the “4 unit” system A mass of stuff accelerating at “standard” sea level gravitational value Weighs 9.8 kg force cgs 1 gram of stuff accelerating at 1 cm/s 2 Weighs 981 dynes mks 1 gram of stuff accelerating at 1 m/s 2 Weighs 1 Newton The amount stuff that has this weight has a mass of 1 kg force-s / meter 2 ( force, length, mass, time ) force system mass system (i) The three “3 unit” systems (ii) 2 “national” systems ( lenght, mass, time ) 1 slug of stuff American Engineering System Object weighs 32.2 pounds force 1 pound of stuff Object accelerates at 1 ft/sec 2 British Mass System 1 lb mass < a < F = ( ) g 1 Common in USA Object weighs 32.2 poundals Weighs 1 lb force 1 lb force will make 1 lb mass accelerate at 32.2 ft/s 2 (mass) 2 s 1 lb force 32.2 lb ft mass

Example using 4 unit system 10 ft 2 Not typical pressure units but they are still pressure units. Z = 100 ft water tower is in Tampa P P 2 = (2,020 x 10 ) 2 sft lb mass water density = Z acceleration ( ) mass ft 3 (a)(ft ) 1 P = g c Pressure difference (top to bottom) Since mass is inlb mass the force islb force < a = = lb force 2 ft 2 s 1 lb force 32.2 lb ft mass < F conversion factor between lb and lb mass force (mass) < a = ( ) g 1 < F P = (62.4 ) ft 3 lb mass 2 ( 10 ft) s 2 ft ( 32.2 ) (2,020 x 10 ) 2 sft lb mass 62.4 x10 2 s 1 lb force 32.2 lb ft mass (mass) ( ) s 1 lb force 32.2 lb ft mass g = Note:The “4 unit” system entertains two density concepts. < a < F = (mass) 2 s ft (force magnitude) ( ) = (mass) ( ) 2 s 1 lb force 32.2 lb ft mass g 2 s 1 lb force 32.2 lb ft mass 3 1 ft 62.4 lb mass mass densityforce density 3 1 ft 62.4 lb force Both look the same (have units of pounds per foot cubed) but each represents a different concept.

  a < F 1 < g z z m

< a < F = ( ) g 1

Math with fixed number of mantissa digits Example 63 digit mantissa numbers x x10 1 Note: Objective is to multiply these two numbers together x x The multiplication with no constraints x x 10 3 Round off to 3 mantissa digits computer does these calculations in the base Example shows the philosophy of how it is done. Alternate approach Truncate least significant digit that is outside the “digit space” as soon as it shows up in the calculation error is created after every arithmetic step error is multiplied if previous value is put back into another multiplication. Bottom Line When computers calculate bad STUFF happens so be skeptical of your numerical method results! (when there isn’t enough digits for intermediate steps) 2 number system so it is not actually done as outlined above.