Engineering Math A Fast Overview By Bill Wittig (Questions? Send me an I’ll be glad to help!)

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License A Simple Problem Design a car to climb a 1 m long hill in 2 seconds

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Linear Motion Variables D – Distance (meters) v – Speed (meters/sec) a – Acceleration (meters/sec 2 ) t – Time (seconds) Example: A car accelerates at 10 m/s 2 for 10 seconds. How far does it travel? D = ½ * 10 m/sec 2 * (10 sec) 2 D = 1000 m Using geometry:D = ½ v * t = ½ (a * t) * t = ½ a t 2 Constant acceleration: v = a * t Distance = The area under the curve D =  v(t) dt t = 0 to 10 sec Using geometry: D = v * t Constant speed:

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Angular Motion Variables v – Speed (meters/sec) ω – Angular speed (radians/sec) r – radius (m) s – arc length (m) Example: If a 5 cm radius wheel turns at 100 rpm, what is the speed of the car? v = 100 rev/min * 2  rad/rev * 1 min/60 sec * 0.05 meters v = 0.52 meters/sec v = 0.52 meters/sec * 3600 sec / 1 hr * 1 mile / 1600 meters = 1.1 mph v r ω v = ω * r Radians are unitless – they convert angles to distance θ = 1 radian = 1 radius of arc 2  radians = 360 degrees θ r s s = r * θ XXXX XXXX

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Angular Force (Torque) T r F T = F * r F = T / r Variables r – radius (m) T – torque (N-m) F – Force (N) In order to convert motor torque to a linear force, the motor acts through a wheel to produce a force. Example: A motor produces a torque of 5 Nm acting through a wheel of 5 cm (0.05 m) What is the force produced ? F = T / r F = 5 Nm / 0.05 m F = 100 N

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Newton’s Laws of Motion Variables F – Force (Newtons – kg-m/s 2 ) m – Mass (kilograms) a – Acceleration (meters/sec 2 ) g – Gravitational acceleration W – Weight (Newtons) Fm F = m * a or a = F / m Newton’s 3 Laws 1.For every action there is an equal and opposite reaction. 2.Force equals mass times acceleration. 3.An object in motion remains in motion, an object at rest remains at rest. W m W = m * g (g = 9.81 m/s 2 ) Example: A 5N force is applied to a 1 kg car. What is its acceleration? a = 5 N / 1 kg after 2 seconds a = 5 m/sec 2 v = a * t v = 10 m/sec d = ½ a * t 2 d = 10 m

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Friction Variables F – Force (Newtons – kg-m/s 2 ) F F – Friction force (Newtons) F N – Normal force (Newtons) c – Coefficient of friction W – Weight (Newtons) F W F F = c * F N (c is experimentally measured) FNFN F W = 2 * F N F F = c*F N 2 * F F = 2 * c * F N FNFN F FNFN F W

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Free Body Diagrams Variables m – Mass (kilograms) a – Acceleration (meters/sec2) g – Gravitational acceleration W – Weight (Newtons) F F – Friction Force (N) F N – Normal Force (N) c – Friction coefficient W FNFN F F m*a Balance the vertical forces: W – F N = 0 W – m*g = 0 Balance the horizontal forces: F – F F – m*a = 0 F – c * F N – m*a = 0 F – c * m*g – m*a = 0

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License More Free Body Diagrams F W FNFN F θ F WN = W * cos (θ) F WT F WN W θ F WT = W * sin (θ) F N – F WN = 0 F – F WT – F F = ma F – mg * sin(θ) – c * mg * cos(θ) = ma ma F WN = mg * cos (θ) F WT F WN W θ

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Electric Motors Variables V bat – Supply Voltage (volts) R m – Motor Resistance (ohms) V m – Motor Voltage (volts) i – Current (Amps) K t – Torque Constant (Nm/A) K e – Voltage Constant (V/rpm) Motor Analysis: T = K t * i (torque) ω = K e * V m (speed) What is the maximum motor torque? at ω = 0, T = 1 Nm What is the maximum motor speed? at T = 0, ω = 200 rpm V bat = 10 volts R m = 1 ohm K t = 0.1 Nm/A K e = 0.05 V/ rpm For our motor: Example: What is the motor torque at 5 Amps? T = 0.1 Nm/A * 5 A = 0.5 Nm What is the motor speed at 3 volts? ω = 3 V / 0.05 V/rpm = 60 rpm

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Electric Circuits iVmVm RmRm V bat Variables V bat – Supply Voltage (volts) R m – Motor Resistance (ohms) V m – Motor Voltage (volts) i – Current (Amps) K t – Torque Constant (Nm/A) K e – Voltage Constant (V/rpm) Ohm’s Law: V = i * R Kirkoff’s Voltage Law:  V = 0 V m = ω * K e V = i * R m (Ohm’s Law) Summing voltages (Kirkoff): V bat – i * R m – ω * K e = 0 Circuit Analysis: Example: What is the motor speed at a current of 5 A from a 10V battery? V bat – i * R m – V m = 0 V m = 10 V – 5 A * 1 ohm V m = 5 V ω = 5 V / 0.05 V/rpm ω = 100 rpm V i R

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Hill Climbing – The Whole Problem Climb the 1 m hill in 2 seconds Choose: Constant Velocity Constant Acceleration Known Values m = 1 kg r = 5 cm c = 0.1 N/N θ = 30 deg R m = 1 ohm K e = 0.05 V/rpm K t = 0.1 Nm/A V bat = 10 V F FWTFWT FWNFWN W FNFN F

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Hill Climbing – The Whole Problem Constant Acceleration D = ½ a t 2 a = 2 ( 1m) / 2 2 = 0.5 m/sec 2 F = m a = 1 kg * 0.5 m/s 2 = 1 N F - F WT – c F WN = ma = 1 N F = 1 N + mg *sin (30) + c * mg * cos (30) = 6.75 N T = F * r = 6.75 N * 0.05 m = 0.34 Nm T = Kt * ii = 0.34 Nm / 0.1 Nm/A = 3.4 A V m = V bat – i * R m = 10 V – 3.4 A * 1 ohm = 6.6 V ω = K e * V m = 6.6 V / 0.05 V/rpm = 133 rpm v = ω * r = 133 rpm * 0.05 m *(2  rad/rev / 60 sec/min) = 0.7 m/s Known Values m = 1 kg r = 5 cm c = 0.1 N/N θ = 30 deg R m = 1 ohm K e = 0.05 V/rpm K t = 0.1 Nm/A V bat = 10 V F FWTFWT FWNFWN W FNFN F v = a * t = 0.5 m/s 2 * 2 s = 1 m/s ω = v / r = 1 m/s / 0.05 m * 60 sec/min / 2  rad/rev = 191 rpm Check calculation: 50% ERROR

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Hill Climbing – The Whole Problem Constant Velocity D = v t, v = 1 m / 2 sec = 0.5 m/sec, a = 0 F = m * a = 1 kg * 0.0 m/s 2 = 0 N (constant speed, no acceleration) F - F WT – c F WN = ma = 0 N F = 0 N + mg *sin (30) + c * mg * cos (30) = 5.75 N T = F * r = 5.75 N * 0.05 m = 0.29 Nm T = Kt * ii = 0.29 Nm / 0.1 Nm/A = 2.9 A V m = V bat – i * R m = 10 V – 2.9 A * 1 ohm = 7.1 V ω = V m / K e = 7.1 V / 0.05 V/rpm = 142 rpm Known Values m = 1 kg r = 5 cm c = 0.1 N/N θ = 30 deg R m = 1 ohm K e = 0.05 V/rpm K t = 0.1 Nm/A V bat = 10 V F FWTFWT FWNFWN W FNFN F v = ω * r = 142 rpm * 0.05 m *(2  rad/rev / 60 sec/min) = 0.75 m/s Check calculation: 40% ERROR

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Reality… The motor will only run at points on its torque-speed curve. At t = 0 s, v = 0 m/s, ω = 0, V m = 0 volts, therefore i = 10 A Some time later, V m = 3 V, i = 7 A, etc Current is changing, therefore torque is changing, therefore acceleration is changing… Differential equations were invented to handle this situation. m * d 2 x/dt 2 = (A – B * dx/dt)D =  dx/dt dt A = K t * N * V bat / r / R m – mg * sin (θ) – c * mg * cos (θ) B = K t * N 2 / Ke / r 2 / R m v = dx/dt a = d 2 x/dt 2

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Difference Equations Transient Steady State

Copyright © 2010 Bill Wittig – Licensed under Creative Commons Attribution-ShareAlike 3.0 Unported LicenseCreative Commons Attribution-ShareAlike 3.0 Unported License Simulation All of the equations and constants previously described on an Excel spreadsheet.