Flipper Math notes
by Dale Heatherington
Oct 6 2003

This is an after the fact attempt to quantify
the flipper performance on Invertabot using
basic physics.



Variables

s distance in meters
a acceleration in meters/sec^2
t time in seconds
F force in Newtons
W = work in joules
P = power in watts

Basic equations

s = 1/2 * a * t^2
t = sqrt(s / (0.5 * a))
a = s / (0.5 * t^2)

F = m * a
a = F / m
m = F / a

Velocity
V = sqrt(2a * s)


Unit conversions

1 newton = 3.596 ounce-force
1 newton = 0.224 pound-force
1 newton = 0.101 kilogram-force

1 ounce-force = 0.278 newton
1 pound-force = 4.448 newton
1 kilogram-force = 9.806 newton

1 kilogram = 2.204 pounds
1 inch = 2.54 cm

[Parameters of 12 pound Invertabot flipper system]

Assume 
	800 psi gas
	.75 bore x 1 inch stroke cylinder
	1:7 leverage on flipper
	6 inch upward travel on flipper
	
	Also assume electric valve orifice is large enough
	to maintain full 800 psi in cylinder over
	its full travel time and distance.  This
	probably is false and actual performance
	will be less than these calculations indicate.

	
Area = pi * r^2

Cylinder piston area = 3.14 * (.75/2)^2 = .442 sq in

800 * .442 = 353 pounds-force from cylinder

353 / 7 = 50.4 pounds-force at tip of flipper

Opponent load = 12 pounds


Convert 50.4 pounds to metric:

Flipper force: 50.4 / 2.204 = 22.9 kg
Opponent load mass: 12 / 2.204 =  5.5 kg
Opposing gravity acting on load: 5.5 kg

net force upward: 22.9 - 5.5 = 17.4 kg

Convert 6 inch flipper travel to meters:
6 * 2.54/100 = .152 meters

Acceleraton:

Force = 17.4 * 9.806 = 170 Newtons

a = 170 / 5.5 = 31 meters/sec^2


Velocity of load at end of flipper travel:

V = sqrt(2a * s)
V = sqrt(2 * 31 * .152) = 3.1 meters/sec

Time flipper takes to fully open:
t = sqrt( s / (0.5 * a))
s = .152 meters
a = 31 m/sec^2
t = Sqrt(.152 / (0.5 * 31)) = .099 seconds


Peak height of launched load above flipper end travel:

Acceleration due to gravity = 9.8 m/sec^2
s = V^2 / (2*9.8)

s = 3.1^2 / 2*9.8 = 0.49 meter

Add in the flipper tip height of .152 meter:
total s = 0.49 + .152 = .642 meter



Work performed:
 
 1 joule = 1 newton exerted through 1 meter

 W = F * s

 Flipper lifted 5.5 Kg to .642 meters
 Force = 9.8 * 5.5 = 53.9 Newtons

 W = 53.9 * .642 = 34.6 joules
 
 Also, as a double check, flipper exerted  224 newtons
 force over a distance of .152 meters.
 W = 224 * .152 = 34.1 joules

 
 Power used:
 Power = Work / t
 1 watt = 1 joule/sec
 
 Flipper opened in .099 seconds and did 34.1 joules of work
 
 P = 34.1 / .099 = 344 Watts
 
Energy used:
 Watt-seconds = 344 * .099 = 34


If electric power was used instead of gas...

At 14 volts, 344 watts requires 24.6 amps, 
ignoring less than 100% motor efficiency.

At 14 volts, milliamp hours used:
 Amps = 344 / 14 = 24.6 amps or 24600 ma
 Time in hours = .099 / 3600 = 27.5E-6 hours
 milliamp hours = 24600 * 27.5E-6 = 0.676 mah
 
 
 ----
 Observation:  It should be feasable to make
 an electric flipper with performance equal to the
 compressed gas flipper described above. Battery
 and motor requirements are reasonable for
 a 12# bot.  Storing energy in a flywheel would
 reduce peak battery and motor loads considerably.
 Experiments need to be done :-)
 
 If you see any errors here let me know at
 dale  at wa4dsy.net
 
 
 end.