Numerical methods for engineers includes dimension systems
inc
Numerical methods for engineers includes dimension systems
(i) The three “3 unit” systems
cgs
1 gram of stuff
2
accelerating at 1 cm/s
Weighs 981 dynes
( lenght, mass, time )
mks
1 gram of stuff
2
accelerating at 1 m/s
mass
system
Weighs 1 Newton
mks
A mass of stuff
accelerating at
“standard” sea level
gravitational value
force
system
Weighs 9.8 kg force
The amount of stuff that has this weight
has a mass of 1 kg force-s 2/ meter
inc
Numerical methods for engineers includes dimension systems
(i) The three “3 unit” systems
( lenght, mass, time )
cgs
mks
1 gram of stuff
2
accelerating at 1 cm/s
1 gram of stuff
2
accelerating at 1 m/s
Weighs 981 dynes
mass
system
Weighs 1 Newton
A mass of stuff
accelerating at
“standard” sea level
gravitational value
force
system
Weighs 9.8 kg force
The amount of stuff that has this weight
has a mass of 1 kg force-s 2/ meter
(ii) 2 “national” systems
British Mass System
1 pound
of stuff
mks
Object
accelerates
American Engineering System
1 slug of
stuff
at 1 ft/sec 2
Object weighs 32.2 poundals
Object weighs 32.2 pounds force
inc
Numerical methods for engineers includes dimension systems
(i) The three “3 unit” systems
( lenght, mass, time )
cgs
mks
1 gram of stuff
2
accelerating at 1 cm/s
1 gram of stuff
2
accelerating at 1 m/s
Weighs 981 dynes
mass
system
A mass of stuff
accelerating at
“standard” sea level
gravitational value
Weighs 1 Newton
force
system
Weighs 9.8 kg force
The amount of stuff that has this weight
has a mass of 1 kg force-s 2/ meter
(ii) 2 “national” systems
British Mass System
1 pound
of stuff
mks
Object
accelerates
American Engineering System
1 slug of
stuff
at 1 ft/sec 2
Object weighs 32.2 poundals
(iii) the “4 unit” system
<
F
= ( 1 ) (mass)< a
g
32.2 lb
ft
mass
1 lb
s2
force
Object weighs 32.2 pounds force
( force, length, mass, time )
Common
in USA
1 lb mass
1 lb
will make 1 lbmass
force
2
accelerate at 32.2 ft/s
Weighs 1 lb force
inc
Example using 4 unit system
Pressure difference =
(top to bottom)
1
(mass )(a)(ft )
ft 3
ft
lb
2
mass
) ( 32.2
)( 10 ft)
P = (62.4
s2
ft3
P = gc
water
density
2
P = (2,020 x 10
lbmass
ft s 2
acceleration
)
Z
water tower
is in Tampa
2
10 ft
Z = 100 ft
Not typical pressure units
but they are still pressure units.
1 lb
will make 1 lbmass
force
Since mass is in lb mass the force is lb force
2
accelerate at 32.2 ft/s
inc
Example using 4 unit system
1
(mass )(a)(ft )
ft 3
ft
lb
2
mass
) ( 32.2
)( 10 ft)
P = (62.4
s2
ft3
P = gc
Pressure difference =
(top to bottom)
water
density
2
P = (2,020 x 10
acceleration
lbmass
ft s 2
)
Z
water tower
is in Tampa
2
10 ft
Z = 100 ft
Not typical pressure units
but they are still pressure units.
1 lb
will make 1 lbmass
force
Since mass is in lb mass the force is lb force
g =
32.2 lb mass ft
1 lb
s2
force
2
accelerate at 32.2 ft/s
conversion factor
between lb force and lb mass
inc
Example using 4 unit system
1
(mass )(a)(ft )
ft 3
ft
lb
2
mass
) ( 32.2
)( 10 ft)
P = (62.4
s2
ft3
P = gc
Pressure difference =
(top to bottom)
water
density
2
P = (2,020 x 10
lbmass
ft s 2
)
Z
g =
force
water tower
is in Tampa
2
10 ft
Z = 100 ft
Not typical pressure units
but they are still pressure units.
Since mass is in lb mass the force is lb force
32.2 lb mass ft
1 lb
s2
acceleration
Remember force is
represented as a vector.
conversion factor
<F
between lb force and lb mass
1 lb
will make 1 lbmass
force
2
accelerate at 32.2 ft/s
= (
1
) (mass) < a
g
< | is symbolism used to
help you remember that
acceleration is a vector.
inc
Example using 4 unit system
P = gc
Pressure difference =
(top to bottom)
Z
acceleration
water tower
is in Tampa
1
(mass )(a)(ft )
ft 3
2
10 ft
Z = 100 ft
g =
conversion factor
<F
between lb force and lb mass
32.2 lb mass ft
1 lb
s2
force
2
(2,020 x 10
lbmass
ft s
2
( 1 lb
)
s2 )
force
32.2 lbmass ft
Pressure (top to bottom) of this
tank in non-conventional units
= (
1
) (mass) < a
g
acceleration
is a vector.
1 lb
will make 1 lbmass
force
2
accelerate at 32.2 ft/s
= 62.4 x10 2
lb
force
ft2
this term also represents the pressure
difference,
P, for this tank in familiar units.
inc
Example using 4 unit system
P = gc
Pressure difference =
(top to bottom)
acceleration
water tower
is in Tampa
1
(mass )(a)(ft )
ft 3
Z
2
10 ft
Z = 100 ft
g =
conversion factor
<F
between lb force and lb mass
32.2 lb mass ft
1 lb
s2
force
2
(2,020 x 10
lbmass
ft s
2
( 1 lb
)
s2 )
force
32.2 lbmass ft
= (
acceleration
is a vector.
1 lb
will make 1 lbmass
force
2
accelerate at 32.2 ft/s
= 62.4 x10 2
Pressure (top to bottom) of this
tank in non-conventional units
Note:
1
) (mass) < a
g
lb
force
ft2
this term also represents the pressure
difference,
P, for this tank in familiar units.
The “4 unit” system entertains two density concepts.
g < F = (mass) < a
(force magnitude) (
32.2 lb
32.2 ft
ft
mass
) = (mass) (
)
2
1 lb
s
1 s2
force
32.2 lb
ft
mass
1 lb
s2
force
force density
62.4 lb force
3
1 ft
Both look the same (have
units of pounds per foot
cubed) but each represents
a different concept.
mass density
(If you do not keep them straight the model of your system might sink.)
62.4 lb mass
3
1 ft
inc
Example using 4 unit system
End of presentation
inc