This document discusses the three factors that determine the holding force of a clutch: force, pressure, and area. It explains that force is measured in pounds and depends on the torque put into the clutch. Pressure is measured in pounds per square inch (PSI) and is controlled by the clutch pump and regulator based on load. Area refers to the surface of the clutch piston measured in square inches, where a larger area or higher pressure results in greater holding capacity. The document uses diagrams and equations to illustrate how multiplying pressure by area determines the output force exerted by the clutch.

2. Force
 The force in pounds
that the piston exerts
on the clutch or band
PSI
 Pounds Per Square
Inch
Area
 The “working” area of
a piston in square
inches
THE 3 THINGS THAT MAKE IT WORK
Force (#)
Pressure
(PSI)
Area
In²

3. FORCE
• Force is what “holds” the clutch from
slipping
• It’s measured in pounds (#)
• The more torque put into the clutch…the
more force is needed
• BUT….
• It’s a Yin & Yang thing
More pressure = More holding capacity
More Area = More holding capacity

4. • Pressure is
created in the
pump
• A regulator
controls
pressure based
on load
• Load is based on
throttle angle
• The pressure is
measured with a
pressure gauge
PRESSURE

5. AREA
 Area refers to the square inches (in²) of surface
that fluid can act on
 When looking at AT clutch pistons, there is a
hole in the center…
 This piston is called an Annulus
 Annulus-A ring-shaped object, structure, or
region.
 To find the area we need only one equation
 A (area) = (3.14) X R²(radius X radius)

6. MULTIPLICATION OF
FORCE
10 PSI
1 in2
Piston Area
10 lb.
Input
Force
20 lb.
Output
Force
10 PSI
10 PSI
10 PSI
10 lb.
Output
Force
10 PSI
5 lb.
Output
Force
2 in2
Piston
Area
Force = PSI x Piston Area
1 in2
Piston
Area
.5 in2
Piston
Area

7. • A Governor
Valve
determines
Shift Points
• Valve
movement is
based on the
“Face” size
• Pressures are
different on
each valve face
SPOOL VALVE MOVEMENT

8. MOTION
10 in.
1 in2
Piston Area
2 in2
Piston
Area
20 lb.
Output
Force
Force = PSI x Piston Area
Distance = PSI  Piston Area
10 PSI
1 in2
Piston
Area
10 lb.
Output
Force
.5 in2
Piston
Area
5 lb.
Output
Force
10 in.20 in. 5 in.
10 lb.
Input
Force