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Comparison summary: Pneumatically vs. Caliper hydraulically actuated brakes Condition Pneumatic Hydraulic Life Up to 10 years Up to 15 years Initial cost, Approx. 50% cost High Energy cost Medium Higher Maintenance Low cost repairs Expensive replacement costs brakes: Noise Quiet – normally remote Loud — power unit nearby Temperature limit Above freezing Sub-zero versus Wet conditions Seal protection to IP67 Underwater compatible Clamp force 2,550 lb 3,143 lb Pneumatically- Broken hoses Clean; hose whip is dangerous Dirty, messy When it comes to choosing the released caliper Controls cost Lower, less complicated Higher, more components right brakes for an application, brake Components of a Pressure, maximum 80 to 100 psi 1,000 to 5,000 psi which is better — pneumatically or pneumatic system Release pressure 72 psi 250 psi include an air hydraulically released ? Friction-facing area 15.70 sq. in. 30.96 sq. in. supply, filter, Learn where each type is most regulator/ Facing life 628 hp-hours 1,238 hp-hours gauge, direc- Safety Non-sparking Flammable appropriate. tional control , and Actuation method Air or are compressible Oil not compressible Edd Brooks quick Control system Open, exhausts air to atmosphere Closed, recirculates oil Nexen Group Inc. exhaust System flexibility Easy to modify and change Not easy to expand valve. Vadnais Heights, Minn. Disc diameter 18 in. 14 in. Disc thickness 0.50 in. 1.0 in. Disc inertia 10 lb-ft2 7 lb-ft2

aliper disc brakes are Brakes can be either hydrauli- function of the brake’s clamp- to perform a full stop. among the most tried and cally or pneumatically released. ing force and the working ra- Other brake considerations Ctrue braking Pneumatic systems (using air) dius of a disc in rotary applica- include the disc diameter, disc in industrial use. Designed to are widely used in manufactur- tions — needed to stop or hold heat sink capacity, and friction clamp onto both sides of discs ing to operate at low power and the load is the most important facing life. (Examples later in or flat rails, these specialized pressures, whereas hydraulic factor when choosing a caliper this article illustrate the steps brakes create friction that ex- systems (using fluids) are used brake. necessary to arrive at these erts a braking force on the disc where greater amounts of pow- Typically, spring-engaged, values.) attached to the rotating shaft, or er and pressure are required. air-released (pneumatic) brakes Brake torque (τ) is equal to on a rail or fin in linear motion appli- The decision to use one are used for stopping and hold- the pressure on the disc (F) Hydraulically- cations. Caliper brakes are well suited brake style over the other is ing in moderate torque range multiplied by the design co- released caliper brake for a range of industrial uses, such as determined by the application’s applications, while spring-en- efficient of friction (r). With Components of a hydraulic construction machinery, material han- torque requirement, system gaged, hydraulically-released r = 0.35 and pressure on two system include an oil tank, strainer, dling equipment, metalworking machin- cost, working environment, brakes are used when higher facing discs, total brake torque pump, pressure gauge, relief valve, and directional control valve. ery, automated assembly , and and safety. torque is needed. equals F × 2 × 0.35. other heavy equipment. In linear motion applica- Spring-engaged caliper brakes use spring forces to se- Determining required tions, a caliper brake clamp Cost comparison cure the load for and static holding appli- torque force value is used to determine While hydraulic systems cations. The return mechanism is where designs diverge: The amount of torque — a the time and distance needed accommodate greater torque

20 APRIL 2011 MOTION SYSTEM DESIGN • motionsystemdesign.com APRIL 2011 MOTION SYSTEM DESIGN • motionsystemdesign.com 21 Brakes

ings at all pivot points reduce ated brakes and clutches, which Application example: Caliper brakes clamping on a rail Caliper brakes: Basic model and working radii maintenance requirements. require a minimum operating Adjustment screws are also range of 32° F. Additionally, available for maintaining the spring-engaged, hydraulically Determine the deceleration rate A in ft/sec2 ... Disc facing gap in vertical shaft ap- released caliper brakes can of- plications, and a built-in man- fer some of the same features as V 2 Pressure on the disc is ual release mechanism allows air-released brakes, including A = Where V = Velocity, ft/sec and S = Required stopping distance, ft Working radius r applied from two sides. safe and easy maintenance. quick-release shoe pins, adjust- 2S F F Typical specifications of ment screws, oil-free bearings spring-engaged, air-released cal- at pivot points, interchange- Then calculate the linear force Fl: iper brakes are as follows: able locations, and a W A Where W = Weight of the load • Spring hold-off chamber manual release accessory. = 2 Examples of working radii of various disc diameters Fl and acceleration due to gravity = 32.16 ft/sec Disc diameter 12 14 16 18 20 22 24 area = 23.758 sq. in. Typical specifications of 32.16 Radii 4.475 5.562 6.625 7.625 8.625 9.625 10.938 • Air pressure needed to fully spring-engaged, hydraulically compress the spring = 70 psi released caliper brakes are as fol- Finally, calculate the clamping force Fc needed to stop the load • Spring force = Area × spring lows: in the distance required. requirements, pneumatic sys- Pressure ranges and force = 23.758 (70) = 1,663 lb • Spring hold-off chamber tems are more cost-effective. safety factors • Caliper mechanical advan- area = 9.62 sq. in. Where: F = linear force Fl l Why? Despite these environmental tage = 2.04 • Air pressure needed to fully Fc = 0.35 = Design coefficient of friction ... And the coefficient is Pneumatic pressure is usual- drawbacks, hydraulic systems • Design coefficient of compress the spring = 250 psi 2(0 .35) multiplied by two, to account for two caliper sides. ly cheaper to obtain — as most are able to develop extremely friction = 0.35 • Spring force = Area x spring industrial facilities already high pressures that produce force = 2,405 lb have readily much higher forces in actuated Spring-engaged, • Caliper mechanical advan- What is the clamping force needed to stop a 1,200-lb load in 4 in. that is traveling at 70 ips? available. components than their pneu- hydraulically released tage = 2.04 And, because many systems matic counterparts. Hydraulic caliper brakes • Design coefficient of Velocity: 70 ips = 70/12 = 5.8 ft/sec don’t require extremely high fluids are not compressible, As mentioned, spring-en- friction = 0.35 4-in. stopping distance = 0.333 ft forces and only need to make so hydraulic systems offer ex- gaged, hydraulically released 5.82 ft movements from one position ceptionally smooth motion of caliper brakes are typically se- Application example: Air A = = 50.5 to another (without making actuated components because lected for applications with and hydraulically actuated 2(0.333) sec2 intermittent stops), pneumatic there is no “bounce” due to greater torque requirements or brakes × systems are often the easier fluid compressing and expand- applications that require brak- Calculate the torque needed 1,200 50.5 Fl = =1,885 lb and less costly choice. ing, as is typical of pneumatic ing between actuator move- to stop a mill roll in 7 sec for the 32.16 systems. ments. While a variety of op- following conditions: 1,885 Environmental factors tions are available, a brake • Roll radius d = 2.083 ft Fc = = 2,692 lb clamping force Along with cost advantages, Spring-engaged, air- should be selected with features 2(0.35) pneumatic systems provide envi- released caliper brakes to simplify setup and enhance ronmental benefits over hydrau- Along with lower cost and performance. lic systems. When a hydraulic environmental advantages, For example, a large fric- system leaks, it leaks hydraulic spring-engaged, air-released tion-facing area lowers fac- fluid that could contaminate the caliper brakes can include sev- ing pressure to provide long environment or product being eral features designed for ease friction-facing life. A nested For this example, a spring-engaged, air-released manufactured. of use and increased accuracy: spring design minimizes spring brake, with a clamp force of 2,550 lb, has rect- When pneumatic systems Rectangular friction facings force loss and allows the brak- angular shoes that are suitable for clamping onto a 0.50-in. thick rail, but will allow a slightly leak, only air escapes, and the provide reliable clamping on a ing torque to be modified based longer stopping distance than specified. In con- air can be kept very clean with rail; quick-release facing shoe on specific application require- trast, the spring-engaged, hydraulically released properly selected and main- pins can facilitate fast brake ments. brake, with a clamp force of 3,143 lb, stops the tained equipment. shoe removal from the actuator Large orifice ports provide load in a shorter distance than specified — but This environmental advan- arms for simple replacement of for fast actuation and reliable the arced shoes must be replaced with the rect- tage can be particularly impor- the friction pad. Interchange- operation — even in tempera- angular shoes from the air-released brake. tant in food, pharmaceutical, able actuator locations may tures as low as -30° F. This printing, and other facilities be mounted on either side of provides a significant environ- where cleanliness is critical to the brake, delivering increased mental advantage compared to the manufacturing process. flexibility, and oil-free bear- traditional pneumatically-actu-

22 APRIL 2011 MOTION SYSTEM DESIGN • motionsystemdesign.com APRIL 2011 MOTION SYSTEM DESIGN • motionsystemdesign.com 23 Brakes

• Roll weight w = 4,410 lb 7.625 in. A brake disc’s thermal horse- • Deceleration rate = 4.92 ft/ Torque τ = 2,550(0.35) power capacity is partially deter- sec2 ×2(7.625) = 13,610 in.-lb mined by velocity V in fps and • Speed = To be calculated • For a spring-engaged, hy- continuous heat dissipation Q: • Line speed = 1,969 fpm draulically-released brake and • Stop time t = To be calcu- 14-in. diameter × 1.00-in. thick lated disc:  Calculate speed. Clamp force = 3,143 lb

14-in. disc working radius = and Q = k1 × ta × A 5.562 in. Where N = Disc rpm

Torque τ = 3,143(0.35) × k1 = 0.00063 √V 2(5.562) = 12,237 in.-lb ta = Allowable temperature rise  Verify the kinetic energy from ambient, use 320° F output per cycle. A = Contact area of the fric- Ec = 0.00017 (WK2) rpm2 tion disc = 0.7854(D2 – d2) For the load energy of our D = Outside diameter of  Calculate inertia. example, Ec = 0.00017(9,567) × contact area 1502 = 36,598 ft-lb. d = Inside diameter of contact  Determine the total area system inertia associated with each disc. Final considerations 14-in. disc: 36,594 + 7 = In the rotary application we 36,601 lb-ft2 explored, both brake-actuator  Calculate the stop time. 18-in. disc: 36,594 + 10 = types perform the stop. The de- 36,604 lb-ft2 cision of which style to use can be determined by other factors Final factors: previously described, including Life and thermal capacity the environment, physical space Friction facing life is the max- for the brake and disc, and over- imum number of cycles based on all system cost. horsepower hours (hp hr) value If the is located inside a  Calculate the torque of the friction facing. building where an air supply is read- needed to stop the roll in 7 Horsepower hours of a fric- ily available and the temperature is sec. tion facing is the usable amount stable, the logical choice is the air- 2 of facing material in cubic inches engaged unit. For outdoor use, at 0.39(WK ) = rpm o = divided by the material’s wear temperatures below freezing, two t rate: types of brake actuation methods 0.039(9,567 ) = 150 are possible — hydraulic or nitro- = gen. Nitrogen is commonly used 7 to replace air in many types of ap- = 8,000 in. - lb plications. Nitrogen is a colorless, odorless, tasteless gas, constituting Applying a service factor of 78.08% of air by volume, which 1.50 gives a final torque = 1.50 × remains stable at any temperature. 8,000 = 12,000 in.-lb. The moisture content in compressed  Compare torque gener- nitrogen is very low compared to the ated by different brakes and moisture content in compressed air. discs. Less moisture (oxygen) means dry- • For a spring-engaged, air- er components and less oxidation released brake and 18-in. diam- (rust). eter × 0.50-in. thick disc: Clamp force = 2,550 lb For more information, visit nexen- 18-in. disc working radius = group.com or call (800) 843-7445.

24 APRIL 2011 MOTION SYSTEM DESIGN • motionsystemdesign.com