Analysis of Small-Scale Hydraulic Systems

Jicheng XiaDepartment of Mechanical Engineering, University of Minnesota

111 Church Street SE, Minneapolis, MN, 55455Email: xiaxx028@umn.edu

William K. Durfee∗

Department of Mechanical Engineering, University of Minnesota111 Church Street SE, Minneapolis, MN, 55455

Email: wkdurfee@umn.edu

We investigated small-scale hydraulic power system usinga system level analysis, where small-scale refers to sys-tems generating 10 to 100 Watts output power, to deter-mine whether the high power density advantage of hydraulicpower system holds at small sizes. Hydraulic system powerdensity was analyzed with simple physics models and com-pared to an equivalent electromechanical system comprisedof off-the-shelf components. Calculation results revealedthat high operating pressures are needed for small-scale hy-draulic power systems to be lighter than the equivalent elec-tromechanical system.

1 IntroductionHydraulic fluid power systems are well known for their

high power density [1, 2]. This advantage is best illus-trated in applications such as excavators and heavy manu-facturing equipment that require extremely large power andforce. Hydraulics is the only practical way to attain theselevels of force and power while at the same time being rel-atively light weight compared to the equivalent electrome-chanical system. One reason for the high power density ofhydraulics is that fluid power cylinders are inherently low-velocity, high-force actuators, which is a good match to therequirements for construction, agricultural and manufactur-ing heavy equipment. Contrast this with electric motors thatare high-velocity, low torque actuators and require a trans-mission such as a gear head or a lead screw to match theiroptimal operating point to the application. At high forcesand torques, the weight of the transmission ends up being asignificant fraction of the actuator package weight. A secondreason for the high power density is that exceptionally highpressures can be generated (excavators run up to 5500 psi).

An advantage of hydraulics is that the source of pressur-ized fluid can be housed in a base station and flexible hosesused to transport the fluid to light weight cylinders located

∗Address correspondence to this author.

at the periphery of the machine. For example, an excava-tor has actuators to control the boom, stick and bucket withbulky power supply, reservoir and accumulators placed in thehouse. The more proximal actuators carry the load of themore distal. When the excavator arm is fully extended, thebucket actuator at the end of the arm causes large moments atthe joint connecting the boom to the house, which requires apowerful boom actuator. If the bucket actuator is a cylinder,the weight of the actuator is small compared to the bucket.If the bucket actuator is electromechanical, the weight of theelectric motor and its associated transmissions, both of whichmust be placed at the joint, can be significant.

The power density of electromechanical systems has anupper limit because of inherent characteristics such as mag-netic saturation. In contrast, the power density of hydraulicsystems has no inherent upper limit and can be increased bysimply increasing the pressure. The only barriers for increas-ing power density in a hydraulic system are the design thecontaining structure and the seals.

There has been recent interest in portable, wearablepowered systems including powered exoskeletons and pow-ered orthotics [3, 4]. Examples of mobile systems in the 10to 100 W range include ankle foot orthotics, small robots andpowered hand tools. These devices are usually powered byelectromechanics, typically a lithium-ion battery, DC electricmotor and transmission. Little work has been done on usinghydraulics for these applications because off-the-shelf tinyhydraulic components (cylinders, valves, power supplies) donot exist.

Designers might chose hydraulics for tiny, mobile pow-ered systems because the same power density advantage ofhydraulics over electromechanical should hold for a poweredorthosis as it holds for an excavator. The story, however, iscomplex because the scaling laws are not intuitive. For ex-ample, in a cylinder, force is proportional to area (L2) whileweight is proportional to volume (L3). Surface effects suchas friction drag of seals and viscous drag of gaps become

significant at small bores and impact overall efficiency. Onthe other hand, the thickness, and thus the weight, of a cylin-der wall required to contain a fixed pressure goes down withbore. The final weight of a hydraulic system at small scalecannot be determined by proportionally scaling the weight ofa large system. Thus the answer to the question, “For equalefficiency, which is lighter a fluid power or an electrome-chanical system?” for a tiny system cannot be answered us-ing intuition or linear scaling.

Love [5] demonstrated an application of small scale hy-draulics by prototyping a prosthetic finger. Pressure as highas 2000 psi (about 13.8 MPa) was used to operate 4 mm hy-draulic cylinders. In this way, much higher power densitywas achieved by hydraulic cylinder than electric motor. An-other example of using small scale hydraulics was illustratedin endoscope research [6]. Two systems were studied, hy-draulics and electric. The results showed that the hydraulicsystem had larger output force for the limited space.

A critical barrier for increased hydraulic power densityat reasonable efficiency is the seals. Too tight and frictiondominates. Too loose and the pressurized fluid will leak pastthe seal. Volder et al. developed a ferrofluid seal for mi-croactuators that was able to seal 1.6 MPa (about 230 psi)pressure was sealed without leakage [7, 8]. While microflu-idics has made tremendous advances in recent years, theycannot inform our problem as microfluidic systems operatewell under 1 W and our systems of interest are in the range of10 to 100 W. Reviews of microfluidics components are givenin [9–11]. As shown in [11], micro fluid power cylinders cangenerate 1 to 10 N but their strokes are under 10 mm .

The aim of this study was to use first principles to un-derstand how the weight and other properties of hydraulicsystems change with size and to answer the question, “Is ahydraulic solution lighter than an electromechanical solutionfor tiny, powered systems?” Our goal was to provide guide-lines that mechanical designers could use at the early stagesof evaluating architectures for small systems. Empirical andanalytical equations were used to model hydraulic and elec-tromechanical systems, connecting the methods to real com-ponents wherever possible.

Our approach was to conduct an analysis for a idealized,linear actuation benchmark system. The analysis only con-sidered the transmission line and actuator, leaving the powersupply and control means for future work. Because the sup-ply and control can be sited anywhere on the machine, thislimitation, while important, does not diminish the utility ofthe study. The result of the analysis showed that for equaloutput power and system efficiency, a hydraulic solution willbe lighter than an electromechanical solution only if the hy-draulics operates at high pressure.

2 Benchmark SystemThe top row of Fig. 1 illustrates the architecture of a

generic mobile actuation system that contains a power sup-ply, a means of control, a transmission line and an actuatorlocated at the end-point. For this study, we considered sys-tems that delivered force and velocity along a linear axis.

Fig. 1. Architcture for powered actuation system. Top row isgeneric, middle row is electomechanical, bottom row is hydraulic.

For example, this could be a powered knee prosthesis withthe joint driven by a linear actuator mounted behind the kneeor a tiny powered gripper driven by a linear actuator.

The electromechanical realization (middle row of Fig.1) includes a battery power supply, a PWM motor controller,wire, a brushed or brushless DC electric motor and a leadscrew or ball screw to convert the high velocity, low torqueoutput of the motor to a low velocity, high force linear output.The ball screw was chosen because it is lighter and moreefficient than the equivalent gear box, and it converts rotaryto linear motion, which provides a fair comparison to thehydraulic system. The hydraulic version (bottom row of Fig.1) includes a battery or internal combustion engine drivenpump to generate pressured fluid, a servovalve, pipe or hoseand a hydraulic cylinder. Other realizations are possible.

Our analysis only considers the transmission line plusactuator system, the circled components in Fig. 1. These arethe parts of the system that must be located at the point of me-chanical output where weight is of greatest concern. For ex-ample, for a portable hand tool, the power supply and controlcan be placed in a backpack or tool belt, but the transmissionline and actuator system must be held in the hand. In a realmobile system, the power supply will contribute significantlyto the weight and in a real system, the control means willcontribute significantly to the efficiency. Comparing elec-tromechanical and hydraulic endpoint components, however,still provides valuable information to the designer looking tominimize weight at the endpoint.

3 Hydraulic System AnalysisThe objective of the hydraulic system analysis was to

estimate the weight of an ideal hydraulic cylinder plus theweight of ideal conduit to predict the total weight for a hy-draulic system that delivers a specified mechanical force andpower output. The weight of components was estimatedfrom a set of theoretical equations developed using basicphysics of fluids and solid mechanics.

3.1 Hydraulic CylinderThe simplified hydraulic cylinder used for analysis is il-

lustrated in Fig. 2 and its associated parameters are definedin Tab. 1. The cylinder is double-ended with bore B, stroke S

Fig. 2. Ideal hydraulic cylinder used for analysis

Table 1. Hydraulic cylinder parameters

VAR DESCRIPTION UNIT

B bore m

S stroke m

l1 cylinder length m

l2 rod length m

t1 piston thickness m

t2 cylinder circumferential wall thickness m

t3 left end wall thickness m

t4 right end wall thickness m

d1 rod diameter m

d2 outer diameter m

Pm maximum allowable fluid pressure Pa

P1 cylinder left chamber pressure Pa

Sy cylinder material yield strength Pa

E cylinder material Young’s modulus Pa

ρ cylinder material density Kg/m3

ν cylinder material poisson’s ratio —

N design safety factor —

and rated maximum pressure Pr. The piston is a disk of uni-form thickness t1 and the cylinder housing is a capped tubewith barrel wall thickness t2 and end cap thicknesses t3 andt4. For simplicity, O-ring seals are assumed for piston androd. Only uni-direction extension motion is considered withcap side pressure P1 and rod side pressure zero.

3.1.1 Cylinder and Piston Wall ThicknessThe pressure loading scenario to calculate the required

cylinder wall and the piston thickness is shown in Fig. 3where the cylinder rated pressure Pr acts everywhere on thewall. The end wall calculations assumed a fixed displace-ment boundary condition along the end wall circumference.

Fig. 3. Wall loading scenario used to calculate wall thickness

The piston thickness calculation assumed that the rod wasfixed and the Pr was distributed uniformly across the cap sideof the piston and zero pressure on the rod side. These are allworst-case loading conditions.

The cylinder circumferential wall thickness was calcu-lated using the equation for a thin-walled pressure vessel [12]

t2 =N ·Pr ·B

2Sy(1)

which is valid for t2 < B/6. The cylinder end wall thick-nesses, t3 and t4, and the piston thickness t1 were calculatedusing thin plate formulas [13]

t1 =

√3NPrG1ν

4Sy(2)

t3 =

√3πB2NPr(1+ν)

32πSy(3)

t4 =

√3NPrG2

4νSy(4)

where G1 and G2 are defined as

G1 =4B4(1+ν)log B

d1+4νB2d2

1 +d41(1−ν)−B4(1+3ν)

4ν(B2 −d21)

G2 =d4

1(1−ν)−4d41(1+ν)log B

d1+B2d2

1(1+ν)

4B2(1−ν)+4d21(1+ν)

+B2

4

− d21

2

The thin plate formulas are valid for plate thickness that areless than 1/4 of the plate diameter. The formula used to de-termine t4 was that for a round plate containing a central hole.

3.1.2 Rod DiameterThe rod must be sized so that it will not buckle under

the maximum compressive load. The required rod diameterwas calculated using Euler and JB Johnson buckling formu-las [13], assuming that the rod was fully extended, loaded incompression and carrying the piston force at the maximumrated pressure. The slenderness ratio l2

ρ1dictates whether the

Euler or the JB Johnson formula is appropriate. The criticalrod slenderness ratio is

( l2ρ1

)crit

=

√2π2E

Sy(5)

where ρ1 = d1/4 for a solid round rod. For a slendernessratio less than the critical value the JB Johnson formula wasused

d1 =

√4NPrπ(B/2)2η f

πSy+

4Syl22

π2E(6)

and for other cases, the Euler formula was used

d1 =(64Nl2

2Prπ(B/2)2η f

π3E

) 14

(7)

The cylinder force efficiency η f , which results from sealingfriction, is calculated in the next section.

3.1.3 Cylinder EfficiencyThe force in the rod is less than the pressure times the

area of the piston because of the friction in the piston and rodseals. The cylinder force efficiency, η f is defined as

η f =Fr

P1A1(8)

where Fr is the rod compressive force, P1 is the cap side pres-sure and A1 is the cap side piston area [14].

Equations (9) and (10) are approximations that describethe seal friction [15] and leakage ( [16, 17]) for a rubber O-ring seal. Variables for these equations appear in Tab. 2.

Fs = fs ·π ·D ·d ·Es · ε ·√

2ε− ε2 (9)Qs = 2.99 ·π ·D ·µ0.71 ·U1.71

hc ·δ−0.71m · s0.29

0 (10)

Applying eqns (9) and (10) for the piston and the rodyields the estimation of the cylinder force efficiency, volu-metric efficiency and overall efficiency

η f =P1A1 −Fsp −Fsr

P1A1(11)

ηq =VrA1

VrA1 +Qsp +Qsr(12)

ηhc = η f ·ηq (13)

where Fsp is piston seal friction force (Newton), Fsr is rodseal friction force (Newton), Vr is rod velocity (meter/sec),Qsp is piston seal leakage (cu-m/sec) and Qsr is rod seal leak-age (cu-m/sec).

Table 2. Symbols used in eqns (9) and (10)

VAR DESCRIPTION UNIT

Fs friction force piston with seal N

fs O-ring seal friction coefficient —

D piston or rod diameter m

d O-ring cross-sectional diameter m

Es O-ring Young’s modulus Pa

ε O-ring squeeze ratio —

Qs leakage across sealed piston or rod m3/s

µ hydraulic fluid dynamic viscosity Pa· s

Uhc piston velocity m/s

δm maximum O-ring contact stress Pa

s0 O-ring contact width m

3.1.4 Cylinder WeightThe volume of the cylinder is

Vcyl =π

4

[(d2

2 −B2)l1 +B2(t3 + t1 + t4) (14)

+d21 l2 −∆V

]where ∆V are the adjustments to the volume due to the inlet,outlet and rod openings. For simplicity, only the rod openingvolume will be included as the inlet and outlet openings willbe balanced by the volume of fittings.

∆V = d21 · t4 (15)

Assuming the same material is used for the cylinder wall,piston and rod, the weight of the cylinder is

Mcyl = ρ ·Vcyl (16)

3.1.5 ValidationTo validate the calculation of estimated cylinder weight

based on the basic theory presented in the previous sec-tions, (16) were used to predict the weight of commerciallyavailable cylinders. Catalog data for 187 hydraulic cylin-ders from several manufacturers (Airpot, Beily, Bimba, Her-cules, Prince) was used to build a database of rated pressure,bore, stroke and weight for real products. For the analysis,the cylinder material was assumed to be 304 stainless steel,which provided the yield strength, Young’s modulus, Pois-son’s ratio and material density for the equations. (A realcylinder would be fabricated from several materials.) Thesafety factor N was set to 2 as this was the value found intwo of the vendor catalogs. Common parameters were usedfor O-ring seal and hydraulic oil: 10% for squeeze ratio,

Fig. 4. Comparison between the actual weight and the weight pre-dicted from the theoretical analysis for 187 commercial cylinders. Thesolid line indicates an exact match between actual and predicted.The inset expands the data for light weight cylinders.

10 MPa for O-ring Young’s modulus, 1 mm for O-ring sealcross-section diameter and 0.1 Pa·s for fluid viscosity. Thepressure, bore, stroke, material properties and safety factorwere used to calculate the theoretical wall thickness, volumeand weight for the cylinder. The theoretical weight was thencompared to the real catalog weight for the cylinder.

Fig. 4 compares the predicted weight to the actualweight. If the theory matched the real cylinders exactly, alldata points would lie on the solid line. Practical cylindersare somewhat lighter than their predicted weight for heaviercylinders, and somewhat heavier than their predicted weightfor lighter cylinders (see the figure inset).

A test to check whether the assumptions underlying cer-tain formulas used were valid was run on the 187 cases. Theassumptions tested were: (1) cylinder wall thickness smallerthan 1/6 of cylinder bore for correct use of the thin-walledcylinder formula [12]; (2) plate thickness smaller than 1/4 ofplate diameter for correct use of the thin plate formula [13];(3) rod diameter smaller than cylinder bore to satisfy physicalfeasibility. With a safety factor of two, some of the theoreti-cal results violated the thin plate formula assumption for thepiston and the rod end cylinder wall. This can be correctedby using a safety factor of less than two, which would resultin thinner pistons and walls. The formula assumptions heldfor the 20 lightest cylinders in the data base with weights lessthan 0.4 kg.

3.2 Hydraulic ConduitFor smooth pipes, the fluid flow equations [18] are

P2 −P1 =fp ·ρf ·V 2

p ·Lp

2 ·Dp(17)

Ap =π ·D2

p

4(18)

Vp =Qp

Ap(19)

Re =ρf ·Dp ·Vp

µ(20)

fp =

64/Re laminar flow

0.316/Re0.25 turbulent flow(21)

where P2 is pipe inlet pressure (Pascal), P1 is pipe outlet pres-sure (Pascal), fp is pipe friction coefficient, ρf is fluid density(kg/cu-m), Vp is pipe flow velocity (m/s), Lp is pipe length(m), Dp is pipe inner diameter (m), Ap is pipe cross-sectionarea (sq-m), Qp is pipe flow rate (cu-m/sec) and Re is theReynolds number.

Using eqn (17) through eqn (21), the pipe efficiency is

ηp =P1

P2(22)

=

1− 128µ

π· Qp·Lp

D4p·P2

laminar

1− 1.79µ0.25·ρ0.75f

π1.75 · Q1.75p ·Lp

P2·D4.75p

turbulent

These equatiions enable calculating the pipe i.d. Dp as afunction of Qp, Lp, P2 and ηp.

The pipe weight can be calculated once the pipe wallthickness is found using the thin-walled cylinder formula[12]

t5 =N ·P2 ·Dp

2Sy(23)

where t5 is wall thickness (m), N is design safety factor, andSy is pipe material yield strength (Pa). For this analysis weassumed that the pipes, like the cylinders, were fabricatedfrom 304 stainless steel.

The weight of the pipe is

Mconduit = π

((

Dp

2+ t5)2 − (

Dp

2)2)

Lpρp (24)

where ρp is the pipe density (kg/cu-m). The weight of the oilin the pipe is

MConduitOil = π(Dp

2)2Lpρ f (25)

Fig. 5. Method for calculating the weight of a hydraulic system

where ρ f is the oil density (kg/cu-m).The pipe efficiency ηp, inlet pressure P2 and fluid flow

rate Qp were calculated using

ηp =ηsys

ηcyl(26)

P2 =P1

ηp(27)

Qp =Fr ·Vr

ηsys ·P2(28)

where ηsys is the desired overall efficiency (Section 4), ηcylis the cylinder efficiency, Fr is rod force (N) and Vr is rodvelocity (m/sec).

3.3 Hydraulic System WeightFig. 5 illustrates how the weight of the hydraulic sys-

tem is calculated. First, the output force, output velocityand stroke length are specified by the application require-ments. Using this information, cylinder weight, efficiency,bore and rod diameter are calculated. Next, using the overallsystem efficiency of the equivalent electromechanical system(Section 4.4), the hydraulic pipe inlet power and efficiency iscalculated. Hydraulic pipe inlet pressure is calculated for agiven cylinder operating pressure. Next hydraulic pipe inletflow rate is calculated using inlet power and pressure, thenthe hydraulic pipe diameter using inlet pressure, inlet flowrate, pipe efficiency and pipe length information, as shown ineqn (22). With these numbers, pipe weight is calculated. Fi-nally, total system weight is calculated by summing weightsof the cylinder, pipe and hydraulic oil contained in the cylin-der and pipe.

4 Electromechanical System AnalysisThe electromechanical system includes wire for the

transmission line, a DC electric motor and a ball screw. Un-like hydraulic components, electromechanical components

Fig. 6. Motor weight vs. output power

Fig. 7. Motor efficiency vs. output power

for small-scale applications are readily available. Therefore,rather than using theoretical methods, the approach to esti-mating the total weight of an electromechanical solution wasto develop a set of empirical equations that captured the scal-ing of component weight and efficiency with load or powerbased on the properties of high-end, commercially availableelectromechanical components captured from company cat-alogs.

4.1 DC Electric MotorThe key system-level parameters for DC electric motors

are weight and efficiency. Brushless, permanent magnet DCmotors were chosen because for small precision applicationsthey have the highest efficiency and highest power density.Power, weight and efficiency data for 192 motors from twomanufacturers (MicroMo Electronics Inc. and Maxon Mo-tor) were collected. The power for a motor was taken as thepeak continuous mechanical output power and the efficiencywas the electrical power in to mechanical power out maxi-mum efficiency at the nominal voltage. Fig. 6 plots motorweight versus motor power and Fig. 7 plots motor efficiencyversus power.

For modeling purposes, empirical equations were cre-ated to bound motor properties. The lower curve in Fig. 6is the lower bound of motor weight. Using this curve in asystem analysis means that one is looking for the lightestcommercially available motor for a given power. The upper

Fig. 8. Ball screw weight vs. rated dynamic load at .01 m (top) and.04 m (bottom) stroke

curve in Fig. 7 is the upper bound of motor efficiency. Usingthis curve in an a system analysis means that one is lookingfor the highest efficiency motor for a given power. The twobounding curves are

Wm =P1.5

m

12(29)

Em = 0.9−0.9 · 0.10.15 ·Pm +0.1

(30)

where Wm is motor weight, ηm is motor efficiency and Pm ismotor power (W).

4.2 Ball ScrewThe ball screw converts the motor rotary power to low

speed, high force linear power. The weight of a ball screwis related to its rated dynamic load and to its stroke length.Weight does not depend on rated velocity assuming the ballscrew operates within its rated velocity. Rated dynamic load,stroke length and weight data were collected from catalogdata for 82 ball screws from one manufacturer (Nook Indus-tries). Fig. 8 shows their weight as a function of rated loadfor two strokes.

An empirical equation for the lower bound of weight asa function of load and stroke was developed from the data

Wbs = Fbs ·180+3000 ·Sbs

10000(31)

where Wbs is ball screw weight, Fbs is ball screw rated dy-namic load, and Sbs is ball screw stroke length. The equationis shown as the solid line in Fig. 8.

The transmission equations for the ball screw are

Tm = Fbs ·Lbs

2π· 1

ηbs(32)

ωm =Vbs

Lbs(33)

where Tm is motor shaft torque (mN·m), Fbs is ball screwforce (N), Lbs is the transmission ratio (mm/rev), ηbs is ballscrew efficiency, ωm is motor shaft velocity (rev/sec) and Vbsis ball screw linear velocity (mm/sec).

The ball screw efficiency was assumed to be 90%, whichis typical for a high performance component. To simplifythe electromechanical systems analysis, a fixed transmissionratio of 1 mm/rev was assumed for the ball screw.

4.3 WireThe weight and efficiency of wire should be considered

when analyzing electromechanical systems. The effects ofthe wire can be important when the wire carries large currentover a signifcant distance, which would be the case whenthe battery is located some distance from the motor. Highefficiency wire is large diameter, but heavy.

The voltage drop across a length of electrical wire is [19]

∆Uw =4Kw

π· Pw

Uw· Lw

D2w

(34)

where Kw is wire specific resistance (Ohms· m), Pw is wireinput power (W), Uw is wire input voltage (V), Lw is wirelength (m) and Dw is wire diameter (m). Thus, wire effi-ciency is

ηw =Uw −∆Uw

Uw

= 1− 4Kw

π· Pw

U2w· Lw

D2w

(35)

High wire efficiency results in a large wire diameter and thusa large wire weight. In contrast, low wire efficiency meansthe wire must dissipate a considerable amount of thermal en-ergy, which can melt the insulation or even the conductor. Toprevent the system level weight optimization algorithm fromsuggesting either extreme, the wire efficiency was fixed at99%, which is realistic for many systems.

Inverting (35), provides an equation for wire diameter

Dw =

√4Kw

π· Pw

U2w· Lw

(1−Ew)(36)

Fig. 9. Method for calculating the weight of an electromechanicalsystem

and the weight of the wire, without considering the insulationlayer, is

Ww =π

4·D2

w ·Lw ·ρw ·1000 (37)

where Ww is wire weight, and ρw is the density of the wirematerial. The analysis only considered copper wire withdensity 8960 kg/m3 and specific resistance 17 nΩm.

4.4 Electromechanical System WeightFig. 9 illustrates the approach for calculating the weight

of the electromechanical solution.The application requirements set the ball screw output

velocity, force and stroke. The ball screw weight is then cal-culated using (31). The electric motor shaft power (in W) iscalculated from

Pm =Tm

1000·2π ·ωm =

Fbs ·Vbs

ηbs(38)

using (32) and (33). This determines the motor weight andefficiency according to (29) and (30). Next, the input powerto the wire is determined from

Pw = Pm · 1ηm

· 1ηw

(39)

and then the wire diameter and wire weight are calculatedfrom (36) and (37). The system weight is the sum of the ballscrew, motor and wire weights. The overall electromechani-cal system efficiency is

ηesys = ηbs ·ηm ·ηw (40)

5 Method to Compare Hydraulic and Electromechani-cal SystemsWith the ability to calculate hydraulic and electrome-

chanical system weight and efficiency for a given application

Fig. 10. Hydraulic and electromechanical system weight at severaloutput velocities. Output power: 10 W, stroke: 0.05 m, transmissionline length: 0.1 m. The 100 psi, 100 mm/s data point is missingbecause there is no low pressure, high speed hydraulic system thatcan match the efficiency of the equivalent electromechanical system.

the two realizations can be compared to determine whichwill be lighter. The method used for the comparison wasto: (1) Establish the design problem by specifying a systemforce and power (or force and velocity), and linear excursion.(2) Design an electromechanical solution using the empiri-cal bounding equations as a stand-in for the best-availableDC brushless motor and ball screw. (3) Calculate the effi-ciency of the resulting electromechanical system. (4) Designa comparable hydraulic system with the same force, powerand stroke design requirements and the same efficiency. (5)Calculate and compare the weights of the electromechanicaland hydraulic solutions. A calculator application was imple-mented in Matlab to facilitate the comparisons.

6 ResultsExamples of comparing solution weights are shown in

the following Figs 10-17. Figs. 10-15 show system weightfor a mechanical output power of 100 W and 10 W for var-ious configurations of velocity, stroke length and transmis-sion line length. The nominal voltage for the motors in thedatabase ranged from 6 to 48 V and for this analysis, 24 Vwas selected. Motor voltage has some, but not a significanteffect on the electromechanical system weight because as thevoltage decreases, the system weight will increase due to thewire diameter increasing to accommodate the increase in cur-rent at the same efficiency.

A significant factor that influences the weight of a hy-draulic system is the operating pressure. Fig. 16 shows theweights of hydraulic systems running at several pressurescompared to the weight of the equivalent electromechanicalsystem for three output power conditions with an output ve-locity of 10 mm/s. The 100 psi hydraulic system will beheavier than the equivalent electromechanical system whilethe 500 psi and 1000 psi systems will be much lighter.

Fig. 17 shows the operating pressure required for thehydraulic system to have the same weight as the equivalent

Fig. 11. Hydraulic and electromechanical system weight at severaloutput velocities. Output power: 100 W, stroke: 0.05 m, transmissionline length: 0.1 m.

Fig. 12. Hydraulic and electromechanical system weight at severalstroke lengths. Output power: 10 W, velocity: 0.01 m/s, transmissionline length: 0.1 m.

Fig. 13. Hydraulic and electromechanical system weight at severalstroke lengths. Output power: 100 W, velocity: 0.01 m/s, transmis-sion line length: 0.1 m.

electromechanical system for three output powers. Pressureshigher than the line will result in a lighter hydraulic systemand pressures below the line will result in a heavier hydraulicsystem.

Fig. 14. Hydraulic and electromechanical system weight at severaltransmission line lengths. Output power: 10 W, stroke: 0.05 m, ve-locity: 0.01 m/s.

Fig. 15. Hydraulic and electromechanical system weight at severaltransmission line lengths. Output power: 100 W, stroke: 0.05 m,velocity: 0.01 m/s.

Fig. 16. Hydraulic and electromechanical system weight at severaloutput powers. Stroke: 0.05 m, velocity: 0.01 m/s, transmission linelength: 0.1 m.

7 DiscussionThe key result of this study, indicated in Figs. 16 and

Fig. 17, is that for low power applications (< 100 W), ahydraulic solution will only be lighter than the equivalentelectromechanical solution if the hydraulics are run at highpressure. Fig. 16 shows that a hydraulic solution can besignificantly lighter than an electromechanical solution if the

Fig. 17. Operating pressure required for the hydraulic system to bethe same weight as the equivalent electromechanical system at sev-eral output powers. Stroke: 0.05 m, velocity: 0.01 m/s, transmissionline length: 0.1 m.

Fig. 18. Cylinder efficiency as a function of bore size. The plotwas generated assuming 500 psi operating pressure and 0.1 m/s rodspeed.

pressure is high. For example, a 100 W electromechanicalsystem is predicted to weigh 428 g while a 100 W hydraulicsystem running at 1000 psi is predicted to weigh 63 g, aboutseven times lighter. While the exact numbers are system de-pendent (for example, as the power source is placed furtheraway, the drag in small hydraulic lines become significant),the conclusion is clear: for tiny, light hydraulic systems theoperating pressure should be high.

On the other hand, there is an upper limit for the op-erating pressure. High operating pressure indicates smallcylinder bore size, which in turn indicates low efficiency, asshown in Fig. 18.

Low cylinder efficiency causes many practical chal-lenges: 1) in the system level comparison, no hydraulic sys-tem that matches the efficiency of the electromechanical sys-tem can be found; 2) too much heat generated in the system,which cannot be dissipated by the tiny amount of fluid insmall scale hydraulic systems; 3) a large and heavy powersource is needed to provide the required energy.

At higher power, the weight advantage of hydraulics isevident even at low pressure. In Fig. 16 a 1,000 W elec-tromechanical system would weigh 6.3 kg and a 100 psi hy-draulic system would also weigh 6.3 kg. Moving to 1,000psi, the hydraulic system would weigh 0.63 Kg, ten times

lighter. The weight difference increases in significance be-yond 1,000 W, which is why hydraulics are the standard forhigh power applications such as excavators. What this studyshows is that hydraulics remain favorable at low power, butonly if the pressure is high. If the pressure is low, the elec-tromechanical system will be lighter.

Because tiny high pressure hydraulic components arenot available off the shelf, most small hydraulic systems arerun at low pressures, often using pneumatic components thatare small and light but generally limited to about 200 psi. Anexception is the work by Love et al. [5] who prototyped tiny,high pressure hydraulics for use in prosthetic fingers. Thereis a clear need for innovation in the development of smallcomponents that operate at high pressures and can meet theweight predicted by Eqns. (16) and (24). In particular, theinset for Fig. 4 shows that at small scales there is room forimproved cylinders that are lighter than what is commerciallyavailable.

The main limitation of this study is that it ignored thepower supply and control means (see Fig. 1), which are sig-nificant components of the complete system. Analyzing onlythe distal components still provides guidance to the designerfor two reasons. First, it is often the distally mounted com-ponents that are most weight sensitive and second, includ-ing the power supply and control would not change the mainconclusion which is that tiny hydraulics should be run at highpressure to minimize weight.

Turning to the complete hydraulic system, hydraulicpower supplies are typically large and heavy and tradi-tional throttling valves are extremely inefficient. For trulylightweight, low power, mobile systems such as poweredhand tools and powered orthotics, there is a need to developcompact sources of high pressure fluid using pumps drivenby battery powered motors or driven by tiny, high powerdensity internal combustion engines. Tiny cartridge pistonpumps are available but have a modest efficiency of about30% at 500 psi.

There is also a need for tiny, high pressure, low flowhydraulic control valves that operate in an efficient on-offswitching mode. PWM drivers for electric motors are a well-developed, highly efficient control means that have no equiv-alent in fluid power. Low-pressure, low flow MEMS valvesare common in micro-fluidics, but not suitable for transmit-ting power in the one to 100 W range.

Other problems with tiny hydraulics for human-scale ap-plications that must be solved include leakage of oil into theenvironment, which calls for developing low friction, leak-less seals; cavitation of the fluid, which may be a significantproblem for oil running through small passages at low pres-sure and high velocity; and creating designs that integratestructure, conduit, valving and cylinders to minimize weightby eliminating fittings.

AchnowledgementsProfessor Chris Paredis and his students at the Georgia

Instittute of Technology created the database of hydrauliccylinder properties. This research was supported by the Cen-

ter for Compact and Efficient Fluid Power, a National Sci-ence Foundation Engineering Research Center, funded undercooperative agreement number EEC-0540834.

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