?

What Do I What Do I use?use?

NumasizingNumasizing

Compressed air is an energy medium thatCompressed air is an energy medium that must be conserved, Numasizing has been must be conserved, Numasizing has been developed to use this resource more developed to use this resource more effectively.effectively.

What Is Numasizing?What Is Numasizing?

What Is Numasizing?What Is Numasizing?

Numasizing is a technique which takes into Numasizing is a technique which takes into account all of the physical specifications account all of the physical specifications of a of a circuit and results circuit and results in a in a tailor made tailor made circuit designed circuit designed

to the customer’s to the customer’s targets.targets.

What Is Numasizing?What Is Numasizing?

Numasizing is not based on theoretical Numasizing is not based on theoretical approach or mathematical model. A data base approach or mathematical model. A data base of over 250,000 test firings of cylinders allows of over 250,000 test firings of cylinders allows Numatics to predict components and Numatics to predict components and pressures with confidence.pressures with confidence.

The Numasizing Process is...The Numasizing Process is...

A + B + C = DA + B + C = D

A.A. Establishing Customer ObjectivesEstablishing Customer Objectives

B.B. Delineating All Circuit SpecificationsDelineating All Circuit Specifications

C.C. Selecting All Circuit ComponentsSelecting All Circuit Components

D.D. Results in Obtaining Customer GoalsResults in Obtaining Customer Goals

The Numasizing Process is...The Numasizing Process is...

A.A. Establishing Customer ObjectivesEstablishing Customer Objectives

1.1. Increase ProductivityIncrease Productivity

2.2. Optimum Energy UtilizationOptimum Energy Utilization

3.3. Minimum Component SizeMinimum Component Size

The Numasizing Process is...The Numasizing Process is...

B.B. Delineating All Circuit SpecificationsDelineating All Circuit Specifications

1.1. Extend Load & Retract LoadExtend Load & Retract Load

2.2. Available PressureAvailable Pressure

3.3. Desired Extend & Retract TimesDesired Extend & Retract Times

4.4. Conductor LengthConductor Length

5.5. Required Cylinder Stroke, etc.Required Cylinder Stroke, etc.

Generating

Numasizing Computer Bank

INTO….The Numasizing Process is...The Numasizing Process is...

C.C. Selecting All Circuit ComponentsSelecting All Circuit Components

All components in the pneumatic system All components in the pneumatic system starting at the valve through the exhaust starting at the valve through the exhaust port such as the valve, fittings, conductor, port such as the valve, fittings, conductor, regulator and actuator.regulator and actuator.

The Numasizing Process...The Numasizing Process...

D.D. Results in Obtaining Customer GoalsResults in Obtaining Customer Goals

1.1. Actuator Response TimeActuator Response Time

2.2. Compressed Air CostCompressed Air Cost

3.3. Air ConsumptionAir Consumption

4.4. Pressure OptimizationPressure Optimization

5.5. Compatible SizesCompatible Sizes

Effecting PneumaticsEffecting Pneumatics

Knowledge of a pneumatic circuit design begins with an understanding of the term

Cv is essentially a dimensionless number used to express the CONDUCTANCE VALUE of a pneumatic device.All fixed orifice devices in a pneumatic system have a conductance value and therefore a certain capability to flow.

The larger the CThe larger the Cvv --- the greater the flow. --- the greater the flow.

Typically, the greater the CTypically, the greater the Cvv of the entire circuit, of the entire circuit,

the faster the devices in the circuit will respond.the faster the devices in the circuit will respond.

CCvv

CVV = Valve CV

CVFC = Flow Control CV

CVO = Orifice CV

CVP = Pipe CV

CVC = Cylinder Port CV

CVF = Fitting CV

CVV = 4.0 CVFC = 4.5

2.99

CVO = 1.12

1.05

CVP = 4.8 CVC = 6.0 CVF = 7.0

1.03

1.01

2.99

2.47

2.19

2.06

2.000

CVV = 4.0 CVFC = 4.5 CVO = 4.37 CVP = 4.8 CVC = 6.0 CVF = 7.0

1.000

CVV CVFC CVS2 2 2

1 1 1 + =

4 4.5 2.992 2 2

1 1 1 + =

Relation Between CRelation Between Cvsvs & Actuator & Actuator Stroke Time (T)Stroke Time (T)

Stroke time is inversely proportional to Cvs

Stroke time is directly proportional to system exhaust volume

1,092 ft/sec (vel of air - unconfined)

400 ft/sec (vel of air - in a straight conductor)

80 in/sec (vel of actuator piston & rod)

Cvs T (sec) 1 4.0 2 2.0 4 1.0 8 0.5

CCvv of Fittings are Established by of Fittings are Established by Calculating an Equivalent Conductor Calculating an Equivalent Conductor

Length Utilizing a Length Utilizing a K Factor Factor

Once the ID and the line length has been determined --- plus all the fittings equivalent length has been added, we can calculate the Cv of the CONDUCTOR.

Formula: Le’ = n K d”

Equivalent Length = number of fittings x K (feet) factor x ID (inches)

Formula: Le + Lc = Lt

Equivalent Length + Conductor Length = Total Length

CCvv of Fittings are Established by of Fittings are Established by Calculating an Equivalent Conductor Calculating an Equivalent Conductor

Length Utilizing a Length Utilizing a K Factor FactorFitting K Values:Device K Factor

Swing Check Valves, fully open 11

90 Degree Standard Elbow 2.545 Degree Standard Elbow 1.390 Degree Long Radius Elbow 1.5Reducer (1 Size) 1.5Enlarger (1 Size) 2“Y” Fitting 1

Standard TeeFlow Through Run 1.5Flow Through Branch 5

ELBOWSBRANCH TEE K = 5

BENDS

10 x I.D. minimumK = 0

K = 1.5RUN TEE

x

K = 0

K = 1K = 1 Y Fitting

R=3 x I.D. Long Radius K =1.5R=1.5 x I.D. Std. Elbow K = 2.5R=0 x I.D. SHARP 90° K = 5.0

The more significant the change in flow direction, the greater the restriction and therefore the greater the K factor.

Formula: Cv (port)= 23 x Ds

Ds = smallest I.D. in a port

2

To calculate Cv we need to determine the smallest I.D. (DS) of whatever component is in the port of the cylinder.

Reducing the port size with a bushing reduces the flow capability of the port.

The smallest orifice will determine the effective Cv .

What is the cylinder’s CWhat is the cylinder’s Cv v ??

C3.0

3/8 NPT

1/4 NPT

ID .364 - 1/4 NPT Pipe, SCH .40vC3.0C5.6v

ID .493 - 3/8 NPT PipeSCH. 40

.406” ID - ControllinControlling g DimensioDimensionn

Ftg. 1/2 Tube to 3/8 NPT ID = .406

1/2 Steel Tube - .43” ID

3/8 NPT

How shall we decide which ID conductor to use?

The CThe Cv v of a cylinder port is based on the of a cylinder port is based on the

smallest I.D. entering that portsmallest I.D. entering that port

.406” ID

3/8 NPT

1/2 Nylon Tube = .38” ID -- NOW becomes the Controlling Controlling DimensionDimension for the CYLINDER PORT.

A fitting that accepts steel tubing will have a larger effective orifice than a fitting for plastic or poly tubing because the wall thickness for plastic / poly tubing is so much greater.

The CThe Cv v of a cylinder port is based on the of a cylinder port is based on the

smallest I.D. entering that portsmallest I.D. entering that portFtg. 1/2 Tube to 3/8 NPT ID = .406

Flow

Flow

Flow

Flow

Flow

Flow

Flow

Flow

CCdd = 0.98 = 0.98 CCvv = 28.8d = 28.8d 22

CCdd = 0.92 = 0.92 CCvv = 27.0d = 27.0d 22

CCdd = 0.82 = 0.82 CCvv = 24.1d = 24.1d 22

CCdd = 0.80 = 0.80 CCvv = 23.5d = 23.5d 22

CCdd = 0.72 = 0.72 CCvv = 21.2d = 21.2d 22

CCdd = 0.65 = 0.65 CCvv = 19.1d = 19.1d 22

CCdd = 0.61 = 0.61 CCvv = 18.0d = 18.0d 22

CCdd = 0.53 = 0.53 CCvv = 15.6d = 15.6d 22

Various orifice discharge coefficients Cd and their related Cv

All openings = d” (diameter)All openings = d” (diameter)

Optimized Drilling for Intersecting Optimized Drilling for Intersecting HolesHoles

There is an optimum conductor ID for each application.

Choose the best commercially available size.

Too small a conductor ID and there is RESTRICTION; too large and it becomes a VOLUME CHAMBER.

The LENGTH of the conductor now must be considered.

ConductorI.D.

Time

0

0

3/16” I.D. Tubing

1/16” I.D. Tubing

1/8” I.D. Tubing

The Bends - a practical example:The Bends - a practical example:Given: 3/8 NPT Sch.40 pipe has an I.D. of 0.493” or essentially 1/2”

Conductor Length Cv

10’ (3/8 NPT) 3.0

20’ (3/8 NPT) 2.1

30’ (3/8 NPT) 1.75

The direct distance between the valve and the cylinder is 10’.

We choose to pipe the circuit with an additional 10’ (because it looks better).

If we used (8) 90 degree standard elbows (K=2.5) between the valve and

cylinder, it would have an equivalent length of 10’:

Le’ = nKd”

Le’ = 8 x 2.5 x 1/2” = 10’

We have penalized the cylinders’ speed by reducing the efficiency of the circuit.

By identifying the weak link, we can improve our circuit.

The most common way of adjusting a cylinder’s speed is with a FLOW CONTROL.

FREE FLOW

C BOTTOM PORTL

TO CYLINDER

SizeModel

Number *Free Flow

Controlled Flow

1/4 NPTF3/8 NPTF1/2 NPTF3/4 NPTF

2FC23FC24FC35FC3

2.32.76.07.5

2.02.45.56.0

MODEL SELECTION AND FLOW CAPACITY (CMODEL SELECTION AND FLOW CAPACITY (CVV) CHART) CHART

** For optional bottom port, add B to model number (e.g..., 4FC3B

A flow control cannot conserve compressed air, reduce force, or speed up a circuit. If utilize bottom port If utilize bottom port

option:option:2.442.866.367.95

Quick Exhaust ValveQuick Exhaust ValveThe most common device selected for improving cylinder speed is a

Quick Exhaust Valve.

PRESSUREAPPLIED

ACTUATOR

EXHAUST PRESSUREBLOCKED

ACTUATOR

EXHAUST

This device has a limited life -- the disc is slammed (full line pressure)

twice every cycle. No breaking or control of the cylinder will occur

either.

Quick exhaust allows the cylinder to exhaust at the cylinder port, not

back through the valve or through a flow control.

E P

LeLxFLf

Pe = 75 Pr = 75

Loads on a CylinderLoads on a Cylinder On any cylinder, there are threeLOADS resisting movement. Thecylinder must overcome each load before it can extend or retract.

Let’s examine each load separately.

Cylinders, typically, must have some friction in sealing the piston against the cylinder wall and also the rod --- Load Lf.

Work that the cylinder performs is expressed as Le for extend and Lr for retract directions of movement.

The crucial load, however, is Lx --- the EXHAUST BACK PRESSURELOAD.

FLf

Lx Lr

Pe = 75 Pr = 75

Return Stroke

LOAD10 # in 0.45 SEC

75 PSIG

Extend Stroke

SupplyExhaust

LOAD60 # in .4 SEC

75 PSIG

SAMPLE PROBLEM:Move a load in the specific times, given 75 psig supply.

Cylinder Response Cylinder Response Time in SecondsTime in Seconds

ExhaustSupply

75psig

75psig

Pdm = 36psi Pressure differential at inception of motionPd = Running pressure differential varies during cycle (see curve)Td = 0.14 Time delay primarily due to exhaust preload, a minimum

delay due to static friction of cylinder seals and a negligible solenoid time delay (Ty).

Tm = 0.26 Time cylinder piston is in motionTe or Tr = 0.40 Time to extend or Time to retract (Td+Tm)Ty = 0.008 Solenoid time delayP = 10psi Pressure drop from Supply Pressure

Pd

TdTy

P

Pdm

mT

En

d o

f S

trok

e

Pre

ssu

re P

SIG

Time - Seconds00

TT re or

LeLxFLf

Pe = 49 Pr = 21

FLf

Lx Lr

Pe = 49 Pr = 21

Pressures Fit The Load Pressures Fit The Load

Select the cylinder pressures based on required force.

Consider what the cylinder must do in each direction and select the required pressure for each individual action.

Extend Stroke LOAD60 # in 0.40 SEC

21 PSIG

49 PSIG

SupplyExhaust

LOAD10 # in 0.45 SEC

Return Stroke

21 PSIG

49 PSIG

Now that the ideal pressures have been selected, what results can beobserved?

Cylinder Response Cylinder Response Time in SecondsTime in Seconds

ExhaustSupply

21psig

49psig

Pdm = 36psi Pressure differential at inception of motionPd = Running pressure differential varies during cycle (see curve)Td = 0.14 Time delay primarily due to exhaust preload, a minimum

delay due to static friction of cylinder seals and a negligible solenoid time delay (Ty).

Tm = 0.26 Time cylinder piston is in motionTe or Tr = 0.40 Time to extend or Time to retract (Td+Tm)Ty = 0.008 Solenoid time delayP = 2 psi Pressure drop from Supply Pressure

Pdm

Pd

En

d o

f S

trok

e

TmTd

Pre

ssu

re P

SIG

Time - Seconds00

P

Ty

TT re or

Dual Pressure Dual Pressure Response Time Response Time in Secondsin Seconds

Pdm = 36psi Pressure differential at inception of motionPd = Running pressure differential varies during cycle (see curve)Td = 0.09 Time delay primarily due to exhaust preload, a minimum

delay due to static friction of cylinder seals and a negligible solenoid time delay (Ty).

Tm = 0.19 Time cylinder piston is in motionTe or Tr = 0.28 Time to extend or Time to retract (Td+Tm)Ty = 0.008 Solenoid time delayP = 0.4 psi Pressure drop from initial Supply Pressure

ExhaustSupply

20psig

75psig

ExhaustSupply

75psig

75psig

Initial:

Pd

En

d o

f S

troke

Td

Ty

Pre

ssu

re P

SIG

Time - Seconds00

00

En

d o

f S

troke

Pdm

Time saved frominitial pressureconditions (0.11)

Tm

Te or Tr

P

Surv

ey #

Surv

ey #

Pe/P

r

Pe/P

r

Bore

Bore

Te/T

r

Te/T

r

CPM

CPM

FC’s

FC’s

$ Co

st/M

$ Co

st/M

$ Ca

pita

l

$ Ca

pita

l

E

xpen

ses

Expe

nses

$ Ope

rating

$ Ope

rating

Cost

Cost

HP

Requi

red

HP

Requi

red

2 75/75 1.50 .40/.45 70 yes 476 257 4.75 1.34

2B 75/75 2.00 .23/.14 163 no 1733 287 2.08 4.89

3 49/21 1.50 .40/.45 70 no 229 230 4.75 0.65

4 61/35 1.50 .40/.45 70 no 289 217 4.75 0.81

5 70/31 1.25 .40/.45 70 no 219 190 4.75 0.62Compressor pressure for all surveys are kept at 100 PSIG

Valve Cv‘s and costs for survey #’s 2, 2B, 3, 4 and 5 are respectively, 2.05 - $85, 2.05 - $85, 0.41 - $58, .25 - $45, 0.25 - $45

3/8” NPT Pipe conductor cost $25 and 1/8” NPT costs $20

Cylinder costs for 2.0”, 1.5” & 1.25” bores are respectively $152, $102 and $75

A pair of flow controls cost $20 and a pair of regulators cost $50

Survey #2 ($85+$25+$102+$20+$25 = $257) O.E.M. would select #5 because of minimum capital investment

Survey #2B ($85+$25+$152+$102+$50 = $287) End user would select #2B because of low cost/pc

Survey #3 ($58+$20+$102+$50 = $230) Facilities would select #3 because of lowest pressure

Survey #4 ($45+$20+$102+$50 = $217) demands or #5 because of minimum HP requirement

Survey #5 ($45+$20+$75+$50 = $190) depending on the major objective of facilities engineer

2-1/2 Seconds 40 Minutes 300 Hours

Dynamic Seals With LubricationFlow of O-Ring into Metallic Surfaces

“The theory has been proposed and generally accepted that the increase of friction on standing is caused by the rubber O-ring flowing into the microfine grooves or surface irregularities of the mating part. As a general rule for a 70º durometer rubber against an 8 micro-inch surface, the maximum break-out friction which will develop in a system is three times the running friction.”

Friction is always a factor with dynamic seals. Obviously, the more dynamic seals --- the greater the friction.

We can assume the cylinder has been sized properly to overcome the Le and Lr loads.

Lx --- the exhaust back pressure load --- is determined by the ability of the cylinder’s ports to allow exhaust to escape; the Cv of the cylinder.

Effective Orifice --- Effective Orifice --- Valves: Valves: As with cylinders, VALVEVALVE CCvv is determined not by the port

of the valve, but by the smaller orifice of the fitting in the port. Port Size (NPT)

10/321/8 1/4 3/8 1/2 3/4 1 1 1/4 Valve Size

MK 3 .18 .35*

MK 7 .2 .4*

MK 8 .3 .8 1.0*

MK 15 .4 1.1 1.4 1.5*

MK 55 .6 2.0 2.854.0 5.0 5.55*

*there is no improvement in the valve’s Cv beyond this point

Consider: System Cv equals the combination ofCylinder Cv ~ Conductor and Fittings Cv ~ Valve Cv

and any additional devices in the circuit

Critical Pressure Ratio: Critical Pressure Ratio:

P

80“““““““““

=1 P+P2

P P2P Q0

102030404550607080

8070605040353020100

028384345.546““““

1At start, there is 80 psig available and the needle valve is closed. As the needle valve is slowly opened, in P increments of 5 and 10 psi, the flow is noted.

OBSERVEEven with a larger P --- FLOW DOES NOT CONTINUE TO INCREASE.

Air has reached CRITICAL FLOW and FLOW cannot increase.Terminal velocity occurs at the CRITICAL PRESSURE RATIO

= .53 P2

P1 P1 = .47P

Q (SCFM)P1 P2

P

Needle Valve

Flow meter

Terminal Velocity: Terminal Velocity:

A jet engine takes air in, compresses the air with fuel, and then ignites the mixture.

At some point, air cannot enter the engine any faster. No matter how much more fuel is added --- the plane cannot go any faster.

The limiting factor is AIR FLOW, supply to the engine.

In a pneumatic circuit, the cylinder reaches TERMINAL VELOCITYwhen the air cannot enter and/or exit the circuit any faster. This occurs at

P1= .47P

Terminal Velocity for a Terminal Velocity for a Cylinder: Cylinder:

IF a properly sized cylinder had • air supplying and • exhausting at the • CRITICAL PRESSURE DROP RATIO THE PNEUMATIC CYLINDER WOULD BE AT

TERMINAL VELOCITY for that TERMINAL VELOCITY for that particular circuit. particular circuit. To improve cycle time we must improve the smallest Cv of the system and maintain the critical pressure ratio.

Therefore:HOW FAST CAN AN AIR CYLINDER CYCLE?

8012

3 4 567

x1000r/min

H

CF

E

If Energy Waste Was This If Energy Waste Was This Obvious,Obvious,

You’d Put A Stop To ItYou’d Put A Stop To It

Benefits of Dual Pressure: Benefits of Dual Pressure:

Percent Cost Savings Attainable

Utilizing Numasizing in

Industrial Pneumatic Systems

Retract Pressure (PSIG)

Exte

nd

Pre

ssu

re (

PS

IG)

90

80

70

60

50

40

30

0% 5.0 9.2 14.8 19.1 24.0 28.9

10.0 14.1 19.8 24.0 28.9 33.9

18.4 24.0 28.2 33.9 38.8

29.6 33.9 38.8 43.7

38.1 43.0 47.9

47.8 52.9

57.8

90 80 70 60 50 40 30

Why NUMASIZE?

Air Leaks: Air Leaks:

Annual Cost of Compressed Air Leakage

Through Orifices of Various Diameters

30 $ 78 $172 $310 $ 689$1239

50 $112 $252 $448 $1009$1784

70 $146 $330 $566 $1316$2337

90 $181 $403 $722 $1611$2891

110$216 $486 $861 $1931$3444

PSIG1/323/64 1/16 3/32 1/8

24 hrs/day - 365 days/yr.

30¢/1000 SCFM (6.75¢/kwh)

Compressed air is ENERGY that must be CONSERVED.

Resource Cost: Resource Cost:

Average Compressor H.P. In Use based upon

4 s.c.f.m./H.P. - 30¢/1000 s.c.f. -

6.75¢/kwh

Compressed Air Cost

4000375035003250300027502500225020001750150012501000750500250

100020003000400050006000

1 shift/day - 250 days/year

2 shifts/day - 2

50 days/year

3 shift

s/day

- 250 d

ays/

year

24 h

ours

/day

- 3

65 d

ays/

year

Cost

to G

en

era

te C

om

pre

ssed

Air

($1,0

00’s

)

Horsepower

Total Energy Production: Total Energy Production:

1990 FUEL SOURCES -- Production of ENERGY

38,470,000 BPDOE (Barrels Per Day Oil Equivalent)

(81.44 Quadrillion BTU's or Quads)

Electrical Segment of Industry

4,860,000 BPDOE (10.26 Quads) 34%

37% 20%16%

COAL 23.3%

8,960,000 BPDOE

NAT. GAS 24.0%

9,230,000 BPDOE

HYDRO 3.5%

1,350,000 BPDOE

NUCLEAR 7.0%

2,690,000 PBDOE

INDUSTRIAL14,200,000 BPDOE

RESIDENTIAL7,700,000 BPDOE

COMMERCIAL6,170,000 BPDOE

OTHER0.3%

120,000 PBDOE

OIL 41.9%

16,120,000 BPD

Transportation

10,400,000 PBDOE

27%

CONSUMPTION

Electricity Production: Electricity Production:

10% Savings 107,000 BPDOE (.22 Quads)20% Savings 214,000 BPDOE (.45 Quads)

compared to

Prudhoe Bay Oil Fields 1,800,000 BPD

1990 FUEL SOURCES -- Production of ELECTRICITY

14,230,000 BPDOE (Barrels Per Day Oil Equivalent)

(30.1 Quadrillion BTU's or Quads - 37% of US. Energy System of

81.26 Quads)

DISTRIBUTION

COAL 56.8%

8,081,000 BPDOE

NAT. GAS 9.8%

1,392,000 BPDOE

HYDRO 9.5%

1,350,000 BPDOE

NUCLEAR 18.9%

2,690,000 PBDOE

INDUSTRIAL4,860,000 BPDOE

(10.26 Quads)34%

RESIDENTIAL4,690,000 BPDOE

(9.96 Quads)33%

COMMERCIAL4,290,000 BPDOE

(9.0 Quads)30%

OTHER0.8%

120,000 PBDOE

OIL 4.2%

597,000 BPD

ELECTRICITY14,230,000 BPDOE

(30.1 Quads)

OTHER390,000 PBDOE

(0.88 Quads)

3%

If Numasized, a minimum savings of 15% would be realized.

Compressed Air Segment

1,070,000 BPDOE (2.26 Quads) 22%

Conceptual View Point: Conceptual View Point:

The preceding two slides/charts are based on Energy Information Administration (EIA) surveys. There were some slight differences between the 1990 and the updated figures for 1990 as published in the October 1991 report. Some variations also showed up when comparing the figures of National Business, Electric World, Pipe and Gas Journal, etc. with those of EIA. Since the deviations were minor (mostly due to independent rounding), we have reconciled all of them as to render uniform results. From a conceptual viewpoint and from the ultimate potential savings and increased productivity aspect, they have no bearing as the differences are miniscule.

Also, we have incorporated in these figures the fact that 1,033,100 BTU’s (EIA from utilities) is required to generate 100 Kwhrs of electricity (due to losses in conversion and transmission), while 300 Kwhrs are needed to produce 1,024,000 BTU’s of heat (1 Kwhrs = 3413 BTU’s).

Formulae: Formulae:

•Flow of electrons is very similar to flow of air molecules•Think of CONDUCTANCE as the reciprocal of RESISTANCE•Whose formula came first?

Those familiar with electricity will recognize Ohm’s Law.

In pneumatics, FLOW is represented by Q, T is temperature in Rankin, G is the specific gravity (assume 1 for air), P1

and P2 are pressure expressed in PSIA.

Same General Formula

E = I R

E

R

I = E Cd

ELECTRICITYOHM’S LAW

I =

Q = P CV

PNEUMATICS

BERNOULLI’S LAW

Q = 22.48 CV P x P2

T1 G

Q = 22.48 CV .47 P1 x .53P1

528 x 1 (AT CRITICAL FLOW)

Q = 0.488 P1CV

Q = (K) P CV

Your Most Important Design Tool