an OverviewMass and Weight

1

by Arif Nurjaya

Definition

The kilogram is the unit of mass; it

is equal to the mass of the

international prototype of the

kilogram (IPK). Its form is a cylinder

with diameter and height of about

39 mm. It is made of an alloy of 90

% platinum and 10 % iridium,

density greater than 21.530 kg/cm3.

The IPK and two other cylinders

were manufactured in 1879 by

Johnson Matthey. Johnson Matthey

made 40 replicas in 1884, which

were calibrated to IPK. 34 were

distributed in 1889 to signatories of

the Meter Convention for use as

national standards. Now the BIPM

has produced more than 100prototypes.

http://en.wikipedia.org/ http://www.bipm.org/

K-46

6

K46

7

Third Periodic Verification 89-92

http://www.bipm.org/en/bipm/mass/verifications.html

OIML Standard

E1 | E2

K 46

K

DISEMINATION

Direct comparison

Sub Division

Direct comparison

Direct comparison

USE MATERIAL MARKING

Platina Iridium

Platina Iridium

Stainless Steel

Stainless Steel

Grey Cast Iron

International Prototype 1 kg

National Prototype 1 kg

Ensure TraceabilityCalibration Weighing

Instruments of Class I

Ensure TraceabilityCalibration Weighing

Instruments of Class I and Class II

Commercial Transaction Calibration Weighing

Instruments of Class III and Class III

F1 | F2

M1|M2|M3

M1-2|M2-3

Without Marking

F1: Nominal ValueF2: Nomilal Value

with Form “F”

Nominal Value with symbol “g” or “kg”

The sign “M1” or “M”,“M2”, “M3” or “X”

OIML R-111-1, 2004

ASTM Standard

Type IThese weights are of one-piece construction and contain no added adjusting material. They should be specified when weights are to be used as standards of the highest order and where maximum stability is required. A precisemeasurement of density can be made only for one-piece weights.Type IIWeights of this type can be of any appropriate design such as screw knob, ring, or sealed plug. Adjusting material can be used as long as it is of a material at least as stable as the base material and is contained in such a way that it will not become separated from the weight.http://www.astm.org/

M-One

http://www.mt.com/

Direct Comparison K46-E0

Sumber: Nugroho Budi Widodo, 2013

Direct Comparison K46-E0

Sumber: Nugroho Budi Widodo, 2013

Sub Division E0-E1

1000 g500 g200 g100 g

1000 g

=

1 kg

10 kg

2 kg 2 kg 5 kg

10 kg

=

Sumber: Usman, 2006

Sub Division E0-E1

1000 g

500 g200 g100 g

1000 g

=

m500 + m200 + m200* + m100 - m1000 = y1

m500 + m200 + m200* + m100* - m1000 = y2

- m500 + m200 + m200* + m100 = y3

- m500 + m200 + m200* + m100* = y4

- m200 - m100 + m200* + m100* = y5

- m200 - m100 + m200* + m100* = y6

- m200 - m100* + m200* + m100 = y7

- m200 - m100* + m200* + m100 = y8

- m200 + m100 + m100* = y9

- m200 + m100 + m100* = y10

- m200* + m100 + m100* = y11

- m200* + m100 + m100* = y12Sumber: Usman, 2006

Sub Division

m500 + m200 + m200* + m100 - m1000 = y1

m500 + m200 + m200* + m100* - m1000 = y2

- m500 + m200 + m200* + m100 = y3

- m500 - m200 + m200* + m100* = y4

- m200 + m200* - m100 + m100* = y5

- m200 + m200* - m100 + m100* = y6

- m200 + m200* + m100 - m100* = y7

- m200 + m200* + m100 - m100* = y8

- m200 + m100 + m100* = y9

- m200 + m100 + m100* = y10

- m200* + m100 + m100* = y11

- m200* + m100 + m100* = y12

1 1 1 1 0

1 1 1 0 1

-1 1 1 1 0

-1 -1 1 0 1

0 -1 1 -1 1

0 -1 1 -1 1

0 -1 1 1 -1

0 -1 1 1 -1

0 -1 0 1 1

0 -1 0 1 1

0 0 -1 1 1

0 0 -1 1 1

m500

m200

m200*

m100

m100*

1

1

0

0

0

0

0

0

0

0

0

0

m1000

- =

y1

y2

y3

y4

y5

y6

y7

y8

y9

y10

y11

y12

A EDB C

A.B - C.D = EA.B = C.D + E

AT.A.B = C . AT.B + AT.E(AT.A)-1 (AT.A) . B = C (AT.A) -1 . AT.B + (AT.A)-1 . AT.E

B = C (AT.A) -1 . AT.B + (AT.A)-1 . AT.E

Operasi Matrik

Sumber: Usman, 2006

Sub Division

Massa Konvensionalm500 = 0.5 m1000 + 0.25y1 + 0.25y2 + 0.25y3 + 0.25y4

m200 = 0.2 m1000 + 0.1y1 + 0.1y2 - 0.1y3 - 0.1y4 + 0.1y5 + 0.1y6+ 0.1y7+ 0.1y8+ 0.1y9 + 0.1y10

m200* = 0.2 m1000 + 0.1y1 + 0.1y2 - 0.1y3 - 0.1y4 - 0.1y5 - 0.1y6 - 0.1y7 - 0.1y8+ 0.1y11 + 0.1y12

m100 = 0.1 m1000 + 0.1y1 - 0.1y3 + 0.1y5 + 0.1y6 - 0.1y7 - 0.1y8 - 0.1y9 - 0.1y10- 0.1y11 - 0.1y12

m100* = 0.1 m1000 + 0.1y2 - 0.1y4 - 0.1y5 - 0.1y6 + 0.1y7 + 0.1y8 - 0.1y9 - 0.1y10- 0.1y11 - 0.1y12

2/1

2222

2

2

1000

2

500 43214

1

2

1

yyyy UUUUUU

2/1

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2

2

1000

2

200 7654321222

10

1

5

1

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2/1

2222222

2

2

1000

2

*200 8654321222

10

1

5

1

yyyyyyy UUUUUUUUU

2/1

222222

2

2

1000

2

100 8765312222

10

1

10

1

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2/1

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2

2

1000

2

*100 8765422222

10

1

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1

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Ketidakpastian

22

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Sumber: Usman, 2006

True Mass

http://ffden-2.phys.uaf.edu/104_2012_web_projects/Daniela_Wilner/Buoyancy.html

True Mass

1000 gO

N

Fg = m.g

Fb = V.ρ.g

I = Fg - Fb= m.g - V.ρ.g

True Mass, therefore, is not what a weight would weigh in a vacuum, but rather, what it would weigh compared against a reference standard mass (with a known value) on a "perfect" equal arm balance inside a "perfect" vacuum chamber.

http://www.icllabs.com/

Conventional Mass

The conventional value of the result of weighing of a body in air is equal to the mass of a standard of density 8000 kg/m3 at 20 0C which balances this weight at this temperature in air of density 1.2 kg/m3.

1000 gO

N

Fg = m.g

I = Fg

rt

aiiC

1

0

03

1

m

UCi

Air buoyancy correction

Negligible if

OIML R-111-1, 2004

Matur Nuwun

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