8/14/2019 Watt Balance http://slidepdf.com/reader/full/watt-balance 1/22 The Equation of the Watt Balance by Parker Emmerson with assistance from Andrew Berisha and Alden Daniels (1) C = 2 pr This is the circumference of our initial circle of radius r C 2 = 2 pr 1 This is the circumference of our second circle, the base of the cone, of radius r 1 r^2 = r 1 ^2 +h ^2 This is the initial radius squared expressed as the slant of the cone in terms of the height of the cone, h, and the radius of the base of the cone, r 1 r = , Hr 1 ^2 +h ^2L s = qr s ê q = r The arc length taken out of a circle at a given time is = t = C - C 2 = 2 pr - 2 pr 1 = qr Ø Equation 7 r 1 ^2 ã r^2 -h ^2 r 1 = , Hr^2 -h ^2L h § r t = time 1 second = 6 degrees t = 6 q I will now do some algebra to conclude what the height of the cone is in terms of the initial parameters. It can eventually be reduced to a single variable. II. Proof Printed by Mathematica for Students
Transcript
8/14/2019 Watt Balance
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The Equation of the Watt Balance
by Parker Emmerson with assistance from Andrew Berisha and Alden
Daniels
(1)
C = 2 pr
This is the circumference of our initial circle of radius r
C2 = 2 pr1
This is the circumference of our second circle,
the base of the cone, of radius r1
r ^ 2 = r1 ^ 2 + h ^ 2
This is the initial radius squared expressed as the slant
of the cone in terms of the height of the cone, h, and the radius
of the base of the cone, r1
r =,Hr1 ^ 2 + h ^ 2L
s = qrs ê q = r
The arc length taken out of a circle at a given time is =
t = C - C2 = 2 pr - 2 pr1 = qr Ø Equation 7
r1^ 2 ã r ^ 2 - h ^ 2
r1 =,Hr ^ 2 - h ^ 2L
h § r
t = time
1 second = 6 degrees
t = 6 q
I will now do some algebra to conclude what the height of the cone is in terms of the initial parameters. It can eventually
be reduced to a single variable.
II. Proof
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Solve@r1 ^ 2 + h ^ 2 ã r^2, hD(2)::h Ø - r2
- r12 >, :h Ø r2
- r12 >>
We say that the amount of q r = s, taken out of the circle is the change in the circle's circumference that is the base of the cone.
The change is equal to s = 2pr-2pr 1.
Notice that q = HH2 p rL ê rL - HH2 p r 1L ê r L, because we divide by r on both sides.
We will focus on the positive solutions for the height of the cone.
SolveBh == r2 - r12 , r1F
(3)::r1 Ø - r2 - h2 >, :r1 Ø r2 - h2 >>This is the change in circumference with the substituted expression for r1 in terms of h and r.
(4)r q == s ã 2 p HrL - 2 p HHrL^ 2 - h ^ 2L = 2 p HrL - 2 p r1
q == H2 p rL ê r - H2 p r 1L ê r L = HH2 p rL ê rL - 2 p - h2
+ r 2 ì r
SolveBq * r == 2 p HrL - 2 p HHrL^ 2 - h ^ 2L , hF
::h Ø -4 p r2 q - r2 q2
2 p>, :h Ø
4 p r2 q - r2 q2
2 p>>
SolveBh ==4 p r2 q - r2 q2
2 p , rF
::r Ø -
2 p h
4 p q - q2 >, :r Ø
2 p h
4 p q - q2 >>(5)In polar coordinates, it can be said that r1 = r Cos@qD, and that h = r Sin@qD
(6)
2 p h
4 p q - q2
= r =2 p * r * Sin@qD
4 p q - q2
2 Watt Balance Accurate True.nb
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PolarPlotB 2 p Sin@qD4 p q - q2
, 8q, - 13, 13<F
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
1.5
c := speed of light := H2.99792458 * 10^8L m ê s
c := H2.99792458 * 10^8LDepending on the conversion I will decide to use for exchanging angle involved in folding up the circle into a cone and the time
that has passed, I will find different solutions and options regarding the system.
First, we will use the following conversion
(7)h := rate * time = c * t = height of cone
t := 6 IqdegreesM = 6 HH180 ê pL qL, because if theta were in radians to begin with,
we would have to convert that number of radians to degrees. This way,
both our thetas are in radians,
but our result accounts for the constant of the degrees involved in measuring time.
H180 ê p Lr =
2 * c * t * p
4 p q - qL2 =
2 * c * 6 HH180 ê pL qL * p
4 p q - HqL2
SolveBr ==2 * c * 6 HH180 ê p L qL * p
4 p q - HqL2
, qF
::q Ø
4 p r2
4 665 600 c2 + r2
>>
Watt Balance Accurate True.nb 3
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PlotB 4 p r2
4 665 600 c2 + r2, 8r, - 13, 13<F
-10 -5 5 10
1.µ10-21
2.µ10-21
3.µ10-21
4.µ10-21
5.µ10-21
SolveBq ==4 p r2
4 665 600 c2 + r2, rF
::r Ø -
2160 c q
4 p - q
>, :r Ø
2160 c q
4 p - q
>>
PlotB 2160 c q
4 p - q
, 8q, - 13, 13<F
-10 -5 5 10
5.0µ1011
1.0µ1012
1.5µ1012
2.0µ1012
2.5µ1012
3.0µ1012
r :=2 p h
4 p q - q2
; h = r Sin@qD
2160 c q
4 p - q
=2 p h
4 p q - q2
=2 p r Sin@qD
4 p q - q2
SolveB 2160 c q
4 p - q
==2 p r Sin@qD
4 p q - q2
, rF
::r Ø
1080 c q H4 p - qL q Csc@qDp 4 p - q
>>
4 Watt Balance Accurate True.nb
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PolarPlotB 1080 c q H4 p - qL q Csc@qDp 4 p - q
, 8q, - 4 p , 4 p <F
-5µ1012
5µ1012
-1µ1012
-5µ1011
5µ1011
1µ1012
(8)r =2 * c * t * p
4 p HqL - HqL2=
1080 c q H4 p - qL q Csc@qDp 4 p - q
= wavelength of light = l
(9)velocity = ln, where n = number of mols of frequency = n * f
(10)f = 1 ê H6 * H180 ê pL qL = 1 ê t
Velocity = speed of light
SolveB 1080 c q H4 p - qL q Csc@qDp 4 p - q
n H1 ê H6 HqLLL == c, nF
(11)::n Øp 4 p - q q Sin@qD
180 H4 p - qL q
>>
(12)E = h * n = h * n * f = m * c ^ 2
(13)E = h * n = h *p 4 p - q q Sin@qD
180 H4 p - qL q
* H1 ê H6 * H180 ê pL qLL = m * c ^ 2
SolveBh *p 4 p - q q Sin@qD
180 H4 p - qL q
* H1 ê H6 * H180 ê p L qLL == m * c^2, m F
:: m Ø
h p2 4 p - q Sin@qD194400 c2
q H4 p - qL q
>>
This equation describes the details of a Watt Balance in theory. We have in the same equation, Planck's constant, time,
and mass. This will give us a standard expression for Planck' s constant and the variables involved in its use.
