The Museum of HP Calculators
The Museum of HP Calculators
This program is Copyright © 2005 by Jean-Marc Baillard and is used here by permission.
This program is supplied without representation or warranty of any kind. Jean-Marc Baillard and The Museum of HP Calculators therefore assume no responsibility and shall have no liability, consequential or otherwise, of any kind arising from the use of this program material or any part thereof.
Overview
-The following programs use Einstein's special relativity to study uniformly
( and non-uniformly ) accelerated linear motion of a spacecraft.
-In special relativity, the speed is also v = dx/dt , but acceleration
is a = ( dv/dt ) ( 1 - v2/c2 )-3/2
where t = time passed on Earth
-The proper Time T ( passed on board ship ) verifies dT = dt
( 1 - v2/c2 )1/2
-In all these programs, we use the following units:
-Times ( t , T ) are expressed in Julian years (
365.25 days )
-Abcissas and distances ( x , D ) are expressed
in light-years ( one light-year = 9.460,730,472,580,8 1015
m. )
-The unit of accelerations ( a ) is the "standard"
gravity on Earth g = 9.80665 m/s2
-The velocity of light = c = 299,792,458 m/s ( from the definition of 1 meter )
Remark: Integrals are denoted §
1°) A Trip to the Stars
-A spaceship travels to a star ( situated at a distance D from the Earth
) at a constant acceleration "a" , in order to simulate an artificial gravity.
-Halfway to the star, the constant acceleration "a" becomes a constant
deceleration "-a"
-The return journey goes off similarly.
Return Journey
<<<<<Deceleration "a" <<<<< |
<<<<<Acceleration "-a" <<<<<<<
Earth=O-------------------------------D/2------------------------------------D=(Star)--------------->
x
>>>>>Acceleration "a" >>>>> | >>>>>Deceleration "-a" >>>>>>>
Outward Journey
-The graphic representation of x(t) looks like this:
x
|
| D
*
|
*
*
|
*
*
| D/2
*
*
|
*
*
|
*
*
|*---------------|----------------|-----------------|-----------------*-----
t
Takeoff
Spacecraft reaches the star
Landing
Formulae: Proper Time on board ship
= TS = (4c/a) Arccosh ( 1 + a.D/(2c2) )
Time passed on Earth = tE = (4c/a) ( aD/c2 + a2D2/(4c4)
)1/2
Maximum speed ( when x =
D/2 ) = vmax/c = Tanh µ where µ = a.TS/(4c)
-Times concern the whole journey: Earth-Star-Earth.
-Synthetic register M may be replaced by register R00.
01 LBL "ETP"
02 X<>Y
03 .2580738189
this constant = 365.25*86400*9.80665/(299792458*4)
04 *
the "standard" gravity 9.80665 doesn't seem quite "standardized"
05 STO M
for instance, I've seen here and there the value 9.80616
06 *
in this case, replace line 02 by 0.2580609239
07 ST+ X
08 ENTER^
09 ENTER^
10 X^2
11 LASTX
12 ST+ X
13 +
14 SQRT
15 STO Z
16 +
17 LN1+X
18 0
19 X<> M
20 ST/ Z
21 /
22 R^
23 1
24 +
25 1/X
26 ASIN
27 COS
28 2
29 ST/ L
30 X<> L
31 SIN
32 X^2
33 ST+ X
34 X<>Y
35 R^
36 R^
37 END
( 64 bytes / SIZE 000 )
| STACK | INPUTS | OUTPUTS |
| T | / | 1-vmax/c |
| Z | / | vmax/c |
| Y | a | tE |
| X | D | TS |
where TS = time passed on board ship
tE = time passed on Earth.
vmax = maximum speed ( when x = D/2 )
Example1: On 2007/12/25 immediate
boarding for the first interstellar flight in the direction of Toliman
( alpha Centauri ) Distance D = 4.343 light-years.
Fox Mulder remains on Earth but Dana Scully's flying saucer takes off at
10h13m TT , acceleration = 0.7g
0.7 ENTER^
4.343 XEQ "ETP" >>>>
TS = 8.837390060 years
RDN tE
= 13.09998313 years
RDN vmax/c = 0.9211383858
vmax = 92.11383858% * 299792458 m/s = 276150341
m/s
RDN 1-vmax/c = 0.07886161418
-When she steps from the ship, Dana's watch shows 2016/10/26 , 6h46m41s
wheras according to Mulder, it's 4h40m08s o'clock and we are
on 2021/01/30
Example2: With a = 1g and D = 2,200,000 light-years ( approximately the distance of the Andromeda galaxy )
1
ENTER^
2200000 R/S
>>>> 56.71150062 years
RDN 4,400,003.874 years
RDN vmax/c = 1
RDN 1-vmax/c = 3.877715920 E-13 whence
vmax/c was actually 0.999,999,999,999,612,228,408
-Caculating 1-vmax/c is only useful if the maximum
v/c is very close to 1
-If you don't want to compute this value, replace lines 28 to 36 by
STO T RDN
-If you prefer (1-v2max/c2)1/2
, replace lines 26 to 34 by ENTER^ ASIN
COS
-If you don't want to know vmax/c and
1-vmax/c , delete lines 22 to 36.
2°) Uniformly Accelerated Motion
-This program converts t ( time on Earth ) , T ( time on the spacecraft
) and x ( distance from the Earth ) for a given constant acceleration "a"
-It also calculates the speed v/c ( and 1-v/c ).
Formulae: v/c = Tanh µ
where µ = a.T/c , v/c = ( at/c ).( 1 + a2t2/c2
)-1/2
t = ( c/a ) Sinh µ
x = ( c2/a ) ( Cosh µ - 1 ) = ( c2/a ) ( (
1 + a2t2/c2 )1/2 - 1 )
-Due to these formulae, uniformly accelerated motion is often called "hyperbolic motion"
-Synthetic register M may be replaced by register R00.
01 LBL "ETP2"
02 LBL A
03 SF 00
04 GTO 01
05 LBL B
06 CF 01
07 GTO 00
08 LBL C
09 SF 01
10 LBL 00
11 CF 00
12 LBL 01
13 X<>Y
14 1.032295276
this constant is 4 times the value appearing line 02 of the "ETP" listing
above.
15 *
16 STO M
17 *
18 FS? 00
19 GTO 00
20 FS? 01
21 GTO 01
22 X^2
23 STO Y
24 1
25 +
26 SQRT
27 1
28 +
29 /
30 LBL 00
31 ENTER^
32 ENTER^
33 X^2
34 LASTX
35 ST+ X
36 +
37 SQRT
38 FS? 00
39 STO Z
40 +
41 LN1+X
42 RCL M
43 ST/ Z
44 GTO 02
45 LBL 01
46 E^X-1
47 LASTX
48 CHS
49 E^X-1
50 +
51 2
52 /
53 STO Y
54 RCL M
55 /
56 RCL Y
57 2
58 +
59 R^
60 *
61 SQRT
62 RCL M
63 LBL 02
64 /
65 R^
66 1
67 +
68 1/X
69 ASIN
70 COS
71 2
72 ST/ L
73 X<> L
74 SIN
75 X^2
76 ST+ X
77 X<>Y
78 R^
79 R^
80 CLA
81 END
( 128 bytes / SIZE 000 )
1°) a , x >>>> T
, t , v/c
| STACK | INPUTS | XEQ A | OUTPUTS |
| T | / | 1-v/c | |
| Z | / | v/c | |
| Y | a | t | |
| X | x | T |
Example: 0.7
ENTER^
10 XEQ "ETP2" or XEQ A or [A] in user mode
>>>> T = 3.870348933
RDN t = 11.29945015
RDN v/c = 0.992583500
RDN 1-v/c = 0.007416499860
2°) a , t >>>> T
, x , v/c
| STACK | INPUTS | XEQ B | OUTPUTS |
| T | / | 1-v/c | |
| Z | / | v/c | |
| Y | a | x | |
| X | t | T |
Example: 0.7
ENTER^
10 XEQ B or [B] in user mode >>>>
T = 3.702700069
RDN x = 8.711423203
RDN v/c = 0.990559778
RDN 1-v/c = 0.009440221722
3°) a , T >>>> t
, x , v/c
| STACK | INPUTS | XEQ C | OUTPUTS |
| T | / | 1-v/c | |
| Z | / | v/c | |
| Y | a | x | |
| X | T | t |
Example: 0.7
ENTER^
10 XEQ C or [C] in user mode >>>>
t = 951.2809258
RDN x = 949.8980538
RDN v/c = 0.999998942
RDN 1-v/c = 0.000001058151320
-If you don't want to compute 1-v/c, replace lines 71 to 79 by
STO T RDN
-If you prefer (1-v2max/c2)1/2
, replace lines 69 to 77 by ENTER^ ASIN
COS
-If you don't want to know vmax/c and
1-vmax/c , delete lines 65 to 79.
3°) An Example of non-uniformly Accelerated Motion
-The spaceship acceleration a(T) is now supposed to be known
as a function of the proper time T ( on board )
-We have:
a(T) = ( dv/dt ) ( 1 - v2/c2 )-3/2 = ( dv/dT ) ( 1 - v2/c2 )-1 since dT = dt ( 1 - v2/c2 )1/2 therefore, dv/( 1 - v2/c2 ) = a(T).dT
whence v/c = Tanh ( (1/c) F(T) ) where F(T) is the primitive of a(T) such that F(0) = 0 ( we assume x = 0 & v = 0 when t = T = 0 )
-Likewise, dt = dT ( 1 - v2/c2 )-1/2
= dT ( 1 - Tanh2(F(T)) )-1/2 = dT Cosh(F(T))
whence t = §0T
Cosh((1/c)F(u)).du
( § = Integral )
and dx = v.dt = c.Tanh(F(T)).Cosh(F(T)).dT = c. Sinh(F(T)).dT
whence x = c §0T
Sinh((1/c)F(u)).du
-"ETP3" evaluates t ( time on Earth ), x ( abscissa ) and v/c
( speed of the spacecraft ) at a given proper time T,
assuming the primitive F(T) may be expressed by elementary
functions
Data Registers:
• R00: F name
( Register R00 is to be initialized before executing "ETP3" )
R01 = 0 R04 thru R08 are used
by "GL3"
R02 = T R12 , R13: temp
R03 = n = number of subintervals used in Gaussian integration.
when
the program stops: R09 = t , R10 = x , R11 = v/c
Flags: /
Subroutines: "GL3" ( cf "Numerical
Integration for the HP-41" )
and a program which calculates F(T) assuming T is in X-register upon entry.
01 LBL "ETP3"
02 STO 02
03 X<>Y
04 STO 03
05 X<>Y
06 XEQ IND 00
07 1.032295276
08 STO 12
09 *
10 ST+ X
11 E^X-1
12 STO Y
13 2
14 +
15 /
16 STO 11
17 CLX
18 STO 01
19 RCL 03
20 LBL 00
21 STO 03
22 RCL 00
23 STO 13
24 "CT"
25 ASTO 00
26 XEQ "GL3"
27 STO 09
28 "ST"
29 ASTO 00
30 XEQ "GL3"
31 STO 10
32 RCL 13
33 STO 00
34 RCL 11
35 RCL 10
36 RCL 09
37 CLA
38 RTN
39 GTO 00
40 LBL "ST"
41 XEQ IND 13
42 RCL 12
43 *
44 E^X-1
45 LASTX
46 CHS
47 E^X-1
48 -
49 2
50 /
51 RTN
52 LBL "CT"
53 XEQ IND 13
54 RCL 12
55 *
56 E^X
57 LASTX
58 CHS
59 E^X
60 +
61 2
62 /
63 RTN
64 END
( 113 bytes / SIZE 014 )
| STACK | INPUTS | OUTPUTS |
| Z | / | v/c |
| Y | n | x |
| X | T | t |
where n is the number of subintervals over which the 3-point Gauss-Legendre formula is applied
Example: acceleration = a(T) = 0.7 cos ( 360°(T/40) ) the whole journey will take 40 years on board.
-Here, we can easily find the required primitive: F(T) = (14/pi) sin ( 360°(T/40) ) = (14/pi) sin ( 9°T ) So we load the following subroutine
01 LBL "FF" ( global label , at most 6 characters
)
02 9
03 *
04 SIN
05 14
06 *
07 PI
08 /
09 RTN
- "FF" ASTO 00
-We set the HP-41 in DEG mode and if we choose T = 1 ( year )
and n = 2
2 ENTER^
1 XEQ "ETP3" >>>> t
= 1.088871751 year = R09 ( the difference
between Earth time and ship time is already about 32 days )
RDN x = 0.376425754 light-year = R10
RDN v/c = 0.616685490 = R11
-With T = 10 and n = 4
4 ENTER^
10 XEQ "ETP3" >>>> t = 190.9695910
( in 1 minute )
RDN x = 189.5025503
RDN v/c = 0.999798046
-To have another approximation for the same T but a different n, simply key in the new n-value and press R/S , for instance:
8 R/S >>>> t = 190.9695910
RDN x = 189.5025369
RDN v/c = 0.999798046
-Note that v/c is obtained exactly since no approximate integration
formula is used.
-In the above example, the whole journey Earth-Star-Earth required
40 years on board ship but almost 764 years have passed on Earth!
-The distance Earth-Star was approximately 379 light-years.
-Results may be wrong if F(T) or its derivatives have singularities.
4°) Non-uniformly Accelerated Motion, a more general
case
-We study the same problem as in paragraph 3 , but we suppose the primitive difficult ( or impossible ) to express analytically. We have to evaluate:
t = §0T
Cosh( (1/c) §0u
a(v).dv ).du
x = c §0T
Sinh( (1/c) §0u
a(v).dv ).du
v/c = Tanh ( (1/c) §0T
a(u).du )
Data Registers:
• R00: "a" name
( Register R00 is to be initialized before executing "ETP4" )
R01 = 0 R04 thru R17 are used
by "DIN2"
R02 = T R18 , R22: temp
R03 = n = number of subintervals used in Gaussian integration.
when
the program stops: R19 = t , R20 = x , R21 = v/c
Flags: F04
Subroutines: "DIN2" ( see the bottom
of this page )
"GL3" ( cf "Numerical Integration for the HP-41" )
and a program which calculates a(T) assuming T is in X-register upon entry.
01 LBL "ETP4"
02 1.032295276
03 STO 18
04 CLX
05 STO 01
06 RDN
07 STO 02
08 X<>Y
09 LBL 00
10 STO 03
11 RCL 00
12 STO 22
13 "T"
14 ASTO 00
15 "U"
16 ASTO 04
17 "V"
18 ASTO 05
19 "CH"
20 ASTO 17
21 SF 04
22 XEQ "DIN2"
23 STO 19
24 "SH"
25 ASTO 17
26 XEQ "DIN2"
27 STO 20
28 RCL 22
29 STO 00
30 XEQ "GL3"
31 RCL 18
32 *
33 ST+ X
34 E^X-1
35 STO Y
36 2
37 +
38 /
39 STO 21
40 RCL 20
41 RCL 19
42 CLA
43 RTN
44 GTO 00
45 LBL "T"
46 X<>Y
47 XEQ IND 22
48 RCL 18
49 *
50 RTN
51 LBL "U"
52 CLX
53 RTN
54 LBL "V"
55 RTN
56 LBL "SH"
57 E^X-1
58 LASTX
59 CHS
60 E^X-1
61 -
62 2
63 /
64 RTN
65 LBL "CH"
66 E^X
67 LASTX
68 CHS
69 E^X
70 +
71 2
72 /
73 RTN
74 END
( 156 bytes / SIZE 023 )
| STACK | INPUTS | OUTPUTS |
| Z | / | v/c |
| Y | n | x |
| X | T | t |
where n is the number of subintervals over which the 3-point Gauss-Legendre formula is applied
Example: Let's take again a(T) = 0.7 cos ( 360°(T/40) ) as if we couldn't calculate the primitive ( it will be a test case for our program )
-We load the following subroutine
01 LBL "AA" ( global label , at most 6 characters
)
02 9
03 *
04 COS
Store 9 and 0.7 into registers R23 and R24 and replace lines
02 & 05 by RCL 23 & RCL 24 respectively
05 .7
It will speed up execution.
06 *
07 RTN
- "AA" ASTO 00
-We set the HP-41 in DEG mode and with T = 1 and n = 2
2 ENTER^
1 XEQ "ETP4" >>>> t
= 1.088871751
RDN x = 0.376425755
RDN v/c = 0.616685490
-With T = 10 and n = 4
4 ENTER^
10 XEQ "ETP4" >>>> t = 190.9695915
( in 8mn17s )
RDN x = 189.5025508
RDN v/c = 0.999798046
-To have another approximation for the same T but a different n, simply key in the new n-value and press R/S , for instance:
8 R/S >>>> t = 190.9695911
RDN x = 189.5025369
RDN v/c = 0.999798046
-Unlike "ETP3" , "ETP4" re-calculates v/c for each new n-value.
-When n is multiplied by 2 , execution time is approximately multiplied
by 4
5°) Double Integrals §ab G ( §u(x)v(x) f(x,y) dy ) dx
-"ETP4" requires to evaluate integrals of this form, where G is a function
of 1 variable.
-Modify the program listed in "Double Integrals for the HP-41" as follows:
-Replace line 01 by LBL "DIN2"
-Delete line 96
-Add 3.6 / FS? 04
GTO IND 17 after line 89
-In other words, the new listing looks like this:
01 LBL "DIN2"
.......................
the same lines as "DIN"
.......................
87 RCL 14
88 RCL 11
89 *
90 3.6
91 /
92 FS? 04
93 GTO IND 17
94 RTN
95 LBL 04
96 RCL 06
97 RCL 09
98 *
99 3.6
100 /
101 STO 06
102 END
( 148 bytes / SIZE 018 )
-Set flag F04, store G name in register R17, and follow the same instructions as "DIN"
Example: Compute §12 Ln( 1 + §xx^2 dy/(x+y) ) dx
-Load these routines in program memory:
01 LBL "T"
02 +
03 1/X
04 RTN
05 LBL "U"
06 RTN
07 LBL "V"
08 X^2
09 RTN
10 LBL "W"
11 LN1+X
12 RTN
"T" ASTO 00 "U" ASTO 04 "V"
ASTO 05 "W" ASTO 17 SF 04
1 STO 01 2 STO 02
-With n = 2 2 STO 03 XEQ "DIN2"
>>>> 0.191218819
-With n = 4 4 STO 03
R/S >>>>
0.191218775
-If flag F04 is clear, "DIN2" performs the same double integral as "DIN"
( G is not taken into account )
Go back to the HP-41 software library
Go back to the general software library
Go
back to the main exhibit hall