HP Forums

Maybe I miss the joke or my calculations went strange wrong but anyway. ;-)

First take a look to the
HP-35 Direct Mail Letter, Page 1, 1972 link.

The calculation is wrong. In the context of this "memo" it's a bit funny. Why put HP this "memo" back to the
2009 press kit?

If I were evil, I could say it fits to the presented calculators on this 2009 press kit. Sorry HP ;-).

Michael

PS: Tested on the following devices:-) Aristo Darmstadt slide rule, HP45, HP28S, HP42S, HP48SX, HP49G, HP15C LE, HP35s, Casio FX-115M, Sharp EL-W506X, Aristo M85

I also calculated a slightly different answer with my HP35s:

1.772825519

Except that the correct answer is 0.1772825509 on an HP 35 or 0.1772825519 on an HP 35s. So the real error is a factor of 10 decimal point shift.

Edited: 9 May 2013, 4:19 p.m.

You're probably in ALL mode. Just scroll the display right (a truly annoying feature of the HP 35s) and see the exponent is -1:

Hmmmh, my result is 0,177282551904 on my WP 34S (what else?). I wonder how you did that on an Aristo Darmstadt.

d;-)

This can be fixed on the HP 33s/35s by typing DISPLAY FIX .0, which gets you 10 decimal point places w/o the exponent.

0.177282551901 on my Corvus 500.

Thanks. I remember I used FIX 8, but I didn't like this either. The HP 33s method (hiding the least significant digits in favor of the exponent) was much better.

ARE YOU NOT ENTERTAINED?

Yeah, I feel like an idiot. I normally keep my 35s in FIX 4. The answer in the letter didn't include a x10^-1 and I saw that E- at the end, but ignored it. I was paying more attention to the differences between the 19/10000000000 and 9/10000000000 partial fractions than seeing his error by a factor of 10. I don't really know what the equation is about, but he said it was a solid angle and didn't include units. Is that radians or degrees? If it's radians, that's enough of an angle to completely miss the moon and probably hit the barn next door.

Quote:

---

Hmmmh, my result is 0,177282551904 on my WP 34S (what else?). I wonder how you did that on an Aristo Darmstadt.

d;-)

---

I calculate (slide) 0,157.
Yes it's not spot on but I'm not grown up with slide rulers.

That's what I've done:

a) 2,5 / 10,3 = (D)/(C) -> (1D) = 0,243

b) 0,243ˆ2 = (D)->(A) = 0,059

c) 0,059 + 1 = 1,059

d) 1/1,059 = (C)->(CI) = 0,95

e) SQR( 0,95 ) = (A)->(D) = 0,975

f) 1 - 0,975 = 0,025

g) 2 * 0,025 = 0,05

h) Pi * 0,05

h.1) Pi * 5 (D)->(C)->(D) = 15,7

h.2) 15,7 / 100 = 0,157

My Aristo Darmstadt is from my father. He studied EE first with slide rules and later with the ARISTO M85. But this Aristo Darmstadt was never used before. It was a gift from a company called Groschopp & Co. GMBH Viersen as printed on the back. I've gotten both for my little calculator collection. ;-)

Edited: 9 May 2013, 6:44 p.m.

Yes I am very well entertained with this advertising memo. Thanks. ;-)

From time to time I'll check all caculators in my collection.

From time to time I'm walking around the internet about historic HP calculator stuff.

In this case I stumbled over this document and think...

Edited: 9 May 2013, 7:08 p.m.

Quote:

---

I don't really know what the equation is about, but he said it was a solid angle and didn't include units. Is that radians or degrees? If it's radians, that's enough of an angle to completely miss the moon and probably hit the barn next door.

---

If it is a solid angle (which makes sense in context), then the units are steradians.

There are 4*pi steradians in a complete sphere (as viewed from inside looking out - i.e. the entire field of view ALL around you.) The "units" of solid angle are radians^2. A piece of the sky 1 degree by 1 degree (about 1/57.3 of a radian in each direction) would subtend an area of the sky of (1/57.3)^2 steradians (neglecting corrections for projecting on to a curved background).

Quote:

---

A piece of the sky 1 degree by 1 degree (about 1/57.3 of a radian in each direction) would subtend an area of the sky of (1/57.3)^2 steradians (neglecting corrections for projecting on to a curved background).

---

d:-?

For small angles, yes, you can pretty much ignore the effect.

Technically, you are integrating angles over a curved surface, and as the angles get bigger, you MUST correct for projection effects.

In the limit, consider a full circle - if you multiply 360 degrees by 360 degrees you get 129600 square degrees, versus the correct value of about 41253.

In spherical coordinates, you are integrating

d(Omega) = sin(theta) d(theta) d(phi)

0.177282551895 on my new toy a Psion Organiser II LZ64.

Dave, thanks for your explanation :-) Once (some decades ago) I knew by heart how that shall be integrated, but I never would have attributed that to a curved surface. Thanks for the refresher.

d:-)

Thanks! This demonstrates the slide rule is good enough for detecting that error at least, although the result is some 11% off.

d:-)

You could even do that in your head using the fact that for small x:

With x = 2.5/10.3 ~ 1/4 and 2*Pi ~ 6 we end up with 6/32 = 3/16 = 0.1875.

Cheers

Thomas

You could avoid cancellation using the identity:

You would end up with:

a) 2,5 / 10,3 = (D)/(C) -> (1D) = 0,243

b) 0,243ˆ2 = (D)->(A) = 0,059

c) 0,059 + 1 = 1,059

...

SQR(1,059) = 1,029

Pi * 0,059 = 0,185

0.185 * 2 = 0,37

0,37 / 1,029 = 0,36

0,36 / 2,029 = 0,177

However I didn't use a slide rule but the results might be close enough.

a) 2,5 / 10,3 = (D)/(C) -> (1D) = 0,243

b) 0,243ˆ2 = (D)->(A) = 0,059

c) 0,059 + 1 = 1,059

...

SQR(1,059) = 1,029

Pi * 0,059 = 0,185

0.185 * 2 = 0,37

1 + 0,059 + 1,029 = 2,088

0,37 / 2,088 = 0,177

Kind regards

Thomas

Edited: 10 May 2013, 3:03 p.m.

Using a Faber Castell 2/83 which offers better accuracy I get 0,176 which is only 0,73% off.

Edit: With a better calibrated 2/83N that has a better cursor and using the W scales only plus LL02 to get the root, I achieved .1772, but this took me some minutes to position the slider as accurate as possible.

I could try a 1914 Nestler 27a, which is 50cm long and has a W scale too (giving it an effective length of 1m), but I am afraid that it is not much more accurate than a 1972 Faber Castell.

Edited: 13 May 2013, 3:48 p.m.

(2,5 / 10,3)² * = 0,185

0,185 * 2 = 0,37

0,37 / 2,08 = 0,178

Cheers

Thomas

Great documentation, Thomas! I'd have prefered an Aristo 969 though.

d:-)

Edited: 11 May 2013, 1:48 a.m.

You could use Stefan Vorkoetter's Aristo Multilog Nr. 970 Simulator
. Is that close enough?

Best regards

Thomas

Great! Your scheme works nicely here as well:

...but I still prefer

2.5 ENTER 10.3 / XEQ S --> 0.17728255191

Cheers,

Gerson.

Hey Gerson

You can't let us hanging here without posting the program.

Best regards

Thomas

Hello Thomas,

Actually I resorted to a simplification you used one of these days, the one involving hyperbolics. That's my way to say "thank you" :-)

Cheers,

Gerson.

----------------------------

HP-42S:

00 { 18-Byte Prgm }

01>LBL "S"

02 ASINH

03 COSH

04 1/X

05 PI

06 STO+ ST X

07 *

08 RCL- ST L

09 +/-

10 .END.

00 { 18-Byte Prgm }

01>LBL "S"

02 ASINH

03 COSH

04 1/X

05 SIGN

06 RCL- ST L

07 PI

08 STO+ ST X

09 *

10 .END.

WP34S:

001 LBL A

002 ASINH

003 COSH

004 1/x

005 +/-

006 INC X

007 # pi

008 STO+ X

009 *

010 END

001 LBL A

002 1

003 cABS

004 1/x

005 +/-

006 INC X

007 # pi

008 STO+ X

009 *

010 END

My first attempts were pretty straight forward but now that I saw your LASTX-juggling I could improve them a little:

00 { 17-Byte Prgm }

01>LBL "S"

02 ASINH

03 COSH

04 PI

05 STO+ ST X

06 ENTER

07 RCL/ ST Z

08 -

09 END

00 { 16-Byte Prgm }

01>LBL "S"

02 1

03 ->POL

04 PI

05 STO+ ST X

06 ENTER

07 RCL/ ST Z

08 -

09 END

And here's the one using the other formula:

00 { 20-Byte Prgm }

01>LBL "S"

02 X^2

03 1

04 RCL+ ST Y

05 SQRT

06 RCL+ ST L

07 /

08 PI

09 STO+ ST X

10 *

11 END

0.17728255190 on my HP50g.

Edited: 13 May 2013, 10:21 a.m.

Great, two bytes and one step less!

Well, not on the HP-35 and not within 20 seconds.

The result shown in that 1972 ad is exactly the same as the user of the 2011 HP-15C Limited Edition would get! So where's the problem? :-)

Such are the limitations of the old 1972 HP-35 with respect to numerical accuracy. They remained (for this problem) until the Saturn-based machines were produced in 1986 (see HP-15C results below). In the USSR, one could take the popular RPN Elektronika MK 61 (1983 to 1994) and get a far less accurate and precise result.

Machine            Introduced   Result

MK 61                 1983      0.177 282 58

HP-35                 1972      0.177 282 550 9

HP-15C (and -15C LE)  1982      0.177 282 550 9

HP-28S and later      1986      0.177 282 551 901

Casio fx-115ES Plus   2011      0.177 282 551 903 861

Free42                 --       0.177 282 551 903 885 91

Wolfram Alpha          --       0.177 282 551 903 885 912 305...

It's really impressive in a negative way how poorly the MK 61 (sold into the mid-1990s in the former USSR) performs, and really impressive in a positive way that Free42 produces a result that agrees with Wolfram Alpha for all 17 digits generated.

It's also notable that the numerical accuracy and precision of the $18 Casio fx-115ES Plus is significantly better than the HP 42S, or even the HP 50G.

On the WP34S:

001:         LBL A          ; this is essentially Thomas Klemm's program above, the most optimized one.

002:         1

003:         [->]POL

004:         # [pi]

005:         ENTER[^]

006:         RCL/ Z

007:         -

008:         STO+ X

009:         END


DBLON


2.5 ENTER 10.3 / A

0.177 282 551 903 885 912 305 974 246 640 144 0

Quote:

---

The result shown in that 1972 ad is exactly the same as the user of the 2011 HP-15C Limited Edition would get! So where's the problem?

---

d;-)

Very interesting to see how far this thread runs.

Here is my little RPL for the HP48SX:

<<                 ; rewritten from Gerson W.Barosa's program above.

-17 SF -18 CF      ; rad mode

-15 CF -16 CF      ; rec mode

[pi] [pi] ROT

1 ->V2 -16 SF      ; ->pol

V-> DROP

/ - DUP +

->NUM

>>

'A' STO

2.5 ENTER 10.3 / A

STD: .1772825519

SCI: 1.77282551900E-1

Michael

-----------------------------------------------

The easy way on HP48SX ;-)

<<

SQ

1 +

INV

SQRT

NEG 1 +

[pi] * 2 *

>>

'A2' STO

2.5 ENTER 10.3 / A2

STD: .177282551901

SCI: 1.77282551901E-1

Edited: 14 May 2013, 1:56 p.m.

Quote:

---

It's really impressive in a negative way how poorly the MK 61 (sold into the mid-1990s in the former USSR) performs

---

Quote:

---

It's also notable that the numerical accuracy and precision of the $18 Casio fx-115ES Plus is significantly better than the HP 42S, or even the HP 50G.

---

Yes, the absolute error usually is lower if the number of digits (precision) increases. But 15-digit precision does not mean you'll also get 15-digit accuracy. ;-)

Dieter

Quote:

---

Very interesting to see how far this thread runs.

---

It's been interesting since the beginning. I wonder how many people saw the ad and never found the mistake. Thanks for posting!

Quote:

---

The easy way on HP48SX ;-)

<<

SQ

1 +

INV

SQRT

NEG 1 +

[pi] * 2 *

>>

'A2' STO

2.5 ENTER 10.3 / A2

STD: .177282551901

SCI: 1.77282551901E-1

---

There was a time even half a byte mattered. Not anymore! But let's save a few of them, just for fun :-)

40 bytes:

\<< SQ 1 + INV \v/ NEG

1 + \pi * 2 *

\>>

37.5 bytes:

\<< ASINH COSH \pi 2 *

SWAP / LASTARG DROP

- NEG

\>>

35 bytes:

\<< ASINH COSH INV

SIGN LASTARG - \pi 2

* *

\>>

35 bytes:

\<< ASINH COSH INV

NEG 1 + \pi * 2 *

\>>

< 35 bytes:

???

Quote:

---

Very interesting to see how far this thread runs.

---

Maybe we can still go a little further. I was just curious on the impact of the accuracy when using different formulas:

I expected the first to be the most accurate since the cancellation is omitted. And then there's a subtle difference between the 2nd and the 3rd formula as we loose a significant digit when calculating the square root before calculating the reciprocal. This can be observed e.g. when calculating [1/x] for 9.999999999E-1 on a calculator that provides 10 digits. The result is just 1.000000000, thus we lost that last digit. Therefore it's better to calculate the square root of a number which is little smaller as 1 than if it is a little bigger that 1.

Here's what I came up with. The first result is the expected result if we trust WolframAlpha:

WolframAlpha

0.17728255190388591230597424664014404052042011206343029588068530548343231405482470150877366060101266263651199682635366833...

MK-61

0.1772825519...

0.17728255

0.17728258

0.17728195

HP-21

0.177282551903...

0.1772825520

0.1772825509

0.1772825478

HP-15C

0.177282551903...

0.1772825520

0.1772825509

0.1772825541

HP-48

0.17728255190388...

0.177282551904

0.177282551901

0.177282551907

Free42

0.177282551903 885912305974 246640...

0.177282551903 885912305974 1674 

0.177282551903 885912305974 2930

0.177282551903 885912305977 4346

WP34S

0.177282551903 885912305974 246640144040...

0.177282551903 885912305974 2466401440

0.177282551903 885912305974 2466401453

0.177282551903 885912305974 2466401434

The intermediate result was multiplied by and then 2 in this order. If available Rectangle->Polar conversion (or something similar) was used to calculate sqrt(1 + x²) in the 3rd formula.

There's only one result that I didn't expect: why is the 2nd formula more accurate than the 1st in case of Free42?

Cheers

Thomas

\<<

  ASINH COSH

  \pi DUP +

  DUP ROT / -

\>>

32.5 bytes

Another 32.5-byte solution, albeit not so elegant:

\<<

  ASINH COSH

  INV \pi *

  \pi - -2 *

\>>

IMHO, to be really fair one needs to evaluate the formula simply as shown in the original advertisement. None of the identities and trigonometric equivalencies seem appropriate since those weren't expressed in the advertisement. :-)

But still...interesting...

Double precision result of the original formula from the WP 34S:

0.177 282 551 903 885 912 305 974 246 640 145 3

FWIW

d:-)

Quote:

---

It's been interesting since the beginning. I wonder how many people saw the ad and never found the mistake. Thanks for posting!

---

That's exactly what I have asked myself.

Thank you back. It's a pleasure to see all the nice solutions and tiny programs. I've learned and refreshed a lot. Thomas K. shows very smart mathematical transformations. I also like the discussion thinking about accuracy and precision and not to forget the slide rulers...

32,5 Byte = unbelievable ;-)

Michael

Edited: 16 May 2013, 4:53 p.m.

Very impressive work on the WP 34S, as every example that comes up shows the numerical analysis expertise of its creators!

I wish the WP-34 were commercially available...I'm personally just not skilled for any local re-purpose project.

Edited: 17 May 2013, 2:31 p.m.

Quote:

---

I wish the WP-34 were commercially available...

---

d:-)

Quote:

---

It is, e.g. here.

---

Thanks Walter. After all this time, I've somehow missed the commercial availability of the WP 34S.

I just ordered one...there appears to be a small backorder period.

Thanks, Mike. And don't forget the manual ...

d:-)