HP Forum Archive 21

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A slide rule to conquer the moon revisited Message #1 Posted by Palmer O. Hanson, Jr. on 18 June 2013, 4:51 p.m. Admittedly, there are a few things that a slide rule can do well; for example, such as solving Ohm's Law problems. But for many problems in the aeronautical engineering discipline the slide rule is essentially useless. If you don't believe me try doing the Mach Number solution with your slide rule.

Re: A slide rule to conquer the moon revisited Message #2 Posted by Paul Dale on 18 June 2013, 5:50 p.m., in response to message #1 by Palmer O. Hanson, Jr. A slide rule is far better than a calculator for helping you draw straight lines however :-) - Pauli

Re: A slide rule to conquer the moon revisited Message #3 Posted by Dave Shaffer (Arizona) on 18 June 2013, 8:20 p.m., in response to message #2 by Paul Dale That's about all I've done with mine for the last twenty years! Use the slide, and you even have the equivalent of a beveled edge that won't suck ink back under the ruler.

Re: A slide rule to conquer the moon revisited Message #4 Posted by Palmer O. Hanson, Jr. on 21 June 2013, 10:15 a.m., in response to message #3 by Dave Shaffer (Arizona) Quote: Use the slide, and you even have the equivalent of a beveled edge that won't suck ink back under the ruler. You won't find beveled edges on the log-log devices. Many of the older Mannheim slide rules not only have beveled edges but also include centimeter or inch scales on the edges.

Re: A slide rule to conquer the moon revisited Message #5 Posted by BobVA on 18 June 2013, 9:08 p.m., in response to message #1 by Palmer O. Hanson, Jr. Quote: If you don't believe me try doing the Mach Number solution with your slide rule. I used to *love* to work that Mach number sample problem in some of the old HP manuals. I always thought it was the RPN equivalent of the aria from "Carmen" :-) Edited: 18 June 2013, 9:08 p.m.

Re: A slide rule to conquer the moon revisited Message #6 Posted by Steve Simpkin on 18 June 2013, 10:57 p.m., in response to message #5 by BobVA Quote: I used to *love* to work that Mach number sample problem in some of the old HP manuals. I always thought it was the RPN equivalent of the aria from "Carmen" :-) Me too!

Re: A slide rule to conquer the moon revisited Message #7 Posted by John Stark on 18 June 2013, 11:08 p.m., in response to message #6 by Steve Simpkin To all: Actually, a pencil, some patience, and an RPN mindset, along with a log-log slide-rule, could solve this formula handily to a few decimal places. Note the "RPN Mindset". This is the greatest advantage the "HP Way" of calculation gives the operator. Sorely lacking in the "Algebraic/Textbook" world. My $0.02. John Stark PS: Looking forward with great anticipation to the "Prime". May finally give up my 45 and 80 for this beast.

Re: A slide rule to conquer the moon revisited Message #8 Posted by Ingo on 19 June 2013, 6:32 a.m., in response to message #6 by Steve Simpkin Result with HP15C LE: 0,835724536 (in 1 minute, due to typing error, repeated in 30 seconds) Result with FC 2/83 using simplification (e.g assuming first inner bracket to be 1.05^3.5, ...) gives 0,814 in 7 minutes Result with FC 2/83 w/o simplification: 0.8365 (read from LL02) in 15 minutes Actually, I repeated the HP15C calculation after finishing the SR calculation and discovered the error. So error of slide rule is 2.67% with simplification and 9,27*10^-4 w/o simplification, the time advantage for the HP15C is factor 7.5 and 15, respectively. So when the HP35 appeared in 1972 I would assume it would have payed back in saved engineering time within one month or so (trained people are most likely much faster with the slide rule than I am) Anyway, the slide rule seems to be accurate to three digits even for this lengthy formula. Otherwise, I think, the SR-71 (note the pun) might have never flown.

Re: A slide rule to conquer the moon revisited Message #9 Posted by Nick_S on 19 June 2013, 8:34 a.m., in response to message #6 by Steve Simpkin There is something to be said for tackling this calculation with a language using a uniform operator precedence. Here it is rendered in LISP style prefix notation: (sqrt (* 5 (- (expt (+ (* (- (expt (+ 1 (* 0.2 (expt (/ 350 661.5) 2))) 3.5) 1) (expt (- 1 (* (* 6.875 (expt 10 -6)) 25500)) -5.2656)) 1) 0.286) 1))) ;Value: 0.8357245351752515 Nick Edited: 19 June 2013, 8:36 a.m.

Re: A slide rule to conquer the moon revisited Message #10 Posted by Gerson W. Barbosa on 21 June 2013, 4:08 p.m., in response to message #6 by Steve Simpkin Result on the wp34s, following the keystrokes sequence in the HP-67 manual: 0.8357245351752513232390389024784470 The last two digits should be 61, according to W|A. Of course these so many digits don't make any sense, since the speed of sound at sea level in knots is given to only four significant digits (also, it's not known to more than the ten the HP-67 can handle).

Re: A slide rule to conquer the moon revisited Message #11 Posted by Matt Agajanian on 30 June 2013, 5:57 p.m., in response to message #6 by Steve Simpkin Hello all. This also, is one of my favourite formulas to work through and demonstrate RPN efficiency. I'm wondering though, as in every manual, the coefficients and variables are already there but, it its generic form, which are the variables and which are the constant coefficients? Are the only variables 350 (CAS) and 25,500 (PALT)? Edited: 30 June 2013, 5:58 p.m.

Re: A slide rule to conquer the moon revisited Message #12 Posted by Gerson W. Barbosa on 30 June 2013, 6:57 p.m., in response to message #11 by Matt Agajanian 661.5 is the speed of sound in knots, at sea level.

Re: A slide rule to conquer the moon revisited Message #13 Posted by Ken Shaw on 19 June 2013, 3:05 p.m., in response to message #5 by BobVA In the current context, this old thread is interesting to read. http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv013.cgi?read=43770

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