Moon Landing

[hibiki]

I was playing with the game Moon Landing for HP25 published here:

http://www.hpmuseum.org/software/25moonld.htm

I adapted it to HP34C, as I do not have the 25. And also for HP32SII.

It started at 500 feet height, -50 feet/s speed and 120 units of fuel. So my best sequence of burns that solve the problem is 0-0-0-0-0-5-5-80 remaining 30 units of fuel and only 8 seconds (steps) with 3 fuel burns. This is my best and faster landing sequence. There are also other solution sequences , with less amounts of fuel remaining and more landing time.

Solving for the equations used, i found the constant amount of fuel consumption to land the spaceship. It is 7.5 units per burn, doing 20 burns. This of course is 30 units more than the available amount of fuel. So this answer does not solve the problem.

So, now I was looking at the HP67 game version. It starts only with 60 units of fuel, and the same height and speed of the HP25 version. Much less fuel than my best solution for the HP25. Is it possilble to solve with this HP67 version ? Is there any algorithm to minimize the fuel consumption besides landing safely the spaceship ?

[Maximilian Hohmann]

(01-28-2017 01:19 PM)hibiki Wrote:  Is it possilble to solve it ?

Yes. From the dark ages of university (I studied aerospace engineering there) I remember that one burns the least amount of fuel in a rocket with the shortest possible burn at the highest possible thrust. Jules Verne's moon launch using a cannon was therefore the most engergetically optimal way of achieving the goal.

For the lunar landing, wait as long as your nerves can stand it and then fire the maximim amount of fuel units until you are down safely. Will take a couple of trials to find the right moment for the first rocket firing... (or alternatively solve the equations on paper first).

PS: Watch the videos of the Space X Falcon 9 rocket landing back on it's barge to see how it is best done!

[SlideRule]

(01-28-2017 01:19 PM)hibiki Wrote:  I was playing with the game Moon Landing for HP25 ... Is there any algorithm to minimize the fuel consumption besides landing safely the spaceship?

From a long ago and unreferenced source (any know the original?)
[attachment=4425]

BEST!
SlideRule

edit: unhappy w formatting, replaced w a PDF of same!

[BarryMead]

(01-28-2017 01:19 PM)hibiki Wrote:  I adapted it to HP34C, as I do not have the 25. And also for HP32SII.

This version is adapted to the WP-34S Enjoy.

Code:

001 LBL A
002 FIX 04
003 GTO 14
004 LBL 02
005 RCL 00
006 EEX
007 4
008 /
009 RCL 01
010 x<0?
011 GTO 11
012 +
013 GTO 13
014 LBL 11
015 x[<->] Y
016 -
017 LBL 13
018 STOP
019 LBL 14
020 RCL 02
021 x<? Y
022 GTO 34
023 R[v]
024 STO- 02
025 5
026 -
027 STO 03
028 2
029 /
030 RCL 00
031 +
032 RCL 01
033 +
034 STO 00
035 x<0?
036 GTO 44
037 RCL 03
038 STO+ 01
039 GTO 02
040 LBL 34
041 RCL 01
042 x[^2]
043 RCL 00
044 1
045 0
046 [times]
047 +
048 [sqrt]
049 +/-
050 STO 01
051 LBL 44
052 RCL 01
053 FIX 00
054 GTO A
055 END

[Paul Dale]

(01-28-2017 07:08 PM)SlideRule Wrote:  From a long ago and unreferenced source (any know the original?)

This version assumes 120 fuel. The first post said 60 fuel for the HP 67 version.
Is the latter even possible?


Pauli

[BarryMead]

(01-29-2017 06:33 AM)Paul Dale Wrote:  

(01-28-2017 07:08 PM)SlideRule Wrote:  From a long ago and unreferenced source (any know the original?)

This version assumes 120 fuel. The first post said 60 fuel for the HP 67 version.
Is the latter even possible?


Pauli

I assume it would be possible to land with only 60 fuel if you start with a much smaller initial velocity and initial altitude.

[Maximilian Hohmann]

(01-29-2017 06:49 AM)BarryMead Wrote:  I assume it would be possible to land with only 60 fuel if you start with a much smaller initial velocity and initial altitude

The original Hp67 program starts at 500ft, with 50ft/s downward velocity and 60 units of fuel.

[hibiki]

I found that the programs are not equal. The HP25 version takes the fuel units burned and subtracts 5 units and divide by 2 to calculate the new speed. The HP67 multiplies by 2, subtracts 5 and divide by 2. So I think it is possible to solve it.

The PDF solution, uses a sequence of 0-0-0-0-0-20-20-20-20-20, i.e. 10 seconds and 100 units fuel consumption. This is a solution, but not the best.

[Dieter]

(01-28-2017 01:19 PM)hibiki Wrote:  I was playing with the game Moon Landing for HP25 published here:
http://www.hpmuseum.org/software/25moonld.htm
I adapted it to HP34C, as I do not have the 25.

FTR: the HP34C program collection that came with the calculator included a 34C version of the moon landing game.

(01-28-2017 01:19 PM)hibiki Wrote:  So, now I was looking at the HP67 game version. It starts only with 60 units of fuel, and the same height and speed of the HP25 version. Much less fuel than my best solution for the HP25. Is it possilble to solve with this HP67 version ?

The HP67FUN ROM for the 41C includes a HP41 version of the HP67 program that was originally included in the HP67/97 Games Pac. I assume this is the program you refer to. Playing the translated game version (with the same formulas) on the HP41 ends in a successful landing with the following strategy:

Simply enter nothing and watch the lander approach the moon. After the first zero burn you will see a speed of 55 and an altitude of 448. Continue to do nothing until you see a speed of 85 (!) and an altitude of merely 28. After the following countdown now burn 45 units of fuel. You should land softly with zero speed.

In your notation I think this would be 0 - 0 - 0 - 0 - 0 - 0 - 0 - 45

Dieter

[ggauny@live.fr]

At SlideRule,


The original is in HP65-V2N2 from 1976. Pages 4 and 5.

[SlideRule]

(03-08-2017 05:00 PM)ggauny@live.fr Wrote:  At SlideRule,
The original is in HP65-V2N2 from 1976. Pages 4 and 5.

Many Thanks!

SlideRule

[ggauny@live.fr]

You can have a look here :
DAVID HAYDEN post 6


LUNAR LANDER