# Understanding the Viewgraph by Dr. John Houbolt and the probability of success calculation of a Lunar Landing

The well known Dr. John Houbolt used a viewgraph to explain the much higher probability of success with LOR in comparison to EOR and Direct Flight for a lunar landing.

I used Python to check the results of the overall probability:

#LOR
#            1         2         3         4         5      6      7      8      9     10     11     12     13     14
prob_DF = 0.98**8 * 0.94**8 * 0.94**2 * 0.94    * 0.94 * 0.90 * 0.98 * 0.98 * 0.98 * 0.99
prob_LR = 0.98**5 * 0.94**5 * 0.94    * 0.99    * 0.99 * 0.95 * 0.99 * 0.95 * 0.99 * 0.99 * 0.99
prob_ER = 0.98**5 * 0.94**5 * 0.98**5 * 0.94**5 * 0.94 * 0.90 * 0.94 * 0.94 * 0.94 * 0.90 * 0.98 * 0.98 * 0.98 * 0.99

print("overall probability ", format(prob_DF, "4.2"), format(prob_LR, "4.2"), format(prob_ER, "4.2"))

I get the same numbers:

overall probability  0.34 0.53 0.26

The numbers for the first step seem to be the reliability of the first stages of a Saturn V with 5 F-1 engines $$.98^5$$ and of a Nova with 8 F-1 engines $$.98^8$$.

But Houbolt assumed a C-3 or C-4 rocket would do, not a C-5 version with 5 F-1 engines in the first stage which was later named Saturn V. So what else could be the reason for the exponents 5 and 8?

The second step are the 5 or 8 J-2 engines of the second stage $$.94^5$$ or $$.94^8$$. The reliability (.94) of the second stage liquid hydrogen engine ignited in flight was assumed a little lower than the RP-1 engine of the first stage with (.98).

The third and seventh step Earth escape may be interpreted as the third stage with two engines for direct flight and a single engine for lunar rendezvous. But the mass is much bigger for both the direct flight and earth rendezvous, so the relative probability should be both $$.94^2$$?

The probability of the midcourse correction is smaller for the way there and larger for the way back because of the significantly smaller mass. At lunar rendezvous, the mass is smallest, hence the highest probability of .99.

Any other thoughts to explain the different probabilities?

• His 1961 paper about LOR is here hq.nasa.gov/alsj/… I skimmed it and saw a few probabilities but they appeared to be different from the ones in the question. I don't have the time to do a detailed reading of the paper right now but maybe the answer is in there. May 2, 2022 at 1:48
• @OrganicMarble Many thanks for the LOR paper. I scrolled through the whole paper and found no table like the one of the viewgraph. I searched for all occurrences of the words probability and reliability but found nothing about an overall success probability.
– Uwe
May 2, 2022 at 4:07
• Some information about the probabilities of success is to be found here ia903206.us.archive.org/10/items/NASA_NTRS_Archive_19960014824/… on page 52
– Uwe
May 2, 2022 at 6:10

$$p = 0.96^8 = 0.72$$