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The introduction to the article contains the following quote: "the bi-elliptic transfer may require a lower amount of total delta-v than a Hohmann transfer in situations where the ratio of final to the initial semi-major axis is greater than 15.58." This implies that the Hohmann transfer requires a lower delta-v than the bi-elliptic transfer at a final to initial semi-major axis ratio of lower than 15.58. This is shown to be incorrect later in the article. The example given at the end uses a ratio of 14:1 (93800:6700) and shows that the bi-elliptic transfer requires a lower delta-v. Either the 15.58 is wrong (or misleading) or the example is wrong, and I don't know which. Also worth noting, the Hohmann transfer article indicates that the bi-elliptic transfer may require less delta-v for ratios exceeding 12. One of the two articles, or both, probably need to be edited to correct this error/discrepancy. 137.240.136.81 22:04, 11 September 2007 (UTC)
Time has passed, and the article in its current form is now correct, but after reading this discussion I wanted to clarify the confusion. The graph from Vallado page 318 is misleading, but it shows that when the ratio R is greater than 11.94, a bi-elliptic transfer with sufficient R* is always theoretically cheaper than a Hohmann transfer. The R*=infinity line on the graph is always below the Hohmann line for R > 11.94 - they converge as R tends toward infinity, but never cross again. For practical reasons, though, it's important to consider finite values for R*, and several are also drawn on the graph. It shows that R* must be at least 15.58, and for any chosen R* value greater than 15.58, the bi-elliptic transfer will be more efficient so long as R* > R > X(R*), for some X in the range (11.94, 15.58) whose value depends on R*. If R* < R then it's not a true bi-elliptic transfer, because the intermediate radius was lower than the target radius. If R < X(R*) but still R > 11.94, then your R* is not high enough to be more efficient than Hohmann.
However, the graph is misleading because in practice you would not choose R* first - you would almost always already know R. In that case, inverting the inequality, bi-elliptic transfer is more efficient so long as R > 11.94, provided you choose a high enough R* (so that X(R*) < R). For R = 15.58, any R* > R is more efficient than a Hohmann transfer, but larger R* is always even more efficient. For 11.94 < R < 15.58, R* needs to be bigger before you break even with Hohmann - much more so at the lower end of the range, as Xinv(R) tends to infinity as R approaches 11.94. The graph of Xinv(R) is more relevant than the graph in Vallado. 217.154.84.20 ( talk) 14:09, 2 January 2013 (UTC)
The intial diagram doesn't seem to make much intuitive sense. It actually looks as though more delta v would be needed to make such a small change to a circular orbit using the bi-elliptic transfer illustrated, compared to the Hohmann transfer. Sketching out the later example to scale looks much more sensible, (if perhaps a little unwieldy). Nonetheless, would it be worth making the diagram match the later example's figures? —Preceding unsigned comment added by 84.69.96.243 ( talk) 01:42, 23 February 2008 (UTC)
I removed a "reference needed" tag since the results can be easily obtained by simple algebraic manipulation of the vis viva equation included in the article. Dauto ( talk) 16:34, 9 January 2013 (UTC)
The delta-v expression for the third impulse, as it is right now, produces a negative value. While this may be correct in the sense that the third burn is a reduction in speed (accomplished with a retrograde burn), it might be misleading if one wants to compute the total delta-v expended during the transfer, for in that case one must remember the switch the sign of the third delta-v before adding it to the first two. Do you think it would be clearer if we give the *positive* value of the expression instead, and specify explicitly that the burn is to be done retrograde? Meithan ( talk) 17:49, 20 August 2013 (UTC)
I think listing negative Δv's in the table of the example may be confusing, as a reader might think that the negative values are to be subtracted from the others in order to obtain the total. I would suggest listing all values as positive, and letting the color alone indicate the direction of the burns (i.e. whether prograde or retrograde). – Meithan ( talk) 05:37, 13 February 2018 (UTC)
I've removed the negative signs for all delta-v's, as the recent edit by 89.107.5.192 makes it clear when they're applied prograde/retrograde. I liked his/her solution! – Meithan ( talk) 00:38, 16 February 2018 (UTC)
I would like to see a derivation of those numbers in the article, and what the exact (in terms of radicals / exponents / fundamental constants etc) values are. Grassynoel ( talk) 12:16, 3 May 2019 (UTC)
![]() | This ![]() It is of interest to the following WikiProjects: | ||||||||||||||||||||
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The introduction to the article contains the following quote: "the bi-elliptic transfer may require a lower amount of total delta-v than a Hohmann transfer in situations where the ratio of final to the initial semi-major axis is greater than 15.58." This implies that the Hohmann transfer requires a lower delta-v than the bi-elliptic transfer at a final to initial semi-major axis ratio of lower than 15.58. This is shown to be incorrect later in the article. The example given at the end uses a ratio of 14:1 (93800:6700) and shows that the bi-elliptic transfer requires a lower delta-v. Either the 15.58 is wrong (or misleading) or the example is wrong, and I don't know which. Also worth noting, the Hohmann transfer article indicates that the bi-elliptic transfer may require less delta-v for ratios exceeding 12. One of the two articles, or both, probably need to be edited to correct this error/discrepancy. 137.240.136.81 22:04, 11 September 2007 (UTC)
Time has passed, and the article in its current form is now correct, but after reading this discussion I wanted to clarify the confusion. The graph from Vallado page 318 is misleading, but it shows that when the ratio R is greater than 11.94, a bi-elliptic transfer with sufficient R* is always theoretically cheaper than a Hohmann transfer. The R*=infinity line on the graph is always below the Hohmann line for R > 11.94 - they converge as R tends toward infinity, but never cross again. For practical reasons, though, it's important to consider finite values for R*, and several are also drawn on the graph. It shows that R* must be at least 15.58, and for any chosen R* value greater than 15.58, the bi-elliptic transfer will be more efficient so long as R* > R > X(R*), for some X in the range (11.94, 15.58) whose value depends on R*. If R* < R then it's not a true bi-elliptic transfer, because the intermediate radius was lower than the target radius. If R < X(R*) but still R > 11.94, then your R* is not high enough to be more efficient than Hohmann.
However, the graph is misleading because in practice you would not choose R* first - you would almost always already know R. In that case, inverting the inequality, bi-elliptic transfer is more efficient so long as R > 11.94, provided you choose a high enough R* (so that X(R*) < R). For R = 15.58, any R* > R is more efficient than a Hohmann transfer, but larger R* is always even more efficient. For 11.94 < R < 15.58, R* needs to be bigger before you break even with Hohmann - much more so at the lower end of the range, as Xinv(R) tends to infinity as R approaches 11.94. The graph of Xinv(R) is more relevant than the graph in Vallado. 217.154.84.20 ( talk) 14:09, 2 January 2013 (UTC)
The intial diagram doesn't seem to make much intuitive sense. It actually looks as though more delta v would be needed to make such a small change to a circular orbit using the bi-elliptic transfer illustrated, compared to the Hohmann transfer. Sketching out the later example to scale looks much more sensible, (if perhaps a little unwieldy). Nonetheless, would it be worth making the diagram match the later example's figures? —Preceding unsigned comment added by 84.69.96.243 ( talk) 01:42, 23 February 2008 (UTC)
I removed a "reference needed" tag since the results can be easily obtained by simple algebraic manipulation of the vis viva equation included in the article. Dauto ( talk) 16:34, 9 January 2013 (UTC)
The delta-v expression for the third impulse, as it is right now, produces a negative value. While this may be correct in the sense that the third burn is a reduction in speed (accomplished with a retrograde burn), it might be misleading if one wants to compute the total delta-v expended during the transfer, for in that case one must remember the switch the sign of the third delta-v before adding it to the first two. Do you think it would be clearer if we give the *positive* value of the expression instead, and specify explicitly that the burn is to be done retrograde? Meithan ( talk) 17:49, 20 August 2013 (UTC)
I think listing negative Δv's in the table of the example may be confusing, as a reader might think that the negative values are to be subtracted from the others in order to obtain the total. I would suggest listing all values as positive, and letting the color alone indicate the direction of the burns (i.e. whether prograde or retrograde). – Meithan ( talk) 05:37, 13 February 2018 (UTC)
I've removed the negative signs for all delta-v's, as the recent edit by 89.107.5.192 makes it clear when they're applied prograde/retrograde. I liked his/her solution! – Meithan ( talk) 00:38, 16 February 2018 (UTC)
I would like to see a derivation of those numbers in the article, and what the exact (in terms of radicals / exponents / fundamental constants etc) values are. Grassynoel ( talk) 12:16, 3 May 2019 (UTC)