50 Mr. Burns on the Double Altitude Problem, 
in the latitude by account ;—that a chronometer has been ge- 
nerally the only means used to determine the interval ;—and 
that the declination must enter the computation before the times 
A.M. and P.M. could be determined? From this we might 
have said at first, Zx uno disce omnes; but we were willing to 
hear all that Mr. R. could say. In his last remarks is given 
a curious explanation of the assertion, ‘* He assumes as known, 
not only the interval of time between the observations, but the 
true apparent time at each observation ;” for it is said, “ I noted 
the assumption in 7¢alics.” Now we are to understand from 
this, henceforth, that “ noting a passage in italics” must clear 
one of the charge of misconception or misconstruction! My 
having changed the order of the words does not make the least 
change in the sense of the passage certainly. In Mr. R.’s last 
paragraph, where he says, “ I failed in giving any solution of 
the problem,” his language is not only inaccurate but uncan- 
did; for only one method had been proposed when his first 
remarks appeared; and hence the phrase “ any solution” is 
inapplicable: and before his last, two other solutions had been 
given. The first of these latter, however, Mr. R. does not 
seem to approve of, though originally proposed by no less an 
astronomer than Lalande; and the second he passes over in 
silence, without a single word, for reasons best known to him- 
self. And to show Mr. R. that our resources are not so con- 
fined as he imagined, we shall now present him with a fourth 
solution, with an example calculated at full length, lest he may 
doubt the truth of the formula itself. It is, perhaps, the most 
convenient of the four, and shorter by nearly half than Dr. 
Brinkley’s correction alone. The four following equations 
very simply and briefly solve the problem. As I have not 
seen the method which Mr. R. has deduced from Mr. Ivory’s 
investigations, I cannot judge whether it is “ the simplest so- 
lution of this useful problem that has yet been given.” 
Let first polar distance = @ 
second, dittqy 2.) . ==20 
first zenith distance = z 
Second ditto Ss" 5.= == = 
VES gc Se re ieee 7 
The rest as in the third method, page 345. Then, 
vers. c = sin? a. vers. m (1) 
sities sin. be Sipe m (2) 
sin. c 
sin? 2 fo sin.5 (x ahora) sin: h(v=—2+c) (3) 
2 sin.¢. SIN. & 
vers. y = sin. a.sin. z. vers. C + vers.(2—2)  — (4) 
Example. 
