1796. | 
_ N.B. Thus is the folution of the queftion 
determined from the theory. But thofe gentle 
men who are in the habit of making experi« 
ments of this kind, will very probably have 
noticed, that when the image of an object is 
obferved upon water, the image generally (per- 
haps, always) appears longer than the object 
itfelf. And here a query naturally arifes upon 
this point; namely, is the theory of catoptrics 
erroneous when applied to cafes of this kind; 
or, is the difference between the fize of the 
objeé& and its image, a mere mental delufion, 
occafioned by a peculiar deception of vifion ? 
I am ijiclined to think that the latter is the 
cafe; but fhould thefe remarks prove an in- 
ducement for any of your correfpondents to 
confider the fubjeét more attentively than it has 
been hitherto, and fhould their reafonings lead 
to a more fatis aétory method of explaining the 
appearance, than can be deduced from confider- 
ing it as a © deceptio wifus,” I fhall be very 
happy to fee a farther elucidation ina future 
Number. 
This quefiion was alfo anfwered by Mr. Gide 
and by Mr. I. Hiceman. 
=a 
Question VIII (No. II]).—Anfwered by 
Mr, Fe Fr, 
Put x =the diameter of the bottom in feet, 
a= 735393, 6=<=147°262125, and c==62°'5, the 
weight of a cubic foot of water in lbs. avoirdu- 
pois, ==1000 avoirdupois oz. 
Then ax?=<the area of the bottom, and 4 ax? 
==the internal area of the fides. Nowe:1::6: 
b é : 
—==the number of the cubic feet of water whofe 
€ 
abfolute weight is==5. But (ax?xX+v==) ax?== 
the number of cubic feet of water preffing on 
the bottom, and (40x? x fv==) 2ex3==the num- 
ber of cubic feet whofe weight is equal to the 
b 
prefflure of the fides, Therefore 3ax3==—- 
c 
| b 
Whence x==34/——==1 foot—the diameter of 
ac 
the bottom and depth of the veffel. 
The fame anfwered by Mr. Wm, Hilten, 
Tt is very evident, from the principles of 
hydroftatics, that the preffure upon the cylin- 
der’s bafe is equal to the whole weight of the 
fluid ; and fince the preffure upon the upright 
furface at any depth is as that depth, it alo 
appears evident, that the whole. preffure upon 
that furface is the fame as it would be upon 
an equal furface immerfed at half the depth of 
the fluid. This premifed, put the veffel’s 
depth and diameter =v feet; -7354==2 Then 
by menfuration, ax*==the veffel’s folidity, or 
preffure upon the bafe; and 4ax3=2ax3=Sthe 
folidity of a prifm whofe bafe is the upright 
furface and height; half the height of the fame== 
_the preffure upon that furface, and both toge- 
ther—=3ax3, It appears by experiment, that 
t foot of water weighs 62!lb. avoirdupois ; 
theteforey we have 62! 3x°==147°26212 ¥; 
Mathematical Correfpondence. 
- common), 
‘to the triangle A BC, being on equal bafes and 
395 
from which equation the value of x is eafily 
found to be 1 foot, as required. 
This queftior was alfoanfwwered by Mr, F. Hart- 
ley and Mr. T. Hickman. 
—— 
Question IX (No. [11) —Anfwered hy Mr. 
T. Hickman. 
It has often been proved, that the greateft 
cylinder that can be in{cribed in a given {phere, 
has its height 4/1, and its diameter 4/2, of the 
fphere’s diameter; and that its folidity is 4/4 
of the fphere’s folidity: alfo, that the greateft 
cone infcribed in the fphere, has its height 2, 
and its bafe diameter 2/2, of the fphere’s dia- 
meter; and that its folidity is 2 of the fphere’s 
folidity. Now the difference of thefe folidities, 
Sem or ov . 
v 
fphere’s content; and 123% +52 36==904°9808, 
is the folidity of the fphere; therefore 904-7808 
X 12810542==1254°2:92414 is the difference bee 
tween the greateft infcribed cone and cylinder, 
as required, 
This queftion was alfo anfwered by Mr. F. 
F———1, Mr. }. Hartiey, and Mr. Wms 
Hilton. 

=='2310542 of the 
== aes 
Question X (No. IIl).—Anfwered by Mr F. 
F. ‘ 
memes 
Let ABC 
be the given 
right - angled E 
triangle, and 
the other lines 
drawn as in 
the queftion ; 
alfo produce B 
C to M, mak- 
ing CM=BC, 
and join AM. 
There the tri- D Me 
angle AGH is equal to the triangle ABC, be- 
caufe AG=AB, and AH=AC, and the in- 
cluded angle G A H=the angle B A C.—Again, 
the triangle KCE is equal to the triangle 
ACM, becaufes KC=AC, and CE=CM, 
and the angle K CE=the angle ACM (ACK 
and MCE being right angles, and KCM 
But the triangle A C M is equal ta 

etween the fame parallels. Therefore the 
triangle KCE is alfo equal to the triangle 
ABC.—And the fame may in like manner be 
proved of the triangle BF D. Therefore, &c. 
The fame otherwife proved by Mr. Wm. Hilton 
Let ABC be any triangle, right-angled at 
A; alfo, AF, AK, CD {quares upon the three 
fides ; the propofition afferts, that if the points 
GH, FD, KE be joined, the three triangles 
AGH, BFD, CKE are each of them equal to 
the triangle ABC: which propofition may ke 
thus demonftrated :—It is fhown by writers on 
menf{uration, that the area of a triangle is equal 
to half the rectangle of any two fides drawn 
into the fine of their included angle. Hence 
then the area of ABC==A Bx AC Hine of 
BAC=BAxXBEC x! fine of ABCSCAX 
-g Ea CB 
