# How To Grind Glasses

##### 7 minutes • 1455 words

After choosing the glass or crystal that you intend to use, find the proportion, which serves as a measure for its refractions.

This can be conveniently found with the help of such an instrument.

`EFI`

is a completely flat and straight board or ruler made of a not too shiny nor transparent material. The light striking it should be easily distinguished from the shadow.

`EA`

and `FL`

are 2 small blades that are not transparent. These are raised perpendicular to `EFI`

.

It has 2 small round holes, `A`

and `L`

, placed exactly opposite each other.

The ray `AL`

passes through is parallel to the line `EF`

.

Then `RPQ`

is a piece of the glass you wish to test. Cut in into a triangle, of which:

- the angle
`RQP`

is right `PRQ`

is more acute than`RPQ`

.

The 3 sides `RQ, QP`

, and `RP`

are 3 completely flat and polished faces.

The face `QP`

is pressed against the board `EFI`

, and the other face `QR`

against the blade `FL`

.

The sun’s ray passes through the two holes `A`

and `L`

penetrating as far as `B`

through the glass `PQR`

without suffering any refraction. This is because it meets its surface `RQ`

perpendicularly.

But having reached the point `B`

where it meets its other surface `RP`

obliquely, it cannot emerge without bending towards some point on the board `EF`

, as for example towards `I`

.

This instrument makes the sun’s ray pass through these holes `A`

and `L`

, in order to know the ratio that the point `I`

(the center of the small oval of light that this ray describes on the board `EFI`

) has with the 2 other points `B`

and `P`

.

`B`

is the one where the straight line passing through the centers of the two holes`A`

and`L`

terminates on the surface RP;`P`

is the one where this surface`RP`

and that of the board`EFI`

are cut, by the plane that is imagined to pass through the points B and I, and also through the centers of the two holes A and L.

Transfer the triangle `BPI`

with a compass onto paper.

Then from the center B, draw the circle `NPT`

through the point `P`

. Having taken the arc `NP`

equal to `PT`

, draw the straight line `BN`

which cuts `IP`

extended at the point `H`

.

Then again from the center B through H, draw the circle `HO`

which cuts `BI`

at the point `O`

.

You now have the proportion between the lines `HI`

and `OI`

as the common measure of all the refractions that can be caused by the difference between the air and the glass you are examining.

If after that, on the straight line `HI`

, you take `MI`

equal to `OI`

, and `HD`

equal to `DM`

, you will have `D`

for the summit, and `H`

and `I`

for the burning points of the hyperbola that this glass must have the figure of to serve for the lenses I have described.

You can make these 3 points `HDI`

more or less distant than they are, simply by drawing another straight line parallel to `HI`

farther or closer to the point `B`

. Drawing 3 straight lines `BH`

, `BD`

, `BI`

from this point B that cut it.

There is the same ratio between the three points HDI and hdi, as between the three `HDI`

.

Having these 3 points, you can trace the hyperbola by planting 2 pegs at the points `H`

and `I`

, and making the cord placed around the peg `H`

be so attached to the ruler that it cannot fold back towards `I`

, farther than up to `D`

.

If you prefer to trace it with the ordinary compass by finding several points through which it passes; place one of the points of this compass at point H;

Having opened it so that its other point passes a little beyond point D, as far as I, from the center H describe the circle I33; then having made M2 equal to HI, from the center I through point 2, describe the circle 233, which cuts the previous one at the points 33, through which this hyperbola must pass, as well as through the point D, which is its summit. Then put the point of the compass back at the point H, and opening it so that its other point passes a little beyond the point I, as far as 4, from the center H describe the circle 466. Then having taken M5 equal to H4, from the center I through 5 describe the circle 566, which cuts the previous one at the points 66 which are in the hyperbola. And thus continuing to place the point of the compass at the point H, and the rest as before, you can find as many points as you please of this hyperbola.

This may not be bad for roughly making some model that roughly represents the shape of the lenses to be cut. But to give them this figure exactly, it is necessary to have some other invention by means of which one can describe hyperbolas in one stroke, as one describes circles with a compass. And I know of no better one than the following. First, from the center T, which is the middle of the line HI, it is necessary to describe the circle HVI, then from the point D to raise a perpendicular on HI, which cuts this circle at the point V. And drawing a straight line from T through this point V, one will have the angle HTV, which is such that if one imagines it rotating around the axis HT, the line TV will describe the surface of a Cone, in which the section made by the plane VX parallel to this axis HT, and on which DV falls at right angles, will be a hyperbola exactly similar and equal to the previous one. And all the other planes parallel to this one will also cut in this Cone hyperbolas all similar, but unequal, and which will have their foci more or less distant according as these planes are distant from this axis.

After which one can make such a machine. AB is a lathe or wooden or metal roller, which turning on the poles I, 2 represents the axis HI of the other figure. CG, EF are two flat and smooth plates or boards, especially on the side where they touch each other, so that the surface that can be imagined between the two, being parallel to the roller AB, and cut at right angles by the plane that can be imagined passing through the points I, 2, and C, O, G, represents the plane VX which cuts the Cone. And NP the width of the upper CG is equal to the diameter of the glass to be cut, or a little larger. Finally, KLM is a rule which, turning with the roller AB on the poles I, 2, so that the angle ALM always remains equal to HTV, represents the line TV which describes the Cone. And one must think that this rule is so passed through this roller, that it can rise and fall by sliding in the hole L, which is exactly of its thickness; and even that there is somewhere, as towards K, a weight or spring, which always presses it against the plate CG, by which it is supported, and prevented from going further. And moreover that its end M is a well-tempered steel point, which has the strength to cut this plate CG, but not the other EF which is underneath. From which it is manifest that if one moves this rule KLM on the poles I, 2, so that the steel point M passes from N through O towards P, and reciprocally from P through O towards N, it will divide this plate CG into two others, CNOP, and GNOP, of which the side NOP will be terminated by a sharp, convex line in CNOP, and concave in GNOP, which will have exactly the figure of a hyperbola. And these two plates, CNOP, GNOP, being of steel or other very hard material, may serve not only as models, but perhaps also as tools or instruments for cutting certain wheels, from which I will soon say that the glasses must derive their figures. However, there is still a defect here in that the steel point M, being a little differently turned when it is towards N, or towards P, than when it is towards O, the wire or the cutting edge that it gives to these tools cannot be everywhere equal. Which makes me think it will be better to use the following machine, although it is a little more complicated.