Frappé effect

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Frappé effect: image of a truncated cone-shaped, three-dimensional glass with markings of the supposed and actual half fill level

As frappe effect , the optical illusion of the actual volume distribution in conical referred vessels. When viewing a conical drinking glass from the side, test subjects mark the dividing line between the upper and lower half of the volume, usually well below their actual position, because the three-dimensional decrease in volume is not sufficiently taken into account.

The term frappé effect comes from the conical glasses in which frappés are often served. For Martinis and other cocktails used cocktail glasses often have a shape that a circular cone is even closer than the glasses of coffee drinks.

calculation

General formulas

• Volume of any circular cone :

${\displaystyle V={\frac {r^{2}\pi h}{3}}}$

• Volume of a truncated cone

${\displaystyle V={\frac {\pi \cdot h}{3}}\cdot (R^{2}+R\cdot r+r^{2})}$

Mathematical derivation

In the following, the calculations of the heights of the surfaces that divide a conical vessel into two, three or any number of equal parts are presented. An upside-down, straight circular cone (like a beverage glass) is always assumed.

Height of the area bisecting the volume

Illustration of a conical body: The volume of volume A and B is the same.

To calculate the height of the volume-halving area in a conical vessel, the two parts are calculated separately. The volume of the lower part (straight circular cone; denoted as in the figure ) corresponds to ${\displaystyle x}$ ${\displaystyle A}$

${\displaystyle (I)V_{\text{A}}={\frac {z^{2}\pi x}{3}}}$

The height is just ; the radius is set as . The volume of the upper part (truncated cone; denoted as in the figure ) corresponds to ${\displaystyle x}$ ${\displaystyle z}$ ${\displaystyle B}$

${\displaystyle (II)V_{\text{B}}={\frac {\pi (h-x)}{3}}(r^{2}+rz+z^{2})}$

The radius of the smaller parallel circular area is again used as a variable; the height corresponds . One of the two equations is sufficient to calculate the volume-halving height . Easier to calculate is the part why the equation is used. The calculation of several volume-dividing areas can also only be done with an equation . ${\displaystyle z}$ ${\displaystyle h-x}$ ${\displaystyle A}$ ${\displaystyle (I)}$ ${\displaystyle (I)}$

The radius corresponds to the angle functions in a right-angled triangle ( tangent function ; is the adjacent side and the opposite side) ${\displaystyle z}$ ${\displaystyle x}$ ${\displaystyle z}$

${\displaystyle (III)z={\frac {x}{\tan(\displaystyle 90^{\circ }-{\frac {\varphi }{2}})}}}$

or more simply according to the 1st ray law :

${\displaystyle (IV)z=x\cdot {\frac {r}{h}}}$

Since the radius is usually easier to measure, the equation is useful . Inserted in results ${\displaystyle (IV)}$ ${\displaystyle (I)}$

${\displaystyle V_{\text{A}}={\frac {(x\cdot {\frac {r}{h}})^{2}\pi x}{3}}}$

In order to determine the height of the area that halves the volume, substitute for half of the total volume, i.e. (according to the general volume equation for circular cones) . Thus it results ${\displaystyle x}$ ${\displaystyle V_{\text{A}}}$ ${\displaystyle \textstyle ({\frac {r^{2}\pi h}{6}})}$

${\displaystyle {\frac {r^{2}\pi h}{6}}={\frac {(x\cdot {\frac {r}{h}})^{2}\pi x}{3}}}$

${\displaystyle r^{2}}$and can be eliminated. The following is available as a simple formula: ${\displaystyle \pi }$

${\displaystyle x={\sqrt[{3}]{\frac {h^{3}}{2}}}}$ or ${\displaystyle x\approx {\frac {h}{1.26}}}$

From this it can be seen that the angle respectively. the radius for the height of the surfaces that divide the volume of a conical vessel into two, three or parts does not play a role (see illustration). ${\displaystyle \varphi }$ ${\displaystyle r}$ ${\displaystyle n}$

Image of two conical cocktail glasses with the same height but different radius

Height of the top surfaces of any number of parts

This principle can also be applied to the calculation of a third (or part) of the filling quantity. For this purpose, the general volume formula for straight circular cones is divided by (instead of by 2 as above) (or it is multiplied by). This formula is then inserted into the equation in the same way as above . Hence one obtains ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle \textstyle {\frac {1}{n}}}$ ${\displaystyle (I)}$

${\displaystyle {\frac {r^{2}\pi h}{3n}}={\frac {(x\cdot {\frac {r}{h}})^{2}\pi x}{3}}}$

The same procedure can then be used to shorten and the result is a simple general formula:

${\displaystyle x={\sqrt[{3}]{\frac {h^{3}}{n}}}}$ or ${\displaystyle x={\frac {h}{\sqrt[{3}]{n}}}}$

The equation can now be used to assign the other surfaces (without the top top surface) to their heights . However, it is easier to use the equation to get the height between the floor and the surface instead of just the height of the space between two surfaces. Thus, the total volume is multiplied by for the next area . You finally get ${\displaystyle n-2}$ ${\displaystyle (II)}$ ${\displaystyle (I)}$ ${\displaystyle \textstyle {\frac {2}{n}}}$

${\displaystyle x={\sqrt[{3}]{\frac {2h^{3}}{n}}}}$

This process is now repeated until all heights are known. The last height (apart from the area on the edge of the glass) has the value accordingly

${\displaystyle x={\sqrt[{3}]{\frac {(n-1)h^{3}}{n}}}}$