# Truncated cone

Truncated cone

In geometry, a truncated cone is the name for a special solid of revolution . A truncated cone is created by cutting a smaller cone from a straight circular cone parallel to the base . This smaller cone is called the complementary cone of the truncated cone.

The larger of the two parallel circular areas is the base area , the smaller the top area . The third of the delimiting surfaces is called the lateral surface . These terms are also common for the areas of these areas. The height of the truncated cone is understood to be the distance between the base and top surface. ${\ displaystyle G}$ ${\ displaystyle D}$ ${\ displaystyle M}$ ${\ displaystyle h}$

The truncated pyramid is closely related to the truncated cone .

## Formulas

With the radius of the top surface, with the radius of the base surface. be the angle between a surface line and the cone axis. ${\ displaystyle r}$${\ displaystyle R}$${\ displaystyle \ varphi}$

Formulas for the truncated cone
volume ${\ displaystyle V = {\ frac {h \ cdot \ pi} {3}} \ cdot (R ^ {2} + R \ cdot r + r ^ {2})}$

Length of a surface line ${\ displaystyle m = {\ sqrt {(Rr) ^ {2} + h ^ {2}}}}$
Outer surface ${\ displaystyle M = (R + r) \ cdot \ pi \ cdot m}$
Top surface ${\ displaystyle D = \ pi \ cdot r ^ {2}}$
Floor space ${\ displaystyle G = \ pi \ cdot R ^ {2}}$
surface ${\ displaystyle O = \ pi \ cdot \ left [r ^ {2} + R ^ {2} + m \ cdot (r + R) \ right]}$
Height of the truncated cone ${\ displaystyle h = {\ frac {Rr} {\ tan \ varphi}}}$

## proofs

### volume

For the calculation of the volume of the truncated cone, the height of the supplementary cone is denoted by. The volume of the truncated cone is then the difference between the volume of the entire circular cone (radius and height ) and the volume of the supplementary cone (radius and height ). Using the ray theorem (four-way theorem) it follows that ${\ displaystyle k}$${\ displaystyle R}$${\ displaystyle h + k}$${\ displaystyle r}$${\ displaystyle k}$

${\ displaystyle {\ frac {h + k} {R}} = {\ frac {k} {r}}}$.

If one calls this quotient , then applies ${\ displaystyle \ lambda}$

${\ displaystyle h + k = \ lambda \ cdot R}$ and
${\ displaystyle k = \ lambda \ cdot r.}$

The height is thus

${\ displaystyle h = \ lambda \ cdot (Rr).}$

The volume of the large cone is

${\ displaystyle V_ {R} = {\ frac {R ^ {2} \ cdot \ pi \ cdot (h + k)} {3}} = \ lambda \ cdot R ^ {3} \ cdot {\ frac {\ pi} {3}},}$

is the volume of the small cone

${\ displaystyle V_ {r} = {\ frac {r ^ {2} \ cdot \ pi \ cdot k} {3}} = \ lambda \ cdot r ^ {3} \ cdot {\ frac {\ pi} {3 }},}$

the volume of the truncated cone is the difference

${\ displaystyle V = V_ {R} -V_ {r} = \ lambda \ left (R ^ {3} -r ^ {3} \ right) {\ frac {\ pi} {3}} = \ lambda (Rr ) \ left (R ^ {2} + R \ cdot r + r ^ {2} \ right) {\ frac {\ pi} {3}} = {\ frac {h \ cdot \ pi} {3}} \ left (R ^ {2} + Rr + r ^ {2} \ right).}$

Alternatively, the volume of a truncated cone can be calculated using an integral, since such a body can be viewed as a body of revolution rotated around the x-axis. The formula for calculating the volume of the body of revolution is: . If you insert for here and calculate the integral within the limits of and , you get the volume of a truncated cone with the corresponding parameters. The following calculation shows that this formula is the same as the above formula: ${\ displaystyle V = \ pi \ cdot \ int \ limits _ {a} ^ {b} f (x) ^ {2} dx}$${\ displaystyle f (x) = {\ frac {Rr} {h}} \ cdot x + r}$${\ displaystyle a = 0}$${\ displaystyle w = h}$

${\ displaystyle V = \ pi \ cdot \ int \ limits _ {0} ^ {h} {\ Big (} {\ frac {Rr} {h}} \ cdot x + r {\ Big)} ^ {2} \ mathrm {d} x = \ pi \ cdot \ int \ limits _ {0} ^ {h} {\ Big (} {\ frac {(Rr) ^ {2}} {h ^ {2}}} \ cdot x ^ {2} +2 \ cdot {\ frac {Rr} {h}} \ cdot x \ cdot r + r ^ {2} {\ Big)} \ mathrm {d} x}$
${\ displaystyle = \ pi \ cdot {\ Big (} {\ frac {(Rr) ^ {2}} {3 \ cdot h ^ {2}}} \ cdot x ^ {3} + {\ frac {Rr} {h}} \ cdot x ^ {2} \ cdot r + r ^ {2} \ cdot x {\ Big |} _ {x = 0} ^ {x = h} {\ Big)} = \ pi \ cdot {\ Big (} {\ frac {(Rr) ^ {2}} {3}} \ cdot h + R \ cdot r \ cdot h {\ Big)}}$
${\ displaystyle = \ pi \ cdot ({\ frac {R ^ {2} + R \ cdot r + r ^ {2}} {3}} \ cdot h) = {\ frac {h \ cdot \ pi} { 3}} \ cdot \ left (R ^ {2} + R \ cdot r + r ^ {2} \ right).}$

### Outer surface

For the calculation of the surface area of ​​the truncated cone, the surface line of the truncated small cone is designated with . According to the ray theorem, ${\ displaystyle n}$

${\ displaystyle {\ frac {R} {r}} \, = \, {\ frac {n + m} {n}}}$,

so

${\ displaystyle n = {\ frac {m \ cdot r} {Rr}}}$.

The surface area is now calculated from the difference between the surface area of the large cone (radius and surface line ) and the surface area of the small cut-away cone (radius and surface line ): ${\ displaystyle M_ {1}}$${\ displaystyle R}$${\ displaystyle m + n}$${\ displaystyle M_ {2}}$${\ displaystyle r}$${\ displaystyle n}$

${\ displaystyle M \, = \, M_ {1} -M_ {2}}$
${\ displaystyle \, = \, \ pi \ cdot R \ cdot (m + n) - \ pi \ cdot r \ cdot n}$
${\ displaystyle \, = \, \ pi \ cdot m \ cdot R + \ pi \ cdot n \ cdot (Rr)}$
${\ displaystyle \, = \, \ pi \ cdot m \ cdot R + \ pi \ cdot {\ frac {m \ cdot r} {Rr}} \ cdot (Rr)}$
${\ displaystyle \, = \, \ pi \ cdot m \ cdot R + \ pi \ cdot m \ cdot r}$
${\ displaystyle \, = \, \ pi \ cdot m \ cdot (R + r)}$

### surface

The surface of the truncated cone is calculated from the sum of the top surface, base surface and lateral surface:

${\ displaystyle D \, = \, \ pi \ cdot r ^ {2}}$
${\ displaystyle G \, = \, \ pi \ cdot R ^ {2}}$
${\ displaystyle M \, = \, \ pi \ cdot m \ cdot (r + R)}$
${\ displaystyle O \, = \, D + G + M}$
${\ displaystyle \, = \, \ pi \ cdot r ^ {2} + \ pi \ cdot R ^ {2} + \ pi \ cdot m \ cdot (r + R)}$
${\ displaystyle \, = \, \ pi \ cdot \ left [r ^ {2} + R ^ {2} + m \ cdot (r + R) \ right]}$

## Application examples

### Drinking glass

A martini glass is approximately the shape of a cone . The unfilled part has the shape of a truncated cone.

Some drinking glasses , for example a martini glass , have approximately the shape of a cone.

A martini glass with a diameter of 103 millimeters and a filling height of 59 millimeters is filled with orange juice to a height of 40 millimeters . As a result , , and from this the volume of the part not filled, which has the shape of a truncated cone: ${\ displaystyle R = 51 {,} 5 \ \ mathrm {mm}}$${\ displaystyle r = {\ frac {40 \ \ mathrm {mm}} {59 \ \ mathrm {mm}}} \ cdot {51 {,} 5 \ \ mathrm {mm}} \ approx 34 {,} 9 \ \ mathrm {mm}}$${\ displaystyle h = 59 \ \ mathrm {mm} -40 \ \ mathrm {mm} = 19 \ \ mathrm {mm}}$

${\ displaystyle V = {\ frac {1} {3}} \ cdot \ pi \ cdot h \ cdot (R ^ {2} + R \ cdot r + r ^ {2}) = {\ frac {1} { 3}} \ cdot \ pi \ cdot 19 \ \ mathrm {mm} \ cdot ((51 {,} 5 \ \ mathrm {mm}) ^ {2} +51 {,} 5 \ \ mathrm {mm} \ cdot 34 {,} 9 \ \ mathrm {mm} + (34 {,} 9 \ \ mathrm {mm}) ^ {2}) \ approx 113 \ cdot 10 ^ {3} \ \ mathrm {mm ^ {3}} = 113 \ \ mathrm {cm ^ {3}} = 113 \ \ mathrm {ml}}$

The unfilled part has a volume of about 113 milliliters .

The proportion of martini glass that is filled is

${\ displaystyle \ left ({\ frac {40 \ \ mathrm {mm}} {59 \ \ mathrm {mm}}} \ right) ^ {3} \ approx 0 {,} 312}$

The martini glass is about 31.2 percent filled with orange juice .