# How tall is Otto Thorwarth

## Internal combustion engine (3): displacement, compression ratio04.01.2010, 12:55

When calculating the combustion chamber, the displacement, the stroke-to-bore ratio and the compression ratio are key terms. They are explained and supported by calculation examples.

### Calculations on the engine

**1. Displacement**

Image: The cylinder space consists of the displacement V_{H} and the compression space V_{c}. The compression ratio ε (epsilon) indicates how many times larger the cylinder space is than the compression space. For example, if ε = 7: 1, then the cylinder space is seven times larger than the compression space (see below).

When the piston moves back and forth, its direction of movement changes at the upper and lower turning point. These reversal points are the top and bottom dead center (TDC and BDC).

Because the piston area is a circular area, the displacement is a cylindrical volume with the calculation parameters: cylinder bore meter d in cm, piston stroke s in cm. This results in the calculation formula:

Cylinder displacement = cylinder cross-section times stroke or

Fo 1:** V. _{H} = A • s** (cm

^{3})

If the cross section A = d

^{2}

**•**If π / 4 is inserted into the formula, then the cylinder displacement is:

Fo 2:

**V.**

_{H}= d^{2}**• π / 4**

**• s**(cm

^{3})

The units of measure for cubic capacity are: cm

^{3}or dm

^{3}(1 dm

^{3 }= 1l (liter)).

Note: The bore diameter d and the stroke s are given in mm; Before inserting them into the calculation formula, they must be converted into the desired unit of measurement, cm or dm.

Both the stroke s and the bore diameter d can be calculated from the cylinder displacement. To do this, the displacement formula must be converted according to the size you are looking for:

Fo 3 and Fo 4:

In the case of multi-cylinder engines, the total displacement of engine V results_{H }from the cylinder displacement V_{H} multiplied by the number of cylinders z, i.e. (Fo 5):

**2. Stroke-to-bore ratio**

In practice, there are three different options for designing the combustion chamber: The specified displacement of an engine can either be achieved with a large cylinder bore and a small stroke (short stroke); if the stroke and bore are the same size, an speaks of a square hub. Finally, a combination of small bore diameters and large strokes (long stroke) is possible. Each of the three different types has its advantages and disadvantages. The ratio between the motor bore and the stroke is referred to as the stroke-to-bore ratio α.

Fo 6:**α = s: d**

Like all ratios, the stroke-to-bore ratio is dimensionless.**3. Compression ratio**

The cylinder space consists of the displacement V._{H} and the compression space V_{c}. The sucked-in air (diesel engine) or the sucked-in fuel-air mixture (gasoline engine) is compressed to the volume of the compression chamber in the compression stroke when the piston has reached top dead center.

**Compression ratio ε = (V _{H}+ V_{c}): V_{c}**

or Fo 7:

**ε = (V**

_{H}: V_{c}) + 1In it are:

**ε**= Compression ratio

V.

_{H}= Cylinder displacement (cm

^{3})

V.

_{c}= Compression space (cm

^{3})

1. What is the cylinder displacement V_{H} (in cm^{3}) of a tractor engine with a bore diameter of d = 114 mm with a stroke of s = 122 mm?

2. The cylinder displacement of a 4-cylinder V_{H} = 1032 cm^{3}. How big is the bore diameter with a stroke of 72 mm?

3. The cylinders of a six-cylinder engine have a bore diameter of 106 mm and a stroke of s = 125 mm. How big is the total displacement in liters?

4. What is the compression ratio of a gasoline engine with a cylinder displacement of V?_{H} = 470 cm^{3} and a compression space of V_{c}= 58 cm^{3}?

5. A cylinder with a bore diameter of 120 mm has a stroke-to-bore ratio of 1.1333. How big is the hub?

6. A gasoline engine has a V_{H} = 230 cm^{3} and an ε = 7.5: 1. How big is the conurbation?

### solutions

1st V._{H} = A • s = d^{2} • π / 4 • s = (11.4 cm)^{2} • π / 4 • 12.2 cm =__V. _{H} = 1245.26 cm__

^{3}

2nd V._{H} = V_{H} : 4 = 1032 cm^{3}: 4 = 258 cm^{3}

V._{H} = d^{2} • π / 4 • s -> d see formula Fo 3 -> __d = 67.55 mm__

3. Calculate in dm to get liters:

V._{H} = d^{2} • π / 4 • s • z = (1.06 dm)^{2} • π / 4 • 1.25 dm • 6 =__V. _{H} = 6,619__ l

4.ε = (V_{H}: V_{c}) + 1 = (470 cm^{3}: 58 cm^{3}) + 1 = 9,1 = __9,1 : 1__

5. α = s: d -> s = d • α = 120 mm • 1.133 = __136 mm__

6.ε = (V_{H}: V_{c}) + 1 –>

V._{c} = Vh: (ε - 1) = 230 cm^{3} : 6,5__V. _{c}__

__= 35.38 cm__

^{3}

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