(1) | n_{1} * rot_{1} = n_{2} * rot_{2}
* (-1) or n_{1} * rot_{1} + n_{2}
* rot_{2} = 0 |

For the first exercise we had held the shafts fixed. Now mentally fix wheel 1 to the table and turn wheel 2 by moving its shaft on a circular path round wheel 1. Does formula (1) still hold true? This depends on the "point of view" (POV): If you define rot as rotation around the shaft measured against a fixed line somewhere on the table, you are stuck as rot

(2) | Formula (1) is valid only if rot is measured relative to the POC. |

(3) | rot = ^{w}rot + ^{d}rot |

__Make sure the three rules given so far are understood. If
insecure grab
some gear wheels and play with them! Or try your skills with example 1
further
down.__

With this configuration we have three basic modes of operation:

**Reversal**: Block wheel 4. If you turn wheel 1, wheel 3 turns on its (now fixed) shaft and drives wheel 2 in the opposite direction.**Reduction**: Block wheel 2. If you turn wheel 1, wheel 3 is turned but has to roll on the blocked wheel 2 as well. Wheel 3's only possibility is to take its shaft along, by this turning wheel 4. Works the other way round too.**Differential**: Wheels 1 and 2 turn in opposite directions. If the speed is identical, wheel 3 will spin but wheel 4 stays put. As soon as the speeds differ, the difference (hence the name) forces wheel 4 to turn. Of course this can be reversed also: If you turn wheels 1 and 4, wheel 2 will spin at a combined speed.

In the following we will prove all this from the three basic rules
given in the beginning and best of all, derive some simple equations to
calculate the behavior from gear sizes only.

In the context considered here (the POV principle, you remember ?) the
horizontal shafts are assumed to be fixed to some structure, i.e. they
can not move round when the wheels are turned. This can be put down
formally
as

(a) ^{d}rot_{4}
= 0

For the reversal mode, wheel 4 is blocked, so wheel 3's shaft is unable
to move as well. We state (note the upper case letter in the bracket)

(A) ^{d}rot_{3}
= 0

The rest is quite simple. Bevel gears very much behave like spur gears:

(b) n_{1} * ^{w}rot_{1}
+ n_{3} * ^{w}rot_{3} = 0

(c) n_{2} * ^{w}rot_{2}
+ n_{3} * ^{w}rot_{3} = 0

But wait - what about the angle between the shafts ? As long as
your POV is relative to each single wheel, that doesn't matter. We are
allowed to calculate

(d) n_{1} * ^{w}rot_{1}
- n_{2} * ^{w}rot_{2} = 0

if we define the direction of rotation by looking down on the wheel,
i.e. from the very right for wheel 1 but from left for wheel 2. If we
normalize our POV to the direction of wheel 1, we have to look at wheel
2 from the other face, which exactly reverses its direction ! So we
finally get:

(B) n_{1} * ^{w}rot_{1}
+ n_{2} * ^{w}rot_{2} = 0

which is nothing else than the formal definition of **Reversal**
!

Now block wheel 2 and do the string trick introduced for rule (2):
Tie the string round the shaft of wheel 2 and the horizontal part of
the kinked shaft of wheel 3. The observation will show (remember to
look at all wheels from the same POV, e.g. far right) that

(e) ^{w}rot_{4}
= ^{d}rot_{2}

(C) n_{3} * ^{d}rot_{3}
+ n_{2} * ^{d}rot_{2} = 0

Same experiment with wheel 1:

(f) ^{w}rot_{4}
= ^{d}rot_{1}

At this point we boldly add up all equations we like (n_{1}*f +n_{2}*e+B+(n_{1}+n_{2})*a):

(g) n_{1} * (^{d}rot_{1}
+ ^{w}rot_{1}) + n_{2} * (^{d}rot_{2}
+ ^{w}rot_{2}) = (n_{1} + n_{2}) * (^{d}rot_{4}
+ ^{w}rot_{4})

Formally applying the superposition rule (3) reveals the universal
equation for differentials:

(4) | n_{1}
* rot_{1} + n_{2} * rot_{2} - (n_{1}
+ n_{2}) * rot_{4} = 0 |

(D) rot

**Reduction** now is a simple to understand: Set rot_{2}
to 0 and (D) becomes

(E) rot_{4} = rot_{1}/2

**Differential** is no harder: Set rot_{2} =
-rot_{1} + delta and you get

(F) rot_{4} = delta/2

Some more observations:

- (A) is the logical consequence of (e):
^{w}rot_{4}= 0 =>^{d}rot_{2}= 0 =>^{d}rot_{3}= 0 - The size of wheel 3 does not influence the behavior of the differential. It can therefore be chosen as space (or torque) dictate.
- Be careful with bevel gears: They may confuse your sense of direction !

(h) n

Merging this with equation (D) from above, we get

(i) rot

For convenience the faktors preceding rot

(G) rot

In the plans on this site we will depict the packaged differential with the green symbol to the left.

(H) rot

The red symbol to the left shall notify this mirrored or inverse differential.

Using rule (1) from above - all shafts are fixed to the table -, we note:

nEliminating n_{1}*^{w}rot_{1}+ n_{2}*^{w}rot_{2}= 0

n_{2}*^{w}rot_{2}+ n_{3}*^{w}rot_{3}= 0

nor in other words, wheels 1 and 3 rotate in the same direction._{1}*^{w}rot_{1}- n_{3}*^{w}rot_{3}= 0

Another useful point to remember is that n2 (the size of wheel 2) is completely irrelevant. It can be chosen freely to best fit size and torque constraints.

Wheels 3 and 6 are both fixed to a common shaft.

(a) rotWheel 1 is the power input and meshes with transfer wheel 2, which runs freely on the axle._{3}= rot_{6}

(b) nBoth wheels' shafts are attached to the vehicle._{1}*^{w}rot_{1}+ n_{2}*^{w}rot_{2}= 0

(c)Wheel 4 is an idler and the left road wheel is attached to wheel 5.^{d}rot_{1}=^{d}rot_{2}= 0

(d) nThe right road wheel is attached to wheel 7._{5}*^{w}rot_{5}- n_{3}*^{w}rot_{3}= 0

(e) nAnd now the trick: The shaft of wheels 3 and 6 is bushed in transfer wheel 2, so when wheel 2 turns with the shaft locked:_{7}*^{w}rot_{7}+ n_{6}*^{w}rot_{6}= 0

(f)The superposition rule (3) decrees:^{d}rot_{5}=^{d}rot_{7}=^{w}rot_{2}

(g) rotPutting all this through the math mill gives:_{5}=^{d}rot_{5}+^{w}rot_{5}

(h) rot_{7}=^{d}rot_{7}+^{w}rot_{7}

(h:) rotSetting all ratios to 1 results in_{7}=^{d}rot_{7}+^{w}rot_{7}

(e:) =^{d}rot_{7}- (n_{6}/n_{7}) *^{w}rot_{6}

(a:) =^{d}rot_{7}- (n_{6}/n_{7}) *^{w}rot_{3}

(d:) =^{d}rot_{7}- (n_{6}/n_{7}) * (n_{5}/n_{3}) *^{w}rot_{5}

(g:) =^{d}rot_{7}- (n_{6}/n_{7}) * (n_{5}/n_{3}) * rot_{5}+ (n_{6}/n_{7}) * (n_{5}/n_{3}) *^{d}rot_{5}

(f:) =^{w}rot_{2}- (n_{6}/n_{7}) * (n_{5}/n_{3}) * rot_{5}+ (n_{6}/n_{7}) * (n_{5}/n_{3}) *^{w}rot_{2}

= ( 1 + (n_{6}/n_{7}) * (n_{5}/n_{3})) *^{w}rot_{2}- (n_{6}/n_{7}) * (n_{5}/n_{3}) * rot_{5}

(b:) = (-1) * {( 1 + (n_{6}/n_{7}) * (n_{5}/n_{3}) * (n_{1}/n_{2}) *^{w}rot_{1}+ (n_{6}/n_{7}) * (n_{5}/n_{3}) * rot_{5}}(c:) (n

_{7}/n_{6}) * rot_{7}+ (n_{5}/n_{3}) * rot_{5}+ {(n_{7}/n_{6}) + (n_{5}/n_{3})} * (n_{1}/n_{2}) * rot_{1}= 0

rotSounds familiar? Don't worry about the sign mismatch to rule (4) - another spur wheel or starting at wheel 2 will fix this easily._{7}+ rot_{5}+ 2 * rot_{1}= 0

Last not least it shall be
mentioned that the spur wheel differential can be built "boxed" as
well: The central drive wheel (2) is replaced by a drum enclosing the
complete mechanism. (2) now being out of the way, wheels (3) and (6)
can be
merged. To cater for higher forces any number of pairs ((4),(3+6)) can be put into the box. |

© odts 2001,2003,2009