We all have an intuitive sense of how our vehicles propel
themselves around. We have an intuitive sense of how to use the gearbox
to make the engine power work at different speeds. But how does engine
torque actually get increased or reduced?
Our previous Horsepower
and Torque article explained horsepower and torque. Using that
knowledge, this article takes a technical view of how gear reduction
works and how torque, a rotational force, gets translated into a linear
force to push our vehicles down the trail.
Wheel
Speed, Engine Speed
We know that our tires have a constant circumference. To
travel a certain distance, they must turn a certain number of times.
Hence, the speed that the vehicle travels at must be directly
proportional to the rotational speed of the tires. No great surprise
here.
Taking an example of 30" tires, they move 7.85 feet every
revolution. At 25 miles an hour, they must rotate at 280 RPM. At 60
miles an hour, they must rotate at 673 RPM.
In our last article, we determined that torque and horsepower are
functions of engine RPM. Since our tires must rotate at quite a range of
speeds, we use transmissions to vary the engine speed to match wheel
speed from a fixed ratio set by the differentials (the final drive
ratio) and perhaps the transfer case. At all but the highest speeds, we
need the engine speed to be higher than the wheel speed. We need the
engine torque to be available at lower wheel speeds.
Gears:
Torque
In just about every vehicle on the road, gears are used to
increase or decrease rotational speeds and torque. Gears come in many
forms such as spur, bevel, helical, or planetary. They may or may not
have teeth. They might not even touch and instead use a chain or belt.
They may even be called pulleys but in the end, they all operate under
the same simple principles.
Let's look at gears and torque first. Recall that torque is just the
rotational equivalent of force. It is measured in units of
force*distance, e.g. lbs-ft, N-m. We know from using a wrench that the
linear force is applied to the end of the wrench to create torque on a
bolt. Similarly, something that has torque applied to it can generate a
linear force. The linear force is just the torque divided by distance, F = T/R;
basic algebra here. The force is always tangential to the circle of
radius R.
So how does this work for gears? The torque on the driven gear can be
converted to a linear force at the point on contact with the next gear
in the chain, F = Tin / Rin where Tin is the input torque and Rin is the radius of the gear. The force generates torque on the driven
gear, Tout = F * Rout.
Mucking around with the equations gives us what we are looking for:
Tout = Tin * Rout / Rin
If the input gear is larger that the output gear, the output torque
is lower. If the input gear is smaller, the output torque is higher. The
ratio of the radii (or diameters) is the gear ratio for torque expressed
to unity: Rout/Rin : 1.
We can just extend this to multiple gears in the chain. If you work it
out, you will see that the ratios multiply out. Shafts connecting gears just transmit the torque to another gear
that may have a different diameter. For example, a driveshaft conducts
the torque to the differential pinion gear.
Gears:
Speed
Two gears of different radii in a chain will also have
different rotational speeds. The circumference of the input gear is Cin = 2 * Rin * pi.
Each time it makes one revolution, it rolls along the circumference of
the output gear. The output gear circumference is Cout = 2 *
Rout * pi. The output may rotate only a
partial revolution or more than 1 revolution to each input gear
rotation. The number of revolutions is just the ratio of the
circumferences yielding:
RPMout = RPMin * Cin / Cout = RPMin * Rin / Rout
The speed ratio is just Rin/Rout : 1. Doesn't this look familiar? It is the inverse of the gear ratio for
torque. When the input gear is smaller, the output rotates slower but
with more torque. When the input is larger, the output spins faster but
with less torque. Makes sense. Lower gears in our rigs (numerically high
ratio) gives us lots of torque but we crawl along.
Now for a revelation of nature. Power is a function of torque and
speed (RPM). If Pin is the input power, Pout is the output power:
Pout = Tout * RPMout = (Tin * Rout / Rin) * (RPMin * Rin / Rout)
= Tin * RPMin = Pin
What? Pout = Pin?This means that by gearing down, the engine does not produce any
more power (in the physics sense). We just get more torque at slower
speeds. If this was not the case, no matter what the motor size, we'd
all have the same horsepower with some just some gearing.
What
about Teeth?
Teeth on gears ensure that the gears do not slip on one
another. Often we derive gear ratios by comparing the tooth count on
gears. This works because a tooth on one gear must fit into the gap on
the other gear, that is, they are the same size on each gear in the
chain. Hence, the number of teeth on a gear is directly proportional to
its circumference. Comparing teeth is just like comparing the
circumference or the radius.
Wheels/Tires
So far we have been looking at everything in the
rotational world. But we know that our trucks move in the linear world.
The final piece of the puzzle is the wheel. We treat it just like a
gear. However, the linear force developed at the tangent (the contact
patch) doesn't drive another gear. We just use this linear force to push
the vehicle. As we know, the taller the tire, the larger the radius. The
larger the radius, the less force at the tangent. We know this because
when we put larger tires on, the vehicle seems sluggish.
The interesting fact is that the radius of a tire on a vehicle is not
constant. The contact area has a shorter radius because the weight
flattens the tire. This is the effective radius or Static
Loaded Radius. Linear force to the
vehicle is actually higher because of this reduced radius. Airing down
increases the effective force on the vehicle.
One benefit of this is that for the same linear force, we need less
torque input. Hence, the stress on U-joints, diffs, and the like can all
be lower because we need less torque from the motor.
Note that this effect does not change the gear ratio exactly. The
actual force being applied is as if the gear ratio had increased.
For an actual change in gearing, the circumference of the tire would
have to change. This does happen somewhat as we air down. The exact
amount of change depends on the tire size, tire type, weight, and
pressure change. To quantify the effect, you measure the linear distance
traveled over a fixed number of tire rotations. Then change the pressure
and remeasure. In an experiment, tires aired down from 35 psi to 15 psi
changed the rolling distance less than 1.5% but the effective radius was
16% shorter.
Vehicle
Gearing
How much gear reduction is needed depends on so many
factors. Tire height, engine type, and the usage of the vehicle. In
general, gasoline engines are most efficient at RPMs below their torque
peak. Diesels are usually most efficient at RPMs near their torque peak.
On the highway approaching a grade, it is best to be above the torque
peak. As we loose speed, the engine torque increases, preventing further
speed loss. If we were below the peak, we have little choice but to
downshift and lose speed. On the trail, it is not so much power but
rather control. We want to operate the engine at an RPM such that it
does not lug but with a slow vehicle speed.
Examples
Let's pick a vehicle with a 3.48:1 first gear, a 0.76:1
fifth gear, a 2.72/1:1 (low/high) transfer case, 4.10 differential
gears, and 33" tires. Parts that have one input and two outputs
(transfer case, differentials) complicate matters. We will assume that
they are locked (forcing equal rotational speeds on the outputs) and
split torque equally (50/50). If we wanted to figure out maximum stress,
we would instead use the most unfavorable torque divisions that would be
possible.
Let's say the engine is putting out 200 lbs-ft of torque at 2000 RPM.
After the transmission (first gear), the input to the transfer case is
200 * 3.48 = 696 lbs-ft at 2000 / 3.48 = 575 RPM. There are
two outputs from the transfer case (locked, low range), each gets half
of the torque. So the rear driveshaft sees 947 lbs-ft at 211 RPM.
Once, again the differential has two outputs, each carrying half at 1940 lbs-ft
at 52 RPM.
Another way to figure this is to multiply out all the reductions:
3.48 * 2.72 * 4.10 = 38.8:1. 200 lbs-ft * 38.8 = 7762 lbs-ft /
4 wheels = 1940 lbs-ft. Similarly, 2000 RPM / 38.8 = 52 RPM.
This calculates out to 5.1 MPH.
Is 38.8:1 a good rack setup? Some would say that's not too bad ...
take a closer look. Let's assume we were at 800 RPM which is high
enough to not bog the motor down too much. This gives us 2 MPH or 3
feet (35 inches) in one second. Looking at it this way, that is kind of
fast for rock. You'd be forced to slip the clutch or bog the motor.
That's what makes setting up a rock crawler different and why we are
always in quest for lower gears. If we had 115:1 gears and 33"
tires, at 800 RPM we get 12 inches/sec.
To calculate highway performance, we start with the overall reduction
which is 0.76 * 1 * 4.10 = 3.12:1. It is usually more useful to
calculate the engine RPM at a given speed (MPH), usually 60 MPH.
Just algebra here:
MPH = Engine RPM / Ratio * Diameter * pi * 5 / 5280 Engine RPM = MPH * Ratio / Diameter / pi / 5 * 5280
where Ratio is the gear reduction, Diameter is the tire diameter in
inches. At 60 MPH, our example rig turns at almost 2100 RPM.
As a comparison a more "stock" rig turning 28" tires with
a 3.07 differential ratio would be at 1700 RPM.
Summary
The purpose of this series of articles was to examine the
mechanics of how our vehicles more around. They are made up of several
interesting elements that combine to give us torque, speed, power, and
linear force. While we all seem to understand the effects of the
combination, we hope that a technical study of the elements leads us to
a better understanding of our vehicles and how to modify them.
Last Modified May 15, 1998 by Gerald Luiz
Untitled Document
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