The Lotus Cars Community banner

1 - 13 of 13 Posts

·
Registered
Joined
·
7,610 Posts
Discussion Starter #1
What does it mean when Car &Driver's article says the Elise's gear ratio's are:


3.35 to 1;
2.05 to 1;
1.48 to 1;
1/36 to 1;
.94 to 1; and
.076 to 1.

What is the first number in relation to the second (or the "1")? Thanks. :confused:
 

·
Registered
Joined
·
12,374 Posts
Allan Gibbs said:
What is the first number in relation to the second (or the "1")? Thanks.
It is an indication of how may revolutions the input makes in relation to the output.

For instance, 3.35 to 1 indicates that the input (engine) rotates 3.35 revolutions to the output's 1 revolution. As the gear selected "go up", the ratio usually "go down", as in the fifth gear ratio of .94 to 1; which indicates that the engine turns a little less than one revolution to the output's one - this is known as an overdrive gear as the output turns more than the input.

A low gear (high number) gives more power to the rear wheels, along with rapid acceleration, but the engine runs out of revs. That's when you shift - less "power" to the wheels, but you can go faster. On some hills, or when you need a burst of speed, you down shift (shift to a lower gear).

Added to the over-all gear ratio is the ratio of the transmission's output gearing to the rear "axle"s. This is usually known as the "final drive" gear ratio and unlike the transmission, is not usually "shift-able" (although you can disassemble the axle/transaxle and replace the gears). On some cars, the transmission and the final drive (rear axle) are two separate units. On others, like the Elise, the transmission and the final drive are combine in one unit and are referred to as a transaxle.

The engine speed maximum RPMs and the transmission ratios, final drive ratio, and tire size determine the car's maximum speed in each of it's "gears".

Tim Mullen
 

·
Registered
Joined
·
1,256 Posts
Tim's explanation is 100% complete and correct but i thought i'd add a bit in case you still don't understand.

When the tach says the engine is going at 8000rpm, clearly the wheels aren't spinning at 8000rpm. This is because there is gear reduction throughout the drivetrain. So, if there were no gear reduction, 8000rpm at the engine would mean 8000rpm at the wheels (imagine there was a steal rod connected from the engine to the wheels.

Now lets add in the transmission but forget about the final drive. Now, if you are in first gear at 8000rpm, the wheels are only spinning at 8000/3.35= 2388rpm. This effectivly multiplies the torque in order to get the car off the line.

Your 5-6 gears are overdrive gears. That means that the wheels are spinning faster than the engine (again we are forgetting about final drive). This is so that at high speeds, your engine doesn't have to rev as high.

In reality there is another gear reduction (sometimes even more than one) in between the engine and the rear wheels. You can think of this like a transmission with only one gear. To find final ratios, you multiply all the ratios together.
 

·
Registered
Joined
·
1,256 Posts
atyclb said:
when are we going to get to the simple terms :(






:eek:
ok real simple...the wheels spin right? This is because things further up the line spin. The thing generating the spinning is the engine, correct?

So, the engine spins and that spinning is transferred through gears. Gears have ratios such as X:1.

So, part A spins X times, then part B that is connected to part A after the gear set spins once.

get it?

-Steve
 

·
Registered
Joined
·
3,927 Posts
With skid pads #'s we're talking about lateral G's. It's typically measured going around a 300 foot circle. It's a indication of the amount of grip you've got in the turns. So in pulling 1 G you and the car in effect have it's / your entire weight pulling on you laterally.
 

·
Registered
Joined
·
4,634 Posts
G's refers to the force of gravity. So if the car weighs 2000 pounds and the magazines say a car's suspension package is capable of 1.2 G's laterally, that's 2400 pounds of lateral force that the car is capable of pulling before the tires break away (i.e., start to lose the ability to hold the desired line of the turn).
 

·
Registered
Joined
·
100 Posts
atyclb said:
when are we going to get to the simple terms :(
:eek:
gears have teeth

when a 10 tooth gear meshes with another 10 tooth gear, the ratio is 10:10, ie 1:1

when a 10 tooth gear meshes with a 20 tooth gear, the ratio is 10:20, ie 1:2. This means the first gear has to go through 2 full revolutions [ie, 2 x 10 teeth = 20 teeth] to turn the second gear once.
 

·
Registered
Joined
·
592 Posts
All of the above is correct. In addition, the gear ratio is torque multiplication. So a 3.35 to 1 gear ratio multiplies the torque of the engine by 3.35 times and provides more force at the tire contact patch. Hence low gears provide better acceleration.
 

·
Registered
Joined
·
12,193 Posts
mikester said:
if the car weighs 2000 pounds and the magazines say a car's suspension package is capable of 1.2 G's laterally, that's 2400 pounds of lateral force that the car is capable of pulling
That's mostly correct, but quantifying the lateral acceleration as a force by applying it to a specific mass makes the number less meaningful.

Example: if the 2000 pound car has a driver weighing 100#, it will max out at !2520# of lateral load. With a driver weighing 200#, it will max out at ~2640# of lateral load. This, of course, is not 100% accurate, because the max. lateral acceleration will vary relative to mass of the vehicle, due to a lot of things, including downforce (or lift) non-variance.

G is a unit of acceleration equal to ~32 ft/s².
 
1 - 13 of 13 Posts
Top