[Quadzilla]

and if you could share it with the rest of us, that would be fabulous.

thanks in advance
Francis

 

[Carl Brune]

Hi Jon. I suspect you more or less know how your transmission and rear gears work, so your task is mainly putting what you already know into physics language. In my opinion, the most important physics concept here is that (power in) = (power out) when power is sent through a set of gears (let's ignore friction). Rotational power is given by (torque) times (roational frequency) times (2pi). In simple terms, ft-lbs times RPM
going in and coming out must be the same.

The main practical use of gears in cars is to keep the motor at an RPM where makes good power. Also the differential gears rotate the rotation axis by 90 degrees. These are some basics. If you have a decent college library near by, I'll bet you could find an engineering book just devoted to gears. It might also be interesting to talk about different types of gears (helical, worm,...).

Nice choice for a topic. If you have any questions I'll try to help. A while back I assigned my class the following problem:
given a torque (ft-lbs) versus RPM curve for a motor, find the horsepower versus RPM curve. Students hated it because it didn't use metric units! --- Carl

 

[Quadzilla]

go Carl! reminds me of the time my math teacher had us doing part of his taxes.



------------------
Francis Taracido
[email protected]
Proud Patron of Quadzilla

 

[Shrub]

Ok, need to get a draft done by friday. So, what makes you go, torque or horsepower, doesnt one make you start and the other helps you keep accelerating? Why does the horespower and torque produced by an engine fall off after a certain RPM, is it better to keep and engine in the good torque range or good horsepower range? Is there an equation for horsepower? How do these relate to gas mileage etc. How is a cars ideal rear end ratio determined?

I know these are a lot of questions but it is very important that i get answeres by tomorrow so i can write it in by fri. E-mail me if you need to, i will but you down as a reference.

Anything and i mean anything that may be related is also appreciated, all good ideas will help, anything i forgot?

Thanks
Jon

 

[Tom Mobley]

Carl, why did they care what units it was in? HP= (trq x rpm)/5250. All they had to do was run this for however many rpm points they wanted to plot. I could do this by hand on a my TI-85 in a hurry or any spreadsheet makes it easy and does the plot for me.

What class are you teaching? Wish I had had Profs that gave interesting stuff like this. All my stuff was boring book stuff until I got into ECE105 and was able to write a little C program that calcs displacement and compression ratio after prompting the user for the necessary info. First fun I had in all that calculus, physics and engineering. Sure wish I could have finished that degree.... ran out of time and money.

Tom (Integration by Trigonometric Substitution, whoever thought of that oughta be taken out and shot.)

[This message has been edited by Tom Mobley (edited 05-26-99).]

 

[Gene Chas]

Jon, there's a site called performanceprofessor.com, that is written at a low ME level ( since I can understand some of it and I'm a finance guy). I'd click on it and see if there's any physics. Also check Fred Aldrich's site. I think Fred is a retired GM engineer.

I just read all the posts, Jon I believe torque is the first derivative of horsepower, that is, the rate of change of horsepower. Not being an engineer I can't answer which does the work, but I would think under conditions of acceleration it would be the first moment of a power function ( i.e. the first dervative of the power equation or torque) that determines the amount of work.By the way, I prepare term papers on the side!

Let me add one more thought. If you conceptualize the relationship between torque and HP as I mentioned above, the maximum amount of work an engine does occurs before the torque curve peaks. Isn't work = force X acceleration. So that maxumim work would be when the engine's HP is increasing rapidly because the area under the torque curve is increasing. And then even though HP keeps increasing, it's increasing at a decreasing rate since the area under the torque curve falls off. This is easy to see if you pick up any of the car mags with a dyno test in them. The "system" of an internal combustion engine works the same way regardless of make, mdel, displacement, etc. The next logical step is how all this relates to engine/transmisioon is beyond me. If it where my paper I center in on a topic such as the torque multiplication of an auto tranny.

Don't quote me on any of this, my master's is in Econ and I earned that a long time ago. Plagaraze me baby! It's all "common knowledge" anyway. I wish I could've played with this stuff more when I was a kid into building motors, but I didn't have a good grasp of Calculus till later on. I still put together some raging engines when I was 16-18. ( Courtesy of a well equipped machine shop and a boss who is an auto genuis).

------------------




[This message has been edited by Gene Chas (edited 05-27-99).]

 

[Wes V]

Jon66;

Here are some general points (that may tend to keep this string growing).

As stated in one of the prior posts; horsepower is a function of torque and engine speed. I'm not real positive if the formula given is correct, but I've no reason to question it.

Gearing is an attempt to keep the engine in it's desired operating range. The wider the torque band, the less selection of gearing is required. A small displacement engine with narrow torque band NEEDS a 6 speed transmission. The ONLY reason for a 6 speed transmission in a car with a V8 is for decreasing fuel consumption to satisfy CAFE standards! (at anywhere near legal speeds, a .5 overdrive as in the new Camaros puts the engine at an idle RPM) The installation of 6 speeds is just to satisy the government and has nothing to do with performance. If this wasn't the case, the spread between gears would be smaller. (I'm talking the factory 6 speeds here!)

There isn't an "optimum" gear ratio due to the dynamic nature of driving around town. Everybody want's that "4.11" launch when taking off from the light, and yet also wants a low crusing RPM. (close to possible with some of the overdrive transmissions)

The subject of rear-end ratios CAN'T be talked about without talking about the gear ratios in the transmissions!

As far as the typical 4 speed transmission; the path of the power through the transmission in the first 3 speeds is; 1. power goes in the "input shaft", 2. it goes through a gear set and into the "lay gear" or "cluster gear", 3. it then goes from the lay gear through another set of gears to the output shaft. This is important due to the fact that EVERY time you transmit power through a set of gears, you lose power and generate heat!

When in final drive in a typical 4 speed, the power goes straight from the input shaft to the output shaft! Less loss of energy! This is ONLY possible when the transmission is in a one to one ratio. In other words, it doesn't apply to an "overdrive" transmission when in overdrive.

An interesting point dealing with energy loss has to deal with the rear-end and it's configuration. Chevy rear-ends are more efficient than Ford rear-ends! (that's not to say that they are as strong) The gears in the rear-end have to do two things; they not only have to provide a gear ratio, they also have to change the direction of the rotation (from the driveshaft to the axles). This is done through what are called a "hypoid" gear set. This type of gearing isn't very efficient and it gets worse the farther the distance is between the axis of the input shaft to the output. Think about a line drawn through the input shaft in relationship to a line drawn through the axles. The two lines DO NOT hit one another and the farther apart, the less efficient the gear set is. Fords are farther apart than Chevys!!!!!!

The most "efficient" driveline would be a Richmond 5 speed with GM rear-end (The richmond's fifth gear is a one to one). By using the Richmond, you would still have a low first gear for a good launch.


You should also go to the Technical Reference section and see the "gearing" section.

Wes. Vann

 

[Gene Chas]

Wes, is it or is it not the case that if you found the engine's "sweet spot", that is the rpm at which it did the most work ( force X acceleration)you would want to run it at only that narrow rpm band? So that in a theoretical world your transmission would keep the engine in a small band of , say, 5000-7000 on a good semi-race motor, and infinetly vary the final ratio. Isn't this the point behind the "infinite ratio" transmissions I've read about?

Jon66, man you can print this thread and have all the info you need to start your paper! I'm bummed to admit my '67 was built in MD.

[This message has been edited by Gene Chas (edited 05-27-99).]

 

[Todd]

For basic physics of gears find a copy of the 'Mark's Standard Handbook for Mechanical Engineers'. This explains a lot about different gears, physics, losses, etc. If you want to find out more about reference material for automotive related technical I believe there is a powertrain/transmission related engineering organization.

A good place to start would be SAE (Society of Automotive Engineers). I'm sure they could provide a wealth of links, references, or other related orgs. You can also call the public relations departments of any of the BIG 3 and they may be able to help as well.

I wrote lots of papers for my ME degree and every company I was in contact with gave me (read free!) lots of info and many times, samples.

 

[Fred Aldrich]

Wow, I just have to jump in here.
First, we need to define some terms so we're all talking the same language. I know this is booooooring but stick with me here. When we're done, you'll only have to remember one simple formula.

Work
Work is done by moving a force through a distance.
W = F x d
If you raise a 10 lb. weight from the floor to your chest, you've done about 40 ft-lb of work.
W = 10 lb x 4 ft = 40 ft-lb

Power:
Power is the rate of doing work or work per unit time. The faster you work, the more power it requires.
P = W/t
For example, if it takes 2 sec. to raise that 10 lb. weight 4 ft. to your chest, on the average, you've used 20 ft-lb/sec. of power.
P = 40 ft-lb/2 sec. = 20 ft-lb/sec

Horsepower:
One Horsepower is defined as 550 ft-lb of work per sec.
Hp = P/550
In the example, we expended an average of 0.036 Hp.
Hp = 20/550 = 0.036 Hp

Horsepower is unit dependent and only has meaning in the English system of units. Power in the Metric system is measured Kilowatts like electricity. 746 Watts = 1 Hp

But what about Torque:
Torque is a twisting motion or a force acting at the end of a lever arm like a torque wrench.
T = F x L
The correct units for torque are lb-ft to differentiate it from Work with units of ft-lb. Work and Torque are not interchangeable!

Engine Example:
Let's assume we have an engine turning at 3000 RPM and producing 300 lb-ft of torque. In this case, think of 300 lb-ft of torque as a 300 lb force acting at the end of a 1 ft lever arm. As the crankshaft rotates, the lever arm rotates 3000 turns in 1 min. For a 1 ft lever arm, the end travels about 6.28 ft. for each revolution (circumference = 2 x Pi x L) or about 18850 ft/min at 3000 rpm.
The Power generated is 300 lb x 18850 ft/min or 5,655,000 ft-lb/min. Using our Horsepower formula, this is about 171 Hp.
Remember this:
Once you have worked through all this, all you really need to remember is that for an engine:
Hp = RPM x Torque/5252.1132 (Hp = RPM x Torque/5250 is close enough.) Torque must be in lb-ft.

Fred Aldrich
www.GeoCities.com/~69_chevelle

[This message has been edited by Fred Aldrich (edited 05-27-99).]

 

[Fred Aldrich]

The Ideal Drag car
The ideal drag car should deliver the maximum horsepower to the track at any given point in the run. To achieve maximum horsepower, the engine needs to be running at maximum horsepower. That's why a 5 speed is always theoretically better than a 4 speed. Engine is running closer to peak horsepower more of the time.
By assuming that a drag car can deliver a constant horsepower to accelerate the car down the track and there is no tire spin it can be proven that:

ET x MPH = 1338.59 (1/4 mile with speed trap in last 66 ft.)

There is more info on this subject at my web site.

------------------
Fred Aldrich
Web Site: www.GeoCities.com/~69_chevelle

 

[Carl Brune]

Ya, Tom, that's it. But you won't find that formula in any freshman physics textbook, at least published in the last 20 years. The essence of the problem was to figure out the formula using stuff given in the book, like 1hp = 746 Watts. I had them do some other stuff like find the RPM where the maximum torque and hp occur. This was for calculus-based freshman physics.

I just scrolled down and see that several more replies appeared, including an excellent one from Fred. Everyday words like power and work have very specific meanings in this context!

Jon, Keep in mind that torque and horsepower are not independent, but are related by the formula given by Tom and Fred.

Let me add one more thing. In an ideal world where there is no friction or other losses, the kinetic energy of a car (=1/2*mass*speed*speed) will increase by an amount equal to the work done on it. In order to maximize the incrase in energy (and hence speed) you need to maximize the rate of doing work (power). In other words acceleration is maximized when at the peak of the horsepower curve.

Finally, you may find some interesting suff at http://members.home.net/rck/phor/index.html

Carl

 

[Shrub]

WOW! you guys are awesome! Thanks for all the info, it all helps alot, thanks for the links too.

Thanks again
Jon

 

[jmw]

I am having serious flashbacks to engineering school days. In 80-81 one of my profs was working on a continuously variable transmission (CVT). He told us about it and we all nodded our heads approvingly and then moved on. Low and behold a year or two ago I see a production unit in Honda Civic in an article in Car and Driver. Floor it from a dead stop, the engine revs to its peak and just stays there while the transmission does its thing. If I remember right, the input and out put shafts are set up as opposing cone shapes with a super strong belt that slides down one and up the other as you accelerate, then vice versa as you slow down. If they only made one that would mate up to a 502!
John Walker