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Most Ridiculous Item of the Day
Have
any of you ever watched Bill O’Rielly on the Fox news channel? You
know, the guy with the huge ego, stating his opinion as fact to "save
the country" from its own evils. Bill has a section of his daily show
devoted ridiculous happenings in the world of politics.
If I were to do my own "Most Ridiculous"
item in the world of racing it would be based on the following
statement. "Horsepower sells motor cars, but torque wins motor races."
This couldn’t be further from the truth.
Like it or not, everything that goes on
around us is governed by the laws of physics, and these laws are
non-negotiable. The good news is that we don’t have to be Einstein to
apply the basic laws of physics to racing. The fact that too few do is
the reason that such ridiculous statements are common in racing.
Let’s start with a few definitions.
Webster’s dictionary describes torque as "A turning or twisting force."
Note that the definition does not imply motion. As applied to an
engine, it is simply a measure of the twisting force at the
crank/eccentric shaft. Torque is normally rated in Lbs.-Ft. Since
pounds feet doesn’t exactly roll off the tongue, most of us refer to it
as foot pounds.
Notice that there are two terms. Force
(In lbs.) and distance (In ft.). At first it may seem strange to
describe a "Turning or twisting force." in terms of distance, but a
more detailed description makes it clear. If I were to put a shaft in a
bench vice, attach a 1-ft. long lever to the end, perpendicular to the
shaft, and then hang a one pound weight off the end of the lever, I
would be applying one ft.-lb. to that shaft. Notice that the shaft is
not rotating even though a torque is applied to the shaft.
If I were to replace the one-foot lever
with a 100-foot lever, I would now be applying 100 ft.-lbs. to the
shaft with the same one lb. weight. As you can see, the amount of
twisting force on the shaft will vary depending on the length of the
arm, and that requires that we specify a measure of distance to
properly describe the force seen at the center of the shaft.
Let’s say for instance that I pull the
shaft from the vice, and ask you to hold it in your hand. If I do this
with a one pound weight hanging from the end of the one foot lever, I
will be applying a force of one ft.-lb. to your hand, and you will have
no problem holding on to it. If I replace the lever with one that is 10
ft. long, with the same 1lb. weight on the end, (For all these
scenarios, we assume that the lever itself is weightless.) You will now
have a force of 10 ft.-lbs. applied to your hand, and it will be much
harder to keep the shaft from rotating, even though you are still only
resisting the one pound weight.
This would have exactly the same effect
as setting a torque wrench to 10 ft.-lbs., attaching it to the end of
the shaft, and applying force until the wrench clicks. Ten ft.-lbs. is
ten ft.-lbs. whether it is applied with a one-foot lever and a
ten-pound weight, or a ten-foot lever and a one-pound weight. Torque is
equal to the weight, or force, times the length of the lever. It’s that
simple.
If a particular engine has a peak torque
rating of 200 ft.-lbs., that force is equivalent to attaching a one
foot lever to the shaft, and hanging a 200 lb. Weight from the end of
it. Or…any other combination of weight and lever length which has the
product 200.
Notice that I can take you from easily
holding on to the one pound weight, to not even having a chance of
holding it just by changing the length of the lever. (Like a one-lb.
weight, and a 100-ft. lever.) Of course you say, that’s just leverage!
Well…you’re right! Keep that in mind, because it is that leverage that
makes all the difference, and a gear is in fact just a clever way to
apply leverage between two or more rotating devices.
Let’s say that the shaft used in our example is the input shaft of the transmission from a 1993 RX-7 with the following ratios.
1st 3.483 to 1
2nd 2.015 to 1
3rd 1.391 to 1
4th 1.0 to 1
5th .719 to 1
If the transmission is in 4th
gear, one complete revolution of the input shaft will result in one
complete revolution of the output shaft, just as if there were a solid
shaft running all the way through. If we attach a 1-ft. lever to the
input shaft with a 10-lb. weight on the end, the torque at the input
shaft will equal 10 ft.-lbs. as we have already determined. Since we
have a 1 to 1 ratio from input to output, we will also have 10 ft.-lbs.
at the output shaft.
If we were to keep the same weight and
lever on the input shaft, but switch the transmission to third gear, we
would still have 10 ft.-lbs. at the input shaft, but we would now have
13.91 ft.-lbs. at the output shaft. This value is the product of the
input torque and the gear ratio. (10-ft.-lbs. times 1.391 gear ratio
equals 13.91 ft.-lbs.) If we were to switch the transmission into 1st gear, the result would be 34.83 ft.-lbs. at the output shaft.
As you can see, a gearbox gives us a
simple way to vary the torque through leverage, and it is equivalent to
changing the length of the lever. Thanks to gears, we can have any
amount of torque that we want! In fact, a bone stock 12A making only
100 ft.-lbs. of torque could be geared to pull an 18-wheeler up a steep
hill, as long as we are not in any big hurry to get the job done.
Let’s say that it takes 10,000 lbs of
force to pull a heavy weight up a hill. No problem! We could even do it
with our stock 1980-GS in 4th gear if we are willing to
build a custom ring and pinion gear with a ratio of 100 to 1.
(100-ft.-lbs. times transmission gear ratio of 1:1 times ring and
pinion gear ratio of 100:1 equals 10,000 ft.-lbs.)
If we are using a tire with a diameter
of 24", the distance from the axle center to the ground is exactly one
foot, and so the force is equal to 10,000 lbs. Remember, the torque is
equal to the lever length times the force. If we re-write that formula
to solve for force, force is equal to torque divided by the lever
length, and so that 10,000 ft.-lbs. at the rear axle results in 10,000
lbs. of force at the tire contact patch.
With this same information, we can also
calculate the acceleration rate of the vehicle, but first we need to
consider Newton’s second law of motion, which states that "Acceleration
is proportional to force." and "Acceleration is inversely proportional
to Mass." This law is normally stated more simply as "Force equals mass
times acceleration." or F=MA. If we rewrite this to solve for
acceleration, we get A=F/M. To find the rate of acceleration for a
vehicle, we simply divide the force (In lbs. at the tire contact
patch.) by the mass (Total weight of the vehicle in lbs.)
Let’s calculate the acceleration rate of a 1st.
gen. RX-7. The engine has a torque peak of 100-ft.-lbs. In fourth gear,
the ratio is 1 to 1, and so the torque at the output shaft is also
100-ft.-lbs. The ring and pinion ratio is 3.909 to 1, and so the torque
at the rear axle will be (100 times 3.909) 390.9-ft.-lbs. The tire
diameter is 24 inches, and so the lever length (Distance from the center of the axle
to the ground.) is 12 inches, or one foot. The resulting force at the
tire contact patch will be (390.9-ft.-lbs. of torque divided by lever
length of one foot.) 390.9-lbs. of force. The total vehicle weight with
a driver is 2600 lbs., and so the acceleration rate in G’s (The force
of gravity.) will be force (390.9) divided by mass (2600) which equals
.15 G’s.
If we do the same calculations for first
gear acceleration, we find that the force at the contact patch is 1,436
lbs., and the acceleration rate is .55 G’s. It’s clear that we have
used the gears for leverage, with the result being a greater rate of
acceleration in 1st gear. Of course you knew that already, but now you know why.
By now it should be clear that the
acceleration rate of a vehicle is determined by the weight, and the
force at the contact patch, which is the result of the torque output of
the engine, and all the levers/gears between it and the ground.
You’re probably thinking that we have
just determined the acceleration rate of the vehicle, and even changed
it with gearing, with no mention of horsepower. So torque really is the
determining factor right? Wrong! We haven’t considered speed.
We can gain acceleration by changing
the gear ratios, but we can’t go very fast in first gear, so what’s the
point? We have effectively changed the amount of torque available to
accelerate the vehicle, but our top speed is limited to about 25 mph.
Confused yet?
Read on.
OK, so we all know what torque is, now let’s get to horsepower.
Referring once again to Webster’s
dictionary, horsepower is defined as "A foot/pound/second unit of
power, equivalent to 550 foot/pounds per second."
Put more simply, horsepower is a measure of work done over time, or the rate at which work is done.
So now we have another term to confuse
things. As if force and distance weren’t enough, we now have time
involved, and the shaft must actually be spinning. Why… Well, if you
are just standing there holding on to a shaft with a lever and a
weight, you are doing no work. If you stand there long enough, you will
feel like you are working, but in fact you are doing no such thing.
Don’t believe me? Clamp the shaft back in the vice, and you can leave
it there indefinitely without having to feed it, add gas, and any other
means of supplying it with energy.
Torque, all by itself does nothing
useful. In fact, the definition of torque does not even require that
the shaft be moving. I am sure that all of you want your car to do
something useful, like take you to the movies, or get you around the
track before the other guy. In other words, you need your car to do
some work, and you want it to do that work in the least amount of time
possible.
If I give you a wagon full of cement
blocks, and ask you to pull it one mile up a hill, you would agree that
I am asking you to do work. If I ask you and your buddy to each pull a
wagon full of cement blocks up the hill, you will both be doing the
same amount of work. But…If it takes you an hour, and your buddy does
it in 30 minutes we have a very different situation. You have both done
the same amount of work, but your buddy, by completing the task in half
the time, has proven that he can develop twice the horsepower that you
can. You both traveled the same distance, and exerted the same amount
of force, but the third term in the definition of horsepower, time, was
different.
For those of you who are sticklers for
details, the force required to pull the wagon up the hill at a steady
speed is equal to the weight of the wagon, times the sine of the angle
of the hill. If the wagon weighed 100-lbs., and the hill was at a
45-degree angle, the required force would be (Sine 45 Times 100 lbs.)
equals 70.7 lbs.
If we were interested in moving the
wagon by driving the wheels rather than pulling it by the handle, we
could convert the force to torque by dividing the required force by the
radius of the driven wheels. Let’s say that we have 6" diameter wheels.
That would give a lever length (Distance from the center of the axle to
the ground.) of 3 inches, or .25 feet. The required torque would then
be (70.7 lbs. times .25 ft.) which equals 17.675 ft.-lbs.
James Watt, who spent the majority of
his life perfecting the steam engine, created the term horsepower. He
was looking for a way to measure the rate of work done by a horse so
that he could make valid comparisons between horses which did most of
the work in those days, and his steam engines which he hoped would do
most of the work in the future.
Watt found that on average, a horse
could lift 330-lbs of coal 100-ft in one minute. He then stated that
the power available from one horse was equal to (330-lbs. times
100-ft.) or 33,000-lbs./ft./min. If you divide that by 60 to convert to
lbs./ft./sec. you get 550-lbs./ft./sec. Watt called this one
horsepower, which leaves most of us wondering why he didn’t call it one
watt. I don’t have the answer to that, but I do know that 746.6 watts
equals one horsepower. If you ever see an engine rated in watts, (This
is still popular in some countries.) you can divide by 746.6 to
determine the horsepower. Or, you can tell you pals that your bone
stock 3rd. gen. RX-7 puts out One hundred ninety thousand, three hundred and eighty three watts.
So if
one horsepower is equal to 33,000 lbs.-ft. per minute, we can rearrange
that to say that horsepower equals torque times rpm, divided by 5252.
How do we get there?
In the above formula, force and distance
are stated in ft.-lbs., and time is stated in RPM, so we need to
convert our terms. First we need to express that 33,000 lbs. of force
as 33,000 ft-lbs. As you now know, that is equivalent to a 33,000-lb.
weight hanging from a 1-ft. lever. Then we need to express the one-foot
per minute as RPM.
The circumference of a circle is defined
as the diameter of the circle times Pi, which is 3.14159. We have a
one-foot lever, so if we were to spin the shaft, the outer edge of the
lever would scribe a 2-foot diameter circle. The circumference of a 2
foot circle is (3.14159 times 2) 6.282 feet. If we divide 1 foot by the
distance traveled in a complete revolution (1 divided by 6.282) we get
.159 revolutions per minute, which is equal to one foot per minute.
So now we have: One horsepower equals 33,000 lbs.-ft. of torque per .159 RPM.
That’s still kind of ugly dealing with
just a fraction of an rpm, so we divide both terms by .159 and we get:
one horsepower equals 5252 lbs.-ft. of torque per 1 rpm.
This can be rewritten a few different ways that are valuable to us.
Horsepower equals torque times rpm divided by 5252. Horsepower = (Torque X RPM) / 5252
Torque equals horsepower times 5252 divided by rpm. Torque = (Horsepower X 5252) / RPM
RPM equals horsepower times 5252 divided by torque. RPM = (Horsepower X 5252) / Torque
If you know any two of the terms, you
can calculate the third. You might also notice that torque and
horsepower will always be equal at 5,252 rpm, horsepower will be
greater than torque above 5252 RPM, and torque will be greater than
horsepower below 5252 RPM. ALWAYS…NO EXCEPTION! Just look at any dyno
sheet, and you will see what I mean. If you see a dyno sheet where this
is not true, you can be sure that someone fudged the numbers to help
sell a product.
Back to the issue at hand, I’m sure that
the coal tugging horse and a wagon full of cement blocks probably
doesn’t seem all that relevant to your racecar. So let’s look at things
another way.
The definition of horsepower includes
three terms. Force, distance, AND time, where torque is simply a force
applied over a distance. In the case of Watt’s experiment, the force
was exerted by the weight of the coal, which was being lifted from the
mine. In a car, we are interested in acceleration, not the ability to
lift an object. In our case, the force is exerted by the inertia of the
vehicle, which resists acceleration.
So now I need to bore you with another
definition. Back to Webster’s dictionary, inertia is defined as "A
property of matter that causes it to resist changes in velocity." In
more simple terms, your car would rather not be accelerated from 30 to
70 mph, and so an external force is required to make this happen. This
force comes from your engine.
To accurately describe the acceleration
capability of your vehicle, we must consider time. If we just
considered force, and distance, we wouldn’t really be saying much about
the car. If I tell you that my car can pull a 3,000-lb. weight 100-ft.
up a hill, would you be impressed? Certainly not, because I haven’t
really told you much. If I told you that I could do it in 10 seconds,
while your car needed 15 seconds to do the same job, you might be
impressed.
After all, what we are really interested
in is the ability to cover distance in a period of time. The distance
from the exit of one corner to the entry of the next, or the quarter
mile, or maybe even the distance from one stoplight to the next.
If we consider the rate of acceleration,
AND miles per hour, we have all three terms included in the definition
of horsepower. Time, distance, and force. Force is the rate of
acceleration, or the force of inertia. Time is in hours, and distance
is in miles.
So now instead of just considering the rate of acceleration arbitrarily, let’s include miles per hour.
And while we’re at it, let’s consider
the acceleration rate of two very different motors to illustrate the
importance of horsepower, and the absolute irrelevance of torque.
At one extreme we have a Honda F1 motor
which revs to 18,000 rpm, makes nearly 800 horsepower, but a measly 281
ft.-lbs. of torque. At the other end of the spectrum we have the
Cummins turbo diesel available in the 2003 Dodge Ram which makes a
whopping 555 ft.-lbs. of torque, but only 305 horsepower. So which one
do you think will accelerate faster?
Everyone that you ask will answer that
the Honda F1 engine will accelerate faster. Even your neighbor with the
big block who claims that torque is the key to going fast. If torque
were the determining factor, the Cummins diesel would win hands down.
So what gives?
Let’s calculate the acceleration rate
for both engines in a hypothetical 2500 lb. car using the transmission
from the 1995 RX-7 with a two-foot diameter tire. Since we know that an
F1 car will go 200 MPH, we will gear the car for that speed with both
motors.
Starting with the F1 engine which
redlines at 18,000 rpm, we need to calculate the required ring and
pinion ratio to achieve 200 mph at redline in 5th gear.
First we convert miles per hour to miles per minute by dividing by 60.
200/60=3.333 miles per minute
The tire diameter is rated in feet, so
we must convert this 3.333 miles per minute into feet per minute. There
are 5280 feet in a mile, so:
3.333 X 5280 = 17,600 feet per minute
If our tire is 2 feet in diameter, the circumference is 2 feet times Pi
2 X 3.14159 = 6.28318 feet per revolution.
Now we divide the feet per minute, by the feet per revolution and we get:
17,600 / 6.28318 = 2801.13 tire revolutions per minute to achieve 200 MPH.
The engine redlines at 18,000 RPM, and in 5th
gear the transmission ratio is .719 to 1. To determine the RPM of the
output shaft at redline, we take the engine RPM divided by the gear
ratio to get:
18,000 / .719 = 25,034 RPM
The output shaft is spinning 25,034 RPM,
and we need the wheel to spin 2801.13 RPM to go 200 MPH. To find the
correct ring and pinion ratio, we divide the output shaft RPM by the
required tire RPM and we get:
25,034 / 2801.13 = 8.937 to 1
Going through all the same boring math
for the Cummins diesel which is only spinning 3000 RPM at redline, we
get a required ring and pinion gear of 1.489 to 1 to go 200 MPH at
redline in 5th gear.
(Note that the F1 engine is spinning 6
times faster than the Cummins, and so the required ring and pinion
ratio is exactly 6 times higher.)
With the transmission in first gear,
both vehicles will be traveling at 13.76 miles per hour at the bottom
of their powerband. (1,000 RPM for the Cummins, and 6,000 rpm for the
Honda.) The following chart shows the acceleration rate of both engines
in our hypothetical vehicle from that point to 200 MPH.
Note that at any point on the
chart, the percent difference in the rate of acceleration is EXACTLY
the difference in horsepower. For instance, at 200 mph, the Honda F1
engine is accelerating at a rate of .572 G’s, while the Cummins diesel
is accelerating at a rate of .228 G’s. If we divide .228 into .572 we
get 2.5, and so the acceleration rate of the Honda is 2.5 times greater
than that of the Cummins.
The Cummins, at 3,000 rpm is making 305
horsepower, while the Honda is making 763 horsepower. The Honda is
making 2.5 times the power of the Cummins, which is exactly the
difference in the rate of acceleration. You can work this out at any
point on the chart, and you will find that this direct relationship
between horsepower and rate of acceleration always holds true.
I’m sure there is someone out there that
still thinks I’m off my rocker, but as I stated earlier, the laws of
physics are non-negotiable. After showing this article to a few people
for proofreading, one person stated that the results aren’t valid
because torque motors are for low rpm grunt, and aren’t meant to run at
200 MPH. Ignoring the fact that this is a ridiculous statement, let’s
consider what would happen if we geared both combinations for a top
speed of 100 MPH. It should occur to you that this would be a simple
matter of doubling the ring and pinion ratio, and that would be correct.
The end result is that the acceleration
figures would simply double across the board for both combinations. The
difference in acceleration would still match the difference in
horsepower, and ultimately the difference in performance would be the
same.
So there it is. Horsepower is the
determining factor in the rate of acceleration of any vehicle. The next
article will go into more detail, and show you how these simple
calculations can be used to choose appropriate gearing for any track.
Below are some useful definitions and formulas.
TORQUE IN LBS./FT. = (WEIGHT IN LBS. X LEVER ARM LENGTH IN FEET.)
1 HP = 550 LBS./FT./SEC.
1 HP = 33,000 LBS./FT./MIN.
HP = (TORQUE X RPM) / 5252
TORQUE = (HP X 5252) / RPM
TORQUE AT THE REAR WHEELS = (ENGINE TORQUE X TRANSMISSION GEAR RATIO X RING AND PINION RATIO)
ACCELERATIVE THRUST = (TORQUE AT THE REAR WHEELS / TIRE RADIUS IN FT.)
RATE OF ACCELERATION = (ACCELERATIVE THRUST / TOTAL VEHICLE WEIGHT.)
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