Why is math important?
Math is important because it is able to explain many things that we observe but are unable to quantify.
While engineering may be simplified down to being little more than applied Math, there are some forms of math that may not necessarily make the full implementation.
The Basics: Distance, Time, and Speed
Drag Racing occurs on two different lengths of track: 1/8 mile (660 feet) and 1/4 mile (1320 feet). The tracks are equipped with timing systems, which are computers connected to sensors at specific places on the track.
1/8 mile tracks typically have four timing blocks in the middle: 60 Feet, 330 Feet, 594 Feet, and 660 Feet.
1/4 mile tracks typically have the same as 1/8 mile tracks, adding timing blocks for 1000 Feet, 1254 Feet, and 1320 Feet.
To start out simple, consider Speed: Distance per Time. Miles per Hour, Feet per Second, and Meters per second are all examples of speed.
Using the Time Slip shown to the right, can you calculate the average speed for each segment?
0 Feet to 60 Feet in 1.0080 Seconds = 60 / 1.0080 = 59.5 Feet per Second
60 Feet to 330 Feet in 1.6506 Seconds = 270 / 1.6506 = 163.5 Feet per Second
330 Feet to 660 Feet in 1.3714 seconds = 330 / 1.3714 = 240.6 Feet per Second
180 Miles Per Hour x [ 5280 Feet per Mile / 3600 Seconds Per Hour ] = 264 Feet per Second
The Speed Trap
When one looks at a time slip, the third thing a racer looks at (and usually the second thing for a spectator) is the speed.
Speed at a drag strip is measured as distance divided by time, and there are typically two speed traps- 594 feet to 660 feet and 1,254 feet to 1320 feet. While this system has been in place since the beginning of the drag racing timing system, it does have one small flaw: distance divided by time is only an average speed.
Math to the rescue!
Given the time slip to the right, the car traveled the 66 foot long speed trap averaging 180.51 Miles Per Hour, which converts to 264.748 feet per second.
Velocity = Distance / Time can be converted to Time = Distance / Velocity
V = d/t , (t)V = (t)d/t , tV = d , t = d / V
66 feet / 264.748 feet per second = 0.249 seconds
4.0300 - 0.249 = 3.781 seconds
It took the car 0.249 seconds to travel the 66 feet, entering the speed trap at 3.781 seconds and exiting at 4.0300 seconds.
When it comes to analyzing data, that is, looking at it up close....one needs to be able to read between the lines. Data is only what you make of it. If you have a table of numbers in 20 columns and 2000+ rows, and that is how you look at it, then what benefit does it serve? If you take the numbers and apply them in a useful form, then they become information. With that said, there is also data that you do not see. It is hidden in the spaces between the numbers, and the only way to access it is by math.
In the kind of racing I participate in, being consistent is more important than going fast. If the engine and driver can do the same thing 7 times in a row, even if the weather changes, the chance for success is far greater than simply trying to go as fast as possible every time. For the driver, it is pure focus. For the engine and crew chief, it is more about reacting to the other conditions: being able to anticipate what may change in the near future, as well as looking at the data to react to things that may have changed previously. What we battle most is changing weather and changing traction. Those issues are taken into detail on the Science and Engineering pages of this website.
Using Driveshaft Data
If you have read to the bottom of the Engineering page, you know about Tire Growth. One of the benefits of tire growth is that it makes you go faster. One of the hindrances of tire growth is that you cannot directly link the rotational velocity of the driveshaft (and therefore, tires) to the speed registered by the speed trap. However, there is a way to approximate tire growth.
Let us assume that the average speed occurred in the middle of the speed trap.
5.96 Seconds: 7435 RPM , 6.06 Seconds: 7457 RPM , 6.15 Seconds: 7466 RPM
The speed as recorded, 227.73 Miles Per Hour, converts to 334.004 Feet Per Second or 240,483 Inches Per Minute (Why this unit? Keep reading!)
Knowing the Linear Velocity and Rotational Velocity, we may build a relationship between the two. First, the Final Gear Ratio (in our case, 3.70:1) must be divided out of the driveshaft speed, to determine axle speed.
5.96 Seconds: 2009 RPM , 6.06 Seconds: 2015 RPM , 6.15 Seconds: 2018 RPM
Now, we can build a formula for the relationship:
- ( Revolutions per Minute ) x ( Inches per Revolution ) = ( Inches Per Minute )
- ( Inches Per Revolution ) = ( Inches Per Minute ) / ( Revolutions Per Minute )
- ( Inches Per Revolution ) = ( 240,483 Inches per Minute ) / ( 2015 Revolutions Per Minute )
- ( Inches Per Revolution ) = 119.35
Interesting fact: If you measure the tire while the car is sitting still, the circumference is only 111.5 inches! This means the tire is growing tall enough to add 7.85 inches of circumference (or approximately 2.5 inches of height).
Now that the circumference of the tire has been found, we may apply it to the known rotational velocity of the axle to approximate the linear velocity of the car entering and exiting the speed trap.
- 2,009 Revolutions Per Minute x 119.35 Inches Per Revolution = 239,774 Inches Per Minute ( 227.058 Miles Per Hour )
- 2,018 Revolutions Per Minute x 119.35 Inches Per Revolution = 240,848 Inches Per Minute ( 228.07 Miles Per Hour )
If you have read the Chemistry section of the Science page, you know that the recipe to power and efficiency is based around the notion of keeping the Air-Fuel ratio balanced.
Fuel is introduced to the engine through 18 very small nozzles as shown by this page in our logbook.
These nozzles, as well as the main bypass pill, control the pressure and volume of fuel and are measured in thousandths of an inch.
Pressure and volume function as an inverse relationship when in a closed medium. Pressure is generated when a fluid is pressed in a restrictive space - the smaller the orifice, the higher the pressure and less fluid that will flow. Likewise, as the orifice is enlarged, pressure will go down, and a higher volume of air will be able to pass.
A demonstration of this is blowing through a drinking straw. As you pinch the end, it is harder to blow (pressure increases) and less air comes out (volume decreases).
Area of a circle: A = π × r²
The nozzles that inject directly to each individual cylinder are staggered to yield the best air-fuel ratio (AFR) for each individual cylinder. One of the mechanical limitations in the design of engine induction systems is poor air distribution. In our case, more air is received by the front 2 cylinders, which then requires more fuel (larger nozzles).
Once the fuel distribution is optimized by the nozzle sizing, the only adjustment made is by the main bypass pill. However, knowing the entire area of fuel passage is required to know how much change to invoke in the pill.
Pill: 0.120 inch = 0.011310 sq in
4 nozzles @ 0.046 inch = 0.001662 sq in each
2 nozzles @ 0.048 inch = 0.001809 sq in each
4 nozzles @ 0.040 inch = 0.001256 sq in each
1 nozzle @ 0.036 inch = 0.0010179 sq in each
3 nozzles @ 0.034 inch = 0.000908 sq in each
4 nozzles @ 0.030 inch = 0.000707 sq in each
Total Nozzle Area: 0.02186 square inch
Total Fuel Orifice Area: 0.03317 square inch
For the weather on that page of the logbook, the air density was at approximately 99.91 percent, which is very good.
However, weather does not hold constant. What if we are at a different track under a meteorological low pressure system, and it is 80 degrees, 75% humidity, and the barometer is 29.50 inHg? The air density is then 97.8 percent - a change of 2.11 percent.
As a result, the desired action is to take approximately 2.11 percent of fuel away from the engine.
As the name implies, the main bypass pill controls how much fuel is bypassed away from the engine, and returned back to the fuel tank. As a result, to take more fuel away, we enlarge the pill.
0.120 × 1.0211 = 0.1225 inches
As a result, we could use either a 0.122 or a 0.123 pill for the new weather conditions.