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Old 03-10-2008, 10:46 AM   #1
My5ABaby
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Default Calculating Drag, Force, Lift, and Downforce

Here's a nifty Excel spreadsheet that will calculate drag power, force, downforce, lift, and aerodynamic drag coefficient. It's fun to play with.

The figures put in are for a S4 N/A (non sport for .31 drag coefficient). I believe all non-Sport models are .31, and Sport is .29.

You can change anything in the yellow boxes. However, I would only change for FC's - top to bottom:
All-up mass (published weight)
Speed (Air)
Altitude
Temperature

I ran across a few interesting Excel spreadsheets that calculate air pressure, air density, aerodynamic drag coefficient, drag power, force, downforce, lift, front/rear tire lift, weight on front/rear tire, total moving weight, weight loss total, and rolling resistance.

Here's my results::

To fight air resistance in my 86 sport I came out with (2600lbs, sealevel, .29 drag coefficient (86 Sport), and a frontal area of 2763in^2):

140 mph = 104.0 hp
150 mph = 127.9
160 mph = 155.3
170 mph = 186.3

Using other spreadsheets (http://www.mayfco.com/aero1.xls and http://www.mayfco.com/lsrts.zip), I found similar results (they added in road resistance).

At sealevel, 16.6 gallons of fuel, 185lb's of me, and 60 degrees out; the top speed for my 86 sport (assuming it was completely stock) is ~139mph.

Interesting information...

Here's a link to the thread http://www.msgroup.org/forums/mtt/to...p?TOPIC_ID=955 and the important part is quoted below.

Here's another interesting site that provides information to put into the Excel sheet http://www.mayfco.com/mazda.htm .

Drag and Lift Model



Quote:
Originally Posted by James R. Davis
The drag and lift model can be accessed and run by clicking on the link below. (Assuming, of course, that you have Microsoft Excel on your system.)


Drag and Lift Model


This is a very complicated piece of code. Not surprising since there are lots of concepts that need to be included which are not straightforward.

There are two kinds of drag: friction and pressure. Friction drag is essentially insignificant compared to pressure drag at motorcycle speeds so all I have calculated with this model is pressure drag.

Calculating the pressure drag force is a fluid dynamics problem. Air is treated as a liquid. The computation involves a standard algorithm (used even by the Wright brothers):

Drag force = (d * V^2 * Cx * S) / 2

Where:
d = air density
V = velocity
Cx = Drag coefficient
S = Frontal surface area

The first thing you should notice from this is that the drag force increases with the square of your speed. (By the way, in my calculations V is the speed of the wind hitting you from directly in front of you. So, if you are riding at 65 MPH and there is a head wind of 10 MPH, V is set at 75.)

The drag coefficient is a dimensionless number that represents, essentially, the form of the object that is being hit by the wind. A parachute would have a Cx on the order of 1.5, a ball would be more like .45, a wing would be on the order of .1 or .2, and a teardrop has a Cx of .05. Sportbikes have Cx values between .3 and .6, touring bikes have values between .4 and .9, and cars tend to have values of between .26 and .36.

The surface area presented to the wind is extremely important in this analysis. If you have a windscreen on your bike that adds a meaningful amount of area but please note that it is only the area of the windscreen that is OUTSIDE of the profile of the bike and rider that is added to the total.

Air density is a very difficult number to come up with because it is a function of altitude, temperature, and moisture content. Air density is highest at sea level and decreases with altitude. It increases as temperature goes down. Standard air pressure assumes sea level at 59 degrees F. Moisture decreases density (odd as that sounds.) I wrote the model assuming normal dry air.

Once the drag force is calculated all you need to do is multiply it by the velocity again and you get the amount of power involved (Power = Force * Velocity) – in other words, I wanted to see how much Horsepower was being consumed fighting wind drag.

Another thing you can do with force is determine the moment amount. That is, if you multiply the force by the length of a lever arm you can determine how hard that force is put to use. In our analysis I multiplied the drag force by the height of the Center of Wind Resistance (where the surface area above that height is exactly the same as the surface area below it) in order to calculate the downforce caused by the wind drag. Downforce acts as a torque. That is, all forward pointing forces are at the contact patch and wind drag is through the Center of Wind Resistance so that, like weight transfer caused by acceleration, we know how much weight is added to the rear tire (and removed from the front one.)

Pressure drag is vectored in line with the direction of the wind. Vectored at 90 degrees to the direction of the wind is another force called Lift. Lift is most difficult to calculate since it depends on a lift-coefficient very similar to the drag-coefficient used to calculate drag. The lift coefficient is a dimensionless number that reflects such things as the aerodynamic shapes involved, surface texture, and the relative height of obstructions compared to the length of the body following that obstruction. A windscreen, for example, creates a significant lift just aft of itself while the farther back you go the less lift is created. Every part of the surface area, from rider’s helmet to rear running lights, creates both drag and lift. The method used by this author to closely estimate the lift-coefficient was to use the drag-coefficient and multiply it by the height of the Center of Wind Resistance then divide by the length of the wheelbase. This number, in essence, indicates the efficiency of drag forces in the creation of lift.

As the drag coefficient already reflects the 'form' of the bike I chose to use it as the lift coefficient as well. Since lift could be larger or smaller than drag (for an airplane wing it is obviously much higher, for a bike - since it stays on the ground for all reasonable speeds - it would be lower. For those of you wondering why I do not use the lateral area for lift instead of the cross-section (frontal) area used for drag, for a car the lateral area is typically three times the cross-sectional area while on a motorcycle they are approcimately 1:1.

As a reasonable approximation of lift force I used the drag force divided by the ratio of height of the Center of Wind Resistance and the wheelbase. I found an equation in Mr. Gaetano Cocco's book, Motorcycle Design and Technology, that totally supports that approximation. He states that Lift for a motorcycle is the downforce divided by the wheelbase. That, as it turns out, is exactly what my calculation says.

The lift formula I used was:

Lift force = Drag force * H / Wheelbase

Where H is height of the Center of Wind Resistance.

Now, application of the downforce merely shifts weight from the front tire to the rear tire. If nothing else were going on the total weight of the motorcycle and rider would be unaffected. But there is a lift force in addition to the downforce. The question is: how should that lift be apportioned between the front and the rear tires? As an arbitrary choice I elected to use the location of the Center of Gravity along the wheelbase in the same way I used it to determine weight distribution of a stationary bike. Thus, for example, in a typical bike that has 62% of its weight on the rear tire, I allocated 62% of the lift to the rear tire as well.

The weight on the front tire is found as follows:

Front tire weight = Static weight on front tire – Downforce – Front tire lift

And the weight on the rear tire is found as follows:

Rear tire weight = Static weight on rear tire + Downforce – Rear tire lift.

One last subtlety was incorporated into the model - the strength of gravity is diminished the closer you are to the equator to account for the centrifugal force of rotation and the fact that the poles are slightly closer to the center of the earth than is the equator. The total difference in weight between an object at the poles and the same object at the equator is .5%. The difference is small, to be sure, but real. An 800 pound bike at the poles weighs 797.32 pounds in Houston and only 796.86 pounds in Mexico City when both latitude and elevation are input into the model.

To input latitude values please note that the model expects degrees only, not degrees and minutes. Thus, Mexico City, at 19 degrees 24 minutes North is input as 19.4 because 24 minutes is only .4 degrees.

[Please note that the model was upgraded on 3/14/2005 to correct the display of some conversion values only. The calculations used by the model and its output were correct.
Attached Files
File Type: zip DragLift.zip (19.5 KB, 7 views)

Last edited by My5ABaby; 02-17-2009 at 07:30 AM..
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