Coefficient of Lift

[Zipper730]

I'm looking for a basic term for coefficient lift in a non dictionary definition. This one I've seen a lot...

Coefficient of Lift: A dimensionless coefficient that relates the lift generated by a lifting body to the fluid density around the body, the fluid velocity and an associated reference area.

Can somebody put this into terms that a guy who dropped out of HS, got a GED, and then barely kept an unremarkable average in math?

Yes, I know I sound stupid, but I'm willing to learn...

 

[XBe02Drvr]

I'm looking for a basic term for coefficient lift in a non dictionary definition. This one I've seen a lot...
Can somebody put this into terms that a guy who dropped out of HS, got a GED, and then barely kept an unremarkable average in math?

Yes, I know I sound stupid, but I'm willing to learn...

Well, from one mathematically challenged DS to another, here's what I make of it in practical terms.
(rushing in where angels fear....) If you know all of the values for the factors affecting the interaction between airflow and an airfoil or other lifting body (such as air density, air velocity, angle of impact, etc), then the coefficient of lift for that airfoil or shape under those conditions is a value that can be inserted into the lift equation along with the other variables and the size of the shape to determine the lift that can be generated by that shape under those conditions. Confused yet?
There are a lot of moving parts here. Imagine an untapered rectangular wing with a specific NACA airfoil shape. For each degree angle of attack there will be a specific C sub L value that together with all the air fluid values and the size of the wing will give you the theoretical lift of the wing. None of these values are static, as they are constantly changing and inluenced by each other.
Reduce power, aircraft slows, lift drops off, aircraft settles, momentarily increasing angle of attack, increasing drag, slows some more, then aircraft pitches down and increases speed slightly as angle of attack decreases reducing drag and gravity starts to provide a little forward impetus. Now the plane is going a little faster than trim speed and the inverted airfoil in the horizontal stabilizer begins to pull the tail down, raising the nose up and pulling the plane into a shallow zoom climb, increasing the AOA and scrubbing off the excess speed to repeat the process. This is called a phugoid oscillation, and is a good illustration of the interplay of the variables in the lift equation. Left to it's own devices and with the trim left at its original cruise setting, but with the trottle reduced, the phugoid oscillations will gradually dampen out and settle down to a gradual descent at the original trimmed speed.
Does this help? I've tried to do this entirely in English and leave the Math out of it. I'm sure some of the engineering types on here will take exception to this approach, as some things (and this is one of them) can only be truly and fully understood in their native language, which is Mathematics. "Le Medicin Malgre Lui" just isn't as funny in English as in French.
Cheers,
Wes

 

[Zipper730]

Well, from one mathematically challenged DS to another, here's what I make of it in practical terms.
(rushing in where angels fear....) If you know all of the values for the factors affecting the interaction between airflow and an airfoil or other lifting body (such as air density, air velocity, angle of impact, etc), then the coefficient of lift for that airfoil or shape under those conditions is a value that can be inserted into the lift equation along with the other variables and the size of the shape to determine the lift that can be generated by that shape under those conditions. Confused yet?

That's kind of like explaining a volt as "the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points". Now a term I'd use, even though I know little is "a volt is sort of like the pressure in the line"

 

[pbehn]

That's kind of like explaining a volt as "the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points". Now a term I'd use, even though I know little is "a volt is sort of like the pressure in the line"

Zipper, if you take the equation V=IR it defines the relationship between voltage, current and resistance. If voltage is constant then and increase in current must mean a reduction in resistance. By substituting values you can see what voltage means and how it relates to current and resistance.

Take the equations involving coefficient of lift and you will see how it relates to the other values.

 

[KiwiBiggles]

OK, I'll give it a go.

When a fluid is moving, it has a dynamic pressure - you can feel this pressure when you stick your hand out of a car window. This pressure can be calculated by:

(pressure) =0.5 x (fluid-density) x (speed) x (speed)
​

In SI units, if you measure the fluid density in kilograms per cubic metre, and speed in metres per second, you end up with a pressure value in Pascals.

When a pressure (actually a pressure difference) is applied to an area, you get a force.

(force) = (pressure) x (area)​

In SI units, pressure in Pascals times area in square metres gives a force in Newtons.

So thinking of the drag on your hand when you stick it out the car window, perpendicular to the flow:

(drag-force) = (pressure) x (area-of-your-hand)
​

But the air is also flowing around your hand, and becoming turbulent, and doing all kinds of strange things. So we add a factor to the equation, which we call the coefficient of drag.

(drag-force) = (drag-coefficient) x (pressure) x (area)
​

Now although with a wing the airflow is more-or-less parallel to the chord, rather than perpendicular to it as it is when you stick your hand out of the car window, the relationship is the same. The lift force generated by the wing is the product of the dynamic pressure, the wing area and the coefficient of lift:

(lift-force) = (lift-coefficient) x (pressure) x (area)
​

Hope this helps.

 

[Zipper730]

Zipper, if you take the equation V=IR it defines the relationship between voltage, current and resistance. If voltage is constant then and increase in current must mean a reduction in resistance. By substituting values you can see what voltage means and how it relates to current and resistance.

That wasn't the point: The point was that the definition is not easily useful to a person who is just starting out and wants to learn.

The explanation I gave is much easier to grasp and wrap one's mind around

When a fluid is moving, it has a dynamic pressure - you can feel this pressure when you stick your hand out of a car window. This pressure can be calculated by: (pressure) =0.5 x (fluid-density) x (speed) x (speed)

Isn't this similar to a formula used to calculate impact forces?

In SI units, if you measure the fluid density in kilograms per cubic metre, and speed in metres per second, you end up with a pressure value in Pascals.

1013 is atmospheric at sea level right?

When a pressure (actually a pressure difference) is applied to an area, you get a force.
(force) = (pressure) x (area)

Gotcha

In SI units, pressure in Pascals times area in square metres gives a force in Newtons.

So pascals x area = newtons; and density in kg/m^3 and speed in m/s = Pa?

So thinking of the drag on your hand when you stick it out the car window, perpendicular to the flow:

(drag-force) = (pressure) x (area-of-your-hand)​

But the air is also flowing around your hand, and becoming turbulent, and doing all kinds of strange things. So we add a factor to the equation, which we call the coefficient of drag.

(drag-force) = (drag-coefficient) x (pressure) x (area)​

Okay

Now although with a wing the airflow is more-or-less parallel to the chord, rather than perpendicular to it as it is when you stick your hand out of the car window, the relationship is the same. The lift force generated by the wing is the product of the dynamic pressure, the wing area and the coefficient of lift:

(lift-force) = (lift-coefficient) x (pressure) x (area)​

So coefficient of lift is basically coefficient of drag turned 90-degrees?​

 

[KiwiBiggles]

No, it's an equivalent concept, applied to lift instead of drag.

 

[XBe02Drvr]

That wasn't the point: The point was that the definition is not easily useful to a person who is just starting out and wants to learn.

The explanation I gave is much easier to grasp and wrap one's mind around

As you have probably noticed by now fluid dynamics is much less susceptible to a simple conceptual explanation than is voltage, as there are so many more variables. That's why my comment at the end of the last post about the native language of aerodynamics being Mathematics, not English.
Cheers,
Wes

 

[Zipper730]

No, it's an equivalent concept, applied to lift instead of drag.

Okay, is the rest right?

That's why my comment at the end of the last post about the native language of aerodynamics being Mathematics, not English.
Cheers

And what's the formula for this?

 

[KiwiBiggles]

I couldn't find any way of putting equations into this editor, anyway.

 

[pbehn]

That wasn't the point: The point was that the definition is not easily useful to a person who is just starting out and wants to learn.

The explanation I gave is much easier to grasp and wrap one's mind around

No, it is easier for you to wrap your mind around, your analogy doesn't work for me because my physics teacher used a model railway. A carriage was a coulomb, the railway is the wire, the station/shunting yard is the resistor, the current is the speed of the train (carriages per second) and the voltage is how many people get off at the station. A burned out circuit is injured people being crushed or thrown off and this can be avoided by using faster more platforms, longer platforms or better platforms, on a wider gauge railway etc etc etc.

 

[Zipper730]

No, it is easier for you to wrap your mind around

That's a good point, I guess different analogies have to be used for different people since we're all different.

your analogy doesn't work for me because my physics teacher used a model railway.

That'll work

A carriage was a coulomb, the railway is the wire, the station/shunting yard is the resistor, the current is the speed of the train (carriages per second) and the voltage is how many people get off at the station. A burned out circuit is injured people being crushed or thrown off and this can be avoided by using faster more platforms, longer platforms or better platforms, on a wider gauge railway etc etc etc.

Thats's an interesting way of looking at it! Honestly I think they should just call a coulomb an ampere second

 

[pbehn]

That's a good point, I guess different analogies have to be used for different people since we're all different.
That'll work
Thats's an interesting way of looking at it! Honestly I think they should just call a coulomb an ampere second

But when teaching you need to teach what is happening, a Coulomb is a number of electrons, when this passes in a second its an Ampere.

 

[MIflyer]

This might help, a classic book that the University Of Southern California put back into print for its aircraft accident investigation course when the USN stopped using it:
Aerodynamics for Naval Aviators : Free Download & Streaming : Internet Archive

 

[XBe02Drvr]

This might help, a classic book that the University Of Southern California put back into print for its aircraft accident investigation course when the USN stopped using it:
Aerodynamics for Naval Aviators : Free Download & Streaming : Internet Archive

EXCELLENT BOOK!! We had to read it in Training Devices School in the Nav, and it demystified the mysteries, and in language we could understand.
Cheers,
Wes

 

[Zipper730]

(pressure) =0.5 x (fluid-density) x (speed) x (speed)

So to recap, pressure is in Pascals, fluid density is in kg/m^3, speed is m/s?

(force) = (pressure) x (area)

And that's Newtons = Pascals x m^2?

BTW: I was thinking about something, when professors or teachers explain something, they actually usually explain the formula and use a step by step example the first time. It's very hard to grasp some formulas unless you see the whole thing drawn out

 

[Zipper730]

So coefficient of lift has to do with the amount of lift produced by moving a given amount of area of wing a certain speed?

From what I found online, the formula is: Cl = L / (q * A), where

• L = Lift
• q = dynamic pressure
• A = Wing-area

or Cl = L / (A * .5 * r * V^2), where

• L = Lift
• A = Area
• r = Air Density
• V = Velocity

In metric, lift is in kilograms, area in cubic meters, air-density in pascals or kg/m^3, and velocity in m/s?
What figures would you use in imperial, since that's often used a lot as well?

 

[KiwiBiggles]

So coefficient of lift has to do with the amount of lift produced by moving a given amount of area of wing a certain speed?

From what I found online, the formula is: Cl = L / (q * A), where

• L = Lift
• q = dynamic pressure
• A = Wing-area

or Cl = L / (A * .5 * r * V^2), where

• L = Lift
• A = Area
• r = Air Density
• V = Velocity

In metric, lift is in kilograms, area in cubic meters, air-density in pascals or kg/m^3, and velocity in m/s?
What figures would you use in imperial, since that's often used a lot as well?

Lift is in Newtons, not kilograms. The kilogram is the SI unit of mass, not force. You really need to learn some basic physics before you can understand any of this.

Wing area is in square metres, not cubic metres.

Air density is in kilograms per cubic metre; this is absolutely not the same thing as a Pascal, which a Newton per square metre.

You got velocity right.

For Imperial, for the limited extent to which it has a consistent set of units, you could use:

Lift in pounds
Area in square feet
Air density in slugs per cubic foot
Velocity in feet per second

The slug is a fairly obscure unit, which is the mass which accelerates at one foot per second when a force of one pound is applied to it. It is equivalent to the kilogram, where the pound is equivalent to the Newton.

 

[swampyankee]

So coefficient of lift has to do with the amount of lift produced by moving a given amount of area of wing a certain speed?

From what I found online, the formula is: Cl = L / (q * A), where

• L = Lift
• q = dynamic pressure
• A = Wing-area

or Cl = L / (A * .5 * r * V^2), where

• L = Lift
• A = Area
• r = Air Density
• V = Velocity

In metric, lift is in kilograms, area in cubic meters, air-density in pascals or kg/m^3, and velocity in m/s?
What figures would you use in imperial, since that's often used a lot as well?

You need to use consistent units: in SI, newtons for force or weight, kg per cubic meter for density, square meters for area, meters per second for velocity. In US Customary, pounds for force or weight, slugs per cubic foot for density, square feet for area, and feet per second for velocity.

 

[pbehn]

Some confusion always cause by kilogrammes being a unit of mass and pounds being a unit of weight.