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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?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.
Well, from one mathematically challenged DS to another, here's what I make of it in practical terms.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...
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"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?
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'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"
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.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.
Isn't this similar to a formula used to calculate impact forces?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)
1013 is atmospheric at sea level right?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.
GotchaWhen a pressure (actually a pressure difference) is applied to an area, you get a force.
(force) = (pressure) x (area)
So pascals x area = newtons; and density in kg/m^3 and speed in m/s = Pa?In SI units, pressure in Pascals times area in square metres gives a force in Newtons.
OkaySo 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)
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.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
Okay, is the rest right?No, it's an equivalent concept, applied to lift instead of drag.
And what's the formula for this?That's why my comment at the end of the last post about the native language of aerodynamics being Mathematics, not English.
Cheers
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.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
That's a good point, I guess different analogies have to be used for different people since we're all different.No, it is easier for you to wrap your mind around
That'll workyour analogy doesn't work for me because my physics teacher used a model railway.
Thats's an interesting way of looking at it! Honestly I think they should just call a coulomb an ampere secondA 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.
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.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
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.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
So to recap, pressure is in Pascals, fluid density is in kg/m^3, speed is m/s?(pressure) =0.5 x (fluid-density) x (speed) x (speed)
And that's Newtons = Pascals x m^2?(force) = (pressure) x (area)
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
or Cl = L / (A * .5 * r * V^2), where
- L = Lift
- q = dynamic pressure
- A = Wing-area
In metric, lift is in kilograms, area in cubic meters, air-density in pascals or kg/m^3, and velocity in m/s?
- L = Lift
- A = Area
- r = Air Density
- V = Velocity
What figures would you use in imperial, since that's often used a lot as well?
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
or Cl = L / (A * .5 * r * V^2), where
- L = Lift
- q = dynamic pressure
- A = Wing-area
In metric, lift is in kilograms, area in cubic meters, air-density in pascals or kg/m^3, and velocity in m/s?
- L = Lift
- A = Area
- r = Air Density
- V = Velocity
What figures would you use in imperial, since that's often used a lot as well?