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HELP!!!! Lift Drag Ratio (1 Viewer)
- Thread starter tooks94
- Start date
jonahpetrie
New Member
Doing these questions is all about balancing the forces. Lift to drag can be complicated (in an accelerating system) or basic in a static system
You will be given a multitude of forces to begin with, usually the weight, angle of attack, and thrust.
Since lift=weight. L=WsinX, where lift = mg sin (angle of attack)
Drag is a little more complex. Rember all the forces need to be balanced so that the total thrust is equal to the drag force plus the drag caused by the angle of attack.
Therefore 0=D-T+WsinX, where 0= drag force minus thrust + mgsinX.
The L
is simply the drag force divided by the lift force to get a number greater than1. i.e 1:1.65
Here is an example: An aircraft is making a decent at 4* to the horizon, at a uniform velocity. The thrust provided by the engines is 110kn and the mass of the plane is 7tonnes. Calculate the lift to drag ratio.
Let's work everything out in Kn to keep it constant.
out weight force = 7000x10=70Kn
Our angle = 90-X (in most cases) = 86* (can also use cos4*)
using the formula lift = 70sin86 = 69.829kn
Using our second formula
0=110(T)+70sin4(WsinX)-D
D=114.88
So, 69.829:114.88 = 1:1.65
You will be given a multitude of forces to begin with, usually the weight, angle of attack, and thrust.
Since lift=weight. L=WsinX, where lift = mg sin (angle of attack)
Drag is a little more complex. Rember all the forces need to be balanced so that the total thrust is equal to the drag force plus the drag caused by the angle of attack.
Therefore 0=D-T+WsinX, where 0= drag force minus thrust + mgsinX.
The L
is simply the drag force divided by the lift force to get a number greater than1. i.e 1:1.65Here is an example: An aircraft is making a decent at 4* to the horizon, at a uniform velocity. The thrust provided by the engines is 110kn and the mass of the plane is 7tonnes. Calculate the lift to drag ratio.
Let's work everything out in Kn to keep it constant.
out weight force = 7000x10=70Kn
Our angle = 90-X (in most cases) = 86* (can also use cos4*)
using the formula lift = 70sin86 = 69.829kn
Using our second formula
0=110(T)+70sin4(WsinX)-D
D=114.88
So, 69.829:114.88 = 1:1.65
Last edited:
jonahpetrie
New Member
I appolgise for any misinfo on the last post. Not that your lift will = WcosX = 70cos4, drag will be using WsinX, 0= T+WsinX- D
wingman
Let's Rock N' Roll
An aerofoil (e.g. airplane wings) generates lift due to its shape and air rushing past it. However, the aerofoil also creates drag. Obviously you would want to create more lift and less drag to stay efficient.
Remember, drag is always the opposite direction of travel (in most exam questions forward direction of plane) for example left or right of the page. But lift is not entirely upwards, it is usually up and forward direction. So the lift is to counteract the weight force (downwards) and also drag. Thrust is also another component that counteracts the drag.
Why is Lift/Drag ratio important? fuel economy. You design the wings on an aeroplane to use least amount of fuel at it's cruising speed, so engineers tweak the lift/drag ratio of the aerofoil to have minimal drag at cruising speed & altitude.
This diagram may help (but its a glider so has no thrust in this example):
http://wright.nasa.gov/airplane/dlrat.html
Remember, drag is always the opposite direction of travel (in most exam questions forward direction of plane) for example left or right of the page. But lift is not entirely upwards, it is usually up and forward direction. So the lift is to counteract the weight force (downwards) and also drag. Thrust is also another component that counteracts the drag.
Why is Lift/Drag ratio important? fuel economy. You design the wings on an aeroplane to use least amount of fuel at it's cruising speed, so engineers tweak the lift/drag ratio of the aerofoil to have minimal drag at cruising speed & altitude.
This diagram may help (but its a glider so has no thrust in this example):
http://wright.nasa.gov/airplane/dlrat.html
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