Calculating Rate of Climb from T/W, L/D and Velocity

[TheArrow]

TL;DR Summary

Solving for RoC of a plane when T/W, L/D, and forward velocity are provided

Hello everyone,

I'm trying to solve a problem from a textbook suggested by a fellow member here:

Calculate the Rate of Climb of an airplane having a thrust-to-weight ($\frac{T}{W}$) ratio of 0.25 and a lift-to-drag ratio ($\frac{L}{D}$) of 15.0 at a forward velocity of 70 m/s (230 fps). Express $V_{c}$ in $m/s$.

To help me visualize the problem, I made the diagram below (not necessarily to scale or accurate, just an organizational aid):

The textbook gives a simplified equation of $\theta_{c}=\frac{T-D}{W}$ (Call it Equation 1.0) to solve for the angle of climb. If I substitute the values from the simplified fractions of $\frac{T}{W}=\frac{1}{4}$ and $D=1$ (from $\frac{L}{D}=\frac{15}{1}$), the climb angle comes out to be 0 degrees. Because $T-D=0$.

Another equation is provided to get the power available: $P_{avail}=TV$ (Equation 1.1). Since $T=1$, $P_{avail.}=70$.

$P_{req'd}=DV$. Substituting in $D=1$, results in $P_{avail.}=70$.

The equation for the rate of climb of theta finally is $W(V\theta{c})=P_{avail.}-P_{req'd}$. The text says this equation is derived from "multiplying through" (some clarification on what this term means in this context would be helpful), Equation 1.0. Which in this case, the solution would be 0 m/s. Which is in line with the solution to Equation1.0 (in that if the angle of climb is 0 degrees, it would have 0 m/s rate of climb since it isn't climbing). But this doesn't seem right (I suspect I need to solve for the real values of T, W, D, and L rather than the simplified fractions, but I'm not sure how). The text doesn't provide a solution so I have nothing to compare my work to so I have no real feedback on whether I'm approaching this correctly.

Hopefully someone with a bit more familiarity with this can give some feedback.

Thanks in advance!

 

[jrmichler]

Start by getting the free body diagram (FBD) correct. A correct FBD is needed in order to understand the equations.

First, translate the ratios into forces. The gravity force (mg) is vertically downward. The vertical components of the thrust and lift forces must equal $mg$ (the vertical forces must sum to zero). The horizontal components of the thrust, drag, and lift forces must sum to zero.

 

[TheArrow]

Okay, so I know that the components are as follows:

Vertical: $T sin(\theta) - D sin(\theta) + L cos(\theta) - mg = 0$
Horizontal: $T cos(\theta) - D cos(\theta) + L sin(\theta) = 0$

And I get why these have to sum to zero (since the plane is in equilibrium).

I'm just a little fuzzy on how I would translate the $\frac{T}{W}$ and $\frac{L}{D}$ ratios into forces. Or what I'm assuming you mean, finding the magnitude of those vectors from the ratios. I just haven't been exposed to working with forces within the context of only their ratios before with unknown magnitudes before (i.e. T = 300N, etc).

 

[Lnewqban]

 

[Lnewqban]

The textbook gives a simplified equation of $\theta_{c}=\frac{T-D}{W}$ (Call it Equation 1.0) to solve for the angle of climb. If I substitute the values from the simplified fractions of $\frac{T}{W}=\frac{1}{4}$ and $D=1$ (from $\frac{L}{D}=\frac{15}{1}$), the climb angle comes out to be 0 degrees. Because $T-D=0$.
...
The equation for the rate of climb of theta finally is $W(V\theta{c})=P_{avail.}-P_{req'd}$. The text says this equation is derived from "multiplying through" (some clarification on what this term means in this context would be helpful)...

You could simplify your equations by aligning the x-axis with the line of application of T and D, as well as by aligning the y-axis with the line of application of L (always perpendicular to the oncoming flow direction).

Note: In real life, T and the center line of the airplane are not always aligned with the oncoming flow direction, because required changes in the angle of attack of the wing, according to the forward velocity and required lift force.

By relocating your x and y axes, you will see that you will only need to calculate the x and y components of the airplane's weight.
Therefore, you will have:

For the x-axis:
$W\sin\Theta_{climb}=T-D$

And for the y-axis:
$W\cos\Theta_{climb}=L$

As you can see, for the case of steady climb, the value of the thrust or propelling force generated by the propeller must be greater than the drag generated by the required lift (induced drag) and by the aerodynamic resistance (parasitic drag).
There is a component of the total weight resisting the forward movement of the airplane.

For practical angles of climb between 1° and 10°, the value of \cos\Theta_{climb} is not greater than 0.98, and the value of \sin\Theta_{climb} is not greater than 0.17.
Because of that, you can establish an estimated relation between the given ratios including W and L and T and D.

Regarding clarification about equation $W\sin\Theta_{climb}V_x=P_{available}-P_{required}$
This equation is derived from "multiplying through" the equation of balanced x-forces by the forward velocity of the airplane along the x-axis.
Thus:
$W\sin\Theta_{climb}V_x=TV_x-DV_x$

In that case, $DV_x$ is the power (force times velocity) that is available from the propeller and that is needed to keep level flight.
In that case, $TV_x$ is the power that is available from the propeller and that is needed to keep a steady non-accelerated climb.

Because a propeller functions as a rotating wing, producing lift (thrust in this case) based on AOA, it also suffers from induced and parasitic drags.
The value of the generated thrust depends on its velocity of advance, which is always smaller than its geometrical pitch of distance per revolution.
That complication makes the propelling capability of propellers regarded in terms of the the HP's delivered by the engine's shaft rather in thrust force, like it is for turbojet engines.

I find this explanation and diagrams very easy to understand:
http://www.free-online-private-pilot-ground-school.com/aircraft_performance.html

"The most immediate interest in the climb angle performance involves obstacle clearance. The most obvious purpose for which it might be used is to clear obstacles when climbing out of short or confined airports.

The maximum angle of climb would occur where there exists the greatest difference between thrust available and thrust required; i.e., for the propeller-powered airplane, the maximum excess thrust and angle of climb will occur at some speed just above the stall speed.

Thus, if it is necessary to clear an obstacle after takeoff, the propeller-powered airplane will attain maximum angle of climb at an airspeed close to—if not at—the takeoff speed.
...
During a steady climb, the rate of climb will depend on excess power while the angle of climb is a function of excess thrust."