Rate of Climb Calculations (w/ 4 engine jet powered aircraft)

waealu
Messages
36
Reaction score
0
I have this problem and it seems like it has a simple solution, however I can't seem to figure it out.

Using the following values and the previous equation calculate the rate of climb for a
typical airplane at an altitude of 8000m.

4 Engines each with 150kN of thrust.
Aircraft mass is 330 tonnes.
Drag is 500kN
Aircraft Mach number is 0.6.

I know I need to use something like dh/dt=((T-D)V)/W , but I seem to have a problem. I can figure out the airspeed, and get something like dh/dt=24,648/W , but I'm not sure where to go from there. I am not sure of the measurement units or if I need to convert anything.

Thank you
 
Physics news on Phys.org
Wellll... your question wasn't given too clearly. For us to help you, you should probably give more details. Regardless, I'm going to try to guess what you mean.

I'm assuming that the plane in facing at a slightly different angle from the one it is actually moving in.

So let \alpha be the angle from the horizontal to the plane nose, and let \theta be the angle from the horizontal to the direction of the planes velocity.

Let F_T be the thrust force (total--sum of forces from engines) and let F_D be the drag force. Then, you have to solve the following equations:
<br /> F_T \cos{\alpha} - F_D \cos{\theta} = 0<br />
<br /> F_T \sin{\alpha} - F_D \sin{\theta} - Mg = 0<br /> <br />
 
Thanks for the help, but I looked over the problem yesterday and figured it out. It was easier that I originally thought. However, I forgot to return to the forum to remove the question. Thank you for the response, though.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top