I am going to present the general analytical approach to this problem. The
book seems to use estimations based on previous figures but this is not
perfect. The general approach would be to make the parabola go through the
point Re, Ln and then to make it tangent to the circle.
Next just look at the parabola. It should be symmetric about the center
line so it will be in the form y = ax^2 + c. The equation of a parabola of
this form that goes through the point (Re, Ln) would be as follows Ln =
a(Re^2) + c => c = Ln - aRe^2. The equation of the parabola then becomes
y = ax^2 + Ln - aRe^2
This means that the parabola is now only dependent on one free parameter a.
We can now take the derivative as follows
dy/dx = 2ax
Write out the equation for the circle with radius .382Rt. To make it easier
flip the picture and rotate it so that the nozzle is pointing straight up
and in the first quadrant. This equation will take the form
(x - 1.382Rt)^2 + y^2 = (.382Rt)^2
You will need to solve for y:
x^2 - 2.764Rtx + 1.909924Rt^2 +y^2 = 0.145924Rt^2
y = sqrt(-x^2 + 2.764Rtx - 1.764Rt^2)
Taking the derivative:
dy/dx = (1.382-x)/sqrt(-x^2 + 2.764Rtx - 1.764Rt^2)
You can view the progress so far in graphical form here:
https://www.desmos.com/calculator/yj3jvlsmfy
We now need to pick an 'a' value such that the circle will be tangent to
the parabola.
Two conditions must be met at point N (I will refer to this point as (x_0,
y_0)):
y_circle = y_parabola => sqrt(-x_0^2 +
2.764Rtx_0 - 1.764Rt^2) = ax_0^2 + Ln - aRe^2
dy/dx_circle = dy/dx_parabola => 2ax_0 = (1.382-x)/sqrt(-x^2
+ 2.764Rtx - 1.764Rt^2)
These two equations can be solved for x_0 and a numerically with Matlab or
Wolfram Alpha I presume and then you have the function of your parabola as
y = ax^2 + Ln - aRe^2 where 'a' was solved for numerically.
Zachary Martinez
Aerospace & Mechanical Engineering
Missouri S&T
On Thu, Dec 1, 2016 at 6:58 PM, Doug Jones <djones@xxxxxxxx> wrote:
I thrashed on that problem off and on for more than a decade, and never
could replicate the method from the published literature. I use ONC from
Sierra Engineering, and I can guarantee you the shapes it generates are not
parabolas in any way, shape or form.
http://www.sierraengineering.com/ONC/onc.html
Doug Jones, Chief Test Engineer
XCOR Aerospace
1325 Sabovich
Mojave CA 93501(661) 824-4714 x117 <(661)%20824-4714>
cell 661 313-0584 <(661)%20313-0584>
On 12/1/2016 4:00 PM, Graham Sortino (Redacted sender gnsortino for DMARC)
wrote:
I was wondering if anyone could point me towards an example of how to fit
a function to G.V.R. Rao’s parabolic approximation of a bell nozzle given
the initial angle θn and corresponding start point as well as final angle
θe and point?
I’ve been following through the example on page 83 of Huzel and Huang
https://books.google.com/books?id=TKdIbLX51NQC&lpg=
PA76&ots=slcXbGfst7&dq=modern%20design%20initial%
20parabolic-contour%20wall%20angles&pg=PA83#v=onepage&q=
modern%20design%20initial%20parabolic-contour%20wall%20angles&f=false
... which uses θn=27.4 (x,y) = 21.9,12.99 and θe=9.8 (x,y)= 102.4,46.7 but
it leaves out an example of how this is actually solved. Presumably because
it is so trivial but I’m getting a bit stuck on it. I believe an example of
this can be found in the Rao paper "Approximation of Optimum Thrust Nozzle
Contour", however, I cannot seem to find a copy online so I was attempting
to work this out on my own.
I understand a parabola can be calculated via several methods including
via 3 points, which I could have if I use the top and bottom portions of
the curve or if I assume the vertex to be the middle of the throat.
However, I’m thinking that what is intended is that I’m supposed to use
only the two points and angles (eg. slopes) to calculate and it wasn’t
clear to me how this should be done.
Kind Regards,
Graham