Graham,
The parabolic nozzle approximation isn't (in general) parallel to the
axis: the parabola is not of the form f(x) = ax^2 + bx + c, but is
effectively rotated. It's probably better to understand it as a
generalized conic section, which has five free variables, and an
additional discriminant constraint (because it's known to be
parabolic). Interestingly, the discriminant constraint has a square
root, so you may find two parabolas depending on your method.
An interested mathematician might take particular note of the
properties of second-order Bézier curves.
- Peter Hokanson
On Thu, Dec 1, 2016 at 8:56 PM, Zachary Martinez <znm3m8@xxxxxxx> wrote:
I realize that I didn't explain the graph very well. Here is another one
https://www.desmos.com/calculator/ojflgzwncf.
The slider 'a' is our free variable that we are trying to adjust so that the
parabola is perfectly tangent with the circle. In the image it never quite
gets there but a should be somewhere between .9 and 1 for the defaults. I
added a few more sliders for convenience. You put the throat radius in the
slider T, the exit radius in the slider E and the Nozzle length in the
slider L. These are set to 1, 2, 3 respectively for the visualization demo.
Zachary Martinez
Aerospace & Mechanical Engineering
Missouri S&T
On Thu, Dec 1, 2016 at 8:45 PM, Zachary Martinez <znm3m8@xxxxxxx> wrote:
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
cell 661 313-0584
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
