Abstract and Applied Analysis

Volume 2014 (2014), Article ID 459790, 7 pages

http://dx.doi.org/10.1155/2014/459790

## An Efficient Computation of Effective Ground Range Using an Oblate Earth Model

^{1}Department of Mathematics, King Abdulaziz University, Jeddah 21589, Saudi Arabia^{2}Department of Mathematics, Government College University, Lahore 54000, Pakistan^{3}Department of Mathematics, Government College University, Faisalabad 38000, Pakistan

Received 12 May 2014; Revised 22 July 2014; Accepted 4 August 2014; Published 27 August 2014

Academic Editor: Dumitru Baleanu

Copyright © 2014 Dalal A. Maturi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

An effcient method is presented to calculate the ground range of a ballistic missile trajectory on a nonrotating Earth. The spherical Earth model does not provide good approximation of distance between two locations on the surface of Earth. We used oblate spheroid Earth model because it provides better approximations. The effective ground range of a ballistic missile is an arc-length of a planner elliptic (or circle) curve which passes through the launch and target points on the surface of Earth model. A general formulation is presented to calculate the arc-length of an elliptic (or circle) curve which is the intersection of oblate Earth model and a plane. Explicit formulas are developed to calculate the coordinates of center of the ellipse as well as major and minor axes which are necessary ingredients for the calculation of effective ground range.

#### 1. Introduction

We developed a method to calculate the distance between two points constrained to lie on the surface of the oblate spheroid. We assume the definition of distance between launch and target points (the launch and target points are just two given arbitrary points) to be the length of the curve resulting from the intersection of the given oblate spheroid with a plane which passes through the launch point and the target point. There are infinite numbers of planes which pass through the launch and target points, for example, (1) the plane which passes through the normal at launch point (trajectory plane) and (2) the plane which passes through the geometrical center of the oblate spheroid (geocentric plane). All these planes can be obtained from a single plane by rotating it about the line joining the launch and target points. Therefore, we take a general plane which passes through the two given points on the oblate spheroid. We see that the trace of the oblate spheroid in the general plane is an ellipse. We find the semimajor axis, semiminor axis, and center of the ellipse. We also find the unit vector along the major axis of the ellipse. By calculating the position vectors of the launch and target point with respect to the center of the ellipse, we calculate the angles which these vectors make with major axis of the ellipse. Finally, we find the smaller arc length of the ellipse between these two angles which is the surface range between the given two points.

Many researchers have investigated numerical methods related to ballistic missiles and satellite launch vehicles. In [1], the authors discussed computation of the different errors in the ballistic missiles range. Estimation and prediction of ballistic missiles are discussed in [2]. Some recent research work about ballistic missiles and satellite launch vehicles can be found in [3–18]. Escobal [19] and Nguyen and Dixson [20] have formulated the problem to calculate range of a ballistic missile over the none-rotating oblate spheroid Earth model. Both authors define the elliptic curve over the surface of oblate Earth model by the intersection of a plane which passes through the launch, target points, and center of the oblate spheroid. Clearly in this case, the center of the ellipse (if the cutting plane is parallel to equatorial plane then resulting intersecting curve is a circle.) is the center of oblate spheroid. Once they know the center, it is easy to use rotational transformation matrix to calculate the orientation of major and minor axis. But in reality, the trajectory plane of ballistic missile passes through the normal vector at launch (or target) point to the surface of Earth model and the target point (or launch point). It is not necessarily true that trajectory plane passes through the center of oblate spheroid. It means, we require a general formulation of the problem to calculate the elements of general intersecting ellipse. Figure 1 shows two intersecting ellipses passing through launch and target points.

#### 2. Trace of the Oblate Spheroid in the Plane

General equation of a plane is given by where is a vector normal to the plane and is the distance of the plane from the origin of coordinates. The equation of the oblate spheroid can be written as where is the semimajor axis and is the semiminor axis of the oblate spheroid. The trace of the oblate spheroid (2) in the plane (1) is calculated as follows. In (1), for we have If , then is parallel to the equatorial plane, that is, polar axis (0, 0, 1). Substituting this value of in (2) which is the equation of a conic. We know that a general equation of a conic is This equation represents an ellipse or a circle if its discriminant . The discriminant of (5) is which is always less than zero. Hence, (5) represents an ellipse or a circle. Thus, we see that the trace of the oblate spheroid (given by (2)) in the plane (given by (1)) is an ellipse.

#### 3. Center of the Ellipse

Let be the position vector of the center of the ellipse. If is any point on the ellipse, then can be parametrized as
where . Since point also lies on the oblate spheroid, it must satisfy (2)
Equation (10) is quadratic in . If is any point on the ellipse, then there must be another point on the ellipse such that
Thus, (10) should give us two values of , both having the same magnitude but opposite in sign. It means that coefficient of in (10) should be zero:
Let be another point on the ellipse such that
and is perpendicular to . Doing the same steps as done for (12), we have
Dividing (14) by (12), we have
which can be written as
Since and lay in the plane of the ellipse, therefore we can write
Using these relations, (16) can be written as
Using (19) in (12), we have
Since is perpendicular to , therefore
Using the above equation in (20), we have
Since center of the ellipse lies in the plane given by (1), therefore it should satisfy (1); that is,
Using (19) and (22), the above equation becomes
since is a unit vector. Therefore
Using this equation in the above equation for
where is the eccentricity of the oblate spheroid. Substituting the value of in (19) and (26), the center of the ellipse is given by
*Special Case*. If the plane given by (1) passes through the center of the oblate spheroid, then
and ; that is, the center of the oblate spheroid will also be the center of the ellipse.

*Semimajor Axis of the Ellipse*. To find the semimajor axis of the ellipse, we take an arbitrary point on the ellipse whose position vector can be parameterized as
where is the unit vector; this implies that
Since point also lies on the oblate spheroid, therefore it must satisfy (2):
The roots of this equation (value of ) will have the same magnitude but opposite in sign; therefore, the coefficient of is zero. Thus, the above equation becomes
For particular, , will be maximum and value of will be semimajor axis of an ellipse. To find the maximum value of with respect to , we differentiate above equation with respect to
and set
Since is perpendicular to , therefore
Since , this implies that
Since
this implies that
Thus,
For caculated , the maximum value of is
Using (27),
This is the semimajor axis and the direction of the major axis is given by
*Semiminor Axis of the Ellipse*. Let be a unit vector along the minor axis of the ellipse; then
Thus, can be written as
Using (42) we have
If is the semiminor axis of the ellipse, then point on the ellipse which is the closest to the center of the ellipse is given by
Since this point also lies on the oblate spheroid, therefore
This equation is quadratic in . Thus, the above equation should give us two values of , both having the same magnitude but opposite in sign. It means that coefficient of in this equation should be zero. The above equation then becomes
which is the semiminor axis of the ellipse. When the semimajor axis, semiminor axis, and center of the ellipse passing through the launch and target points are known, we can calculate the distance between launch and target points (which will be equal to the arc length between these two points of the ellipse) in the following manner. The equation of the ellipse having semimajor axis and semiminor axis is given by
The unit vectors along the major and minor axis are and , respectively, and the centre of the ellipse with respect to Earth center Earth fixed frame (ECEF) is . Let be the position vector of launch point and let be the position vector of target point. Here is the geodetic latitude, is the longitude of the launch point, is the geodetic latitude, and is the longitude of the target point; then the Cartesian coordinates of the launch and target points in ECEF frame are
where and are reduced latitudes of the launch and target points, respectively, given by
Also the position vectors and can be written as
where , are the reduced latitudes of launch and target points, respectively, with respect to ellipse:
Dividing (54) by (55), we have
Similarly
To calculate arc length of the ellipse given by (49) between and , any point on the ellipse can be written as
The arc length between and on the ellipse is given by
which is the required distance between launch point and target point. There are two possible distances between and depending on the direction. In order to take the shorter distance, we replace the greater angle by if the difference between and is greater than .

#### 4. Numerical Simulations

In numerical experimentation, we conducted several tests to calculate the effective ground range between launch and target points using oblate Earth model. There are infinite number of ellipses that can pass through target and launch points. So we need an additional information to fix the position of an ellipse. In articles [1, 2], the authors assumed that ellipse passes through the center of Earth but in reality the trajectory plane either contains the normal at the launch or target. If ellipse plane contains the normal at launch or target, there is no compulsion to contain the center of Earth. We will address the calculation of effective ground range for both the above-described conditions; namely, an ellipse passes through the center of Earth or from the normal at launch or target. In order to perform the ground range calculations, we fix the geocentric longitude and latitude of a launch site A (−100°, 30°) and change the location of target site. The effective ground ranges between launch and targets are calculated by using two methods ND method [2] and proposed MF method. In ND method, the elliptic plane passes through launch point, target point, and center of Earth and for MF method elliptic plane is defined by using the launch point, target point, and normal at the launch site which is normal to oblate spheroid Earth model. Table 1 shows the ranges difference in meters because the generated ellipses for both cases are different. One can expect the change range in couple of kilometers. We do emphasis the fact that the defining ellipse should contain the local normal at launch site or target site. The assumption that the elliptical plane should pass through the center of Earth provides good results but true condition generates slightly different values of effective ground range. Tables 2, 3, and 4 depict majors axis, minor axis, and center of ellipses for both methods. Clearly in our case, the intersecting ellipse has center different than oblate spheroid Earth model center which is .

#### 5. Conclusion

We have constructed explicitly the elements of intersecting ellipse, namely, the center, major, minor axis lengths, and their directions. The proposed method also provides the straight forward mechanism to plot the trace of intersecting ellipse on the oblate spheroid Earth model. The methods proposed in [19, 20] are the subcases of our developed method because ellipse always has to pass through launch and target sites and the third information one can define according to requirement which helps to define the normal to elliptical plane. The trajectory of ballistic missile lies in plane which contains the local normal to oblate spheroid Earth model either at launch site or target site. The trajectory plane which passes through the center of Earth, generally, does not contain normal at launch site or target site. We have shown that our numerical simulations provide the effective ground ranges in the case of normal trajectory plane which are different in couple of kilometers from Earth center passing trajectory plane.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under Grant no. (130-098-D1434). The authors, therefore, acknowledge with thanks DSR technical and financial support.

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