Could you please help me to find the interpolation

[other-side-of-d-moon]

Hello there!!

Could you please help me to find the value for such a point with respect of the distance. please check the picture bellow


I want find the value of point F respecting the distance between the other points.

Note: all points (a b c d ) for the 4 cases has magnitude, so i need to get the magnitude for point F .

deeply regards

 

[hshah8970]

Re: ## HELP ## interpolation

From what I see, you can make a very simple linear system out of the distances. Here is how:

Let us consider four variables:

a = distance of point F from point A
b = distance of point F from point B
c = distance of point F from point C
d = distance of point F from point D

Now we want to develop a relationship between these distances. I shall assume that you can find the values for a, b, c, and d from the four diagrams given. Simply solve the system as simultaneous equations.

Consider the following general equation:

k₁a + k₂b + k₂c + k₄d = 0 ; where k1, k2, k3, and k4 are constants of which the values we want to find out in order to develop a linear relationship between the distances. One by one, enter the values of a, b, c, and d from the four diagrams into these equations and you shall have four equations and four unknown constants which you can solve by method of elimination, matrices, or substitution - whichever you prefer.

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By the "value of point F," I have assumed you simply mean the distance of point F from other points. After developing a linear relationship between a, b, c, and d, you will be able to input values of any 3 distances and compute the fourth one.

 

[enjunear]

Hello there!!

Could you please help me to find the interpolation for such a point with respect of the distance. please check the picture bellow

**broken link removed**

I want find the value of point F respecting the distance between the other points

You can determine the location of F by using any of the other points using a method called triangulation.

1) You know the distance from point A to F, so you can draw a circle around A with a known distance (radius). YOu now can say that F will lie at some point on that circle (all points on the circle are the correct distance away from A).

2) You know the distance from B to F. Draw a circle around B using that distance. The circle for A and B should meet at two points. F will exist at one of those two intersections, since F must lie on both of the circles for A and B.

3) You know the distance from C to F. Draw a circle around C using that distance. The A, B, and C circles should intersect at one point. That is where F is located. If you drew the distance-circle for the last point (D), it should also intersect at point F.

Mathematically, you could write the equation for a circle with the known distance from three points, then solve the equations simultaneously (read: algebra workout).


Amateur radio operators (HAM's) use a similar method in fox-hunt competitions. In that competition, you use a receiver and antenna to locate a hidden transmitter. By determining the signal power and approximate direction from several sample points, the operator can do triangulation and determine where the transmitter is located (power is related to distance from the source, and the antennas will give you a reasonably narrow region (say 10-20 degree arc).

GPS uses a similar technique, expect that it needs four reference points because each reference point creates a 3D sphere w/ constant radius. 2 satellites give you the intersection of two spheres, which is an ellipse. 3 satellites will give you two points, and 4 satellites gives you the single common point. More satellite signals just help improve the accuracy because of propagation delay and other signal-degrading effects caused by the atmosphere.

 

[other-side-of-d-moon]

You can determine the location of F by using any of the other points using a method called triangulation.

1) You know the distance from point A to F, so you can draw a circle around A with a known distance (radius). YOu now can say that F will lie at some point on that circle (all points on the circle are the correct distance away from A).

2) You know the distance from B to F. Draw a circle around B using that distance. The circle for A and B should meet at two points. F will exist at one of those two intersections, since F must lie on both of the circles for A and B.

3) You know the distance from C to F. Draw a circle around C using that distance. The A, B, and C circles should intersect at one point. That is where F is located. If you drew the distance-circle for the last point (D), it should also intersect at point F.

Mathematically, you could write the equation for a circle with the known distance from three points, then solve the equations simultaneously (read: algebra workout).


Amateur radio operators (HAM's) use a similar method in fox-hunt competitions. In that competition, you use a receiver and antenna to locate a hidden transmitter. By determining the signal power and approximate direction from several sample points, the operator can do triangulation and determine where the transmitter is located (power is related to distance from the source, and the antennas will give you a reasonably narrow region (say 10-20 degree arc).

GPS uses a similar technique, expect that it needs four reference points because each reference point creates a 3D sphere w/ constant radius. 2 satellites give you the intersection of two spheres, which is an ellipse. 3 satellites will give you two points, and 4 satellites gives you the single common point. More satellite signals just help improve the accuracy because of propagation delay and other signal-degrading effects caused by the atmosphere.

Tnx enjunear for help but i wanna say " i am not expert in mathematical ", i am only 18 years old. tnx

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From what I see, you can make a very simple linear system out of the distances. Here is how:

Let us consider four variables:

a = distance of point F from point A
b = distance of point F from point B
c = distance of point F from point C
d = distance of point F from point D

Now we want to develop a relationship between these distances. I shall assume that you can find the values for a, b, c, and d from the four diagrams given. Simply solve the system as simultaneous equations.

Consider the following general equation:

k₁a + k₂b + k₂c + k₄d = 0 ; where k1, k2, k3, and k4 are constants of which the values we want to find out in order to develop a linear relationship between the distances. One by one, enter the values of a, b, c, and d from the four diagrams into these equations and you shall have four equations and four unknown constants which you can solve by method of elimination, matrices, or substitution - whichever you prefer.

- - - Updated - - -

By the "value of point F," I have assumed you simply mean the distance of point F from other points. After developing a linear relationship between a, b, c, and d, you will be able to input values of any 3 distances and compute the fourth one.

Tnx a lot hshah8970 for your help.

i want mention that a,b,c,d,k1,k2,k3 and k4 are known. can you please tell me how to find F?? sir i am not expert in Math so pls can you make it simply. Tnx

 

[hshah8970]

If you know the values of a, b, and c then you will be able to accurately plot the location of F using the locus-drawing method that enjunear outlined. I'm guessing that the fourth distance - 'd' - will dictate the z-axis distance (into / out of the page) from point D.

 

[other-side-of-d-moon]

Thank you guys so much indeed, really you r very kind, However i want mention that a,b,c,d,k1,k2,k3 and k4 are known. can you please tell me how to find F? please help me guys

 

[albbg]

If we represent each ponit in a XY cartesian plane we can write the coordinates:

a --> (ax,ay)
b --> (bx,by)
c --> (bx,cy)
d --> (bx,dy)
F --> (Fx,Fy)

let the distances from each point to F be:

da=distance(a,F)
db=distance(b,F)
dc=distance(c,F)
dd=distance(d,F)

From the formula of the distance between two points:

(Fx-ax)^2 + (Fy-ay)^2 = da^2
(Fx-bx)^2 + (Fy-by)^2 = db^2
(Fx-cx)^2 + (Fy-cy)^2 = dc^2
(Fx-dx)^2 + (Fy-dy)^2 = dd^2

developing the squares:

1. Fx^2+ax^2-2axFx+Fy^2+ay^2-2ayFy=da^2
2. Fx^2+bx^2-2bxFx+Fy^2+by^2-2byFy=db^2
3. Fx^2+cx^2-2cxFx+Fy^2+cy^2-2cyFy=dc^2
4. Fx^2+dx^2-2dxFx+Fy^2+dy^2-2dyFy=dd^2

subtracting now 2. from 1. and 4. from 3

ax^2-bx^2-2(ax-bx)Fx+ay^2-by^2-2(ay-by)Fy=da^2-db^2
cx^2-dx^2-2(cx-dx)Fx+cy^2-dy^2-2(cy-dy)Fy=dc^2-dd^2

this is a simple system 2 equation 2 unknown (Fx and Fy) you should be able to solve by yourself.

 

[other-side-of-d-moon]

Tnx a lot albbg for your huge effort for helping me and what you have done is fully clear but you forget that i need the magnitude of pint F not distance.

please help. thank you a lot

 

[hshah8970]

You cannot define a point in space with a single value. The "Value of point F" by itself does not make much sense. Please clarify your question.

Otherwise, this becomes a matter of F depending on a total of 8 variables:
1. Value of Point A
2. Distance of Point F from Point A
3. Value of Point B
4. Distance of Point F from Point B
5. Value of Point C
6. Distance of Point F from Point C
7. Value of Point D
8. Distance of Point F from Point D

If you are going to relate F to these variables in the simplest possible manner, then F will be a linear combination of these 8 variables; you will make a general equation as before but with 8 variables and their 8 constant co-efficients.

But that would be useless since you will get only 4 system equations (4 diagrams with values).

 

[albbg]

As said by hshah8970 your question is not clear. Could you please clarify better what are you asking ?

 

[enjunear]

Tnx a lot albbg for your huge effort for helping me and what you have done is fully clear but you forget that i need the magnitude of pint F not distance.

please help. thank you a lot

The location of F is given as the Cartesian coordinate pair (Fx, Fy). All of the points are converted to their X-Y pairs, such that you could plot them on a graph. Recall:

a --> (ax,ay)
b --> (bx,by)
c --> (bx,cy)
d --> (bx,dy)
F --> (Fx,Fy)

Using the two simultaneous equations that albbg derived:

ax^2-bx^2-2(ax-bx)Fx+ay^2-by^2-2(ay-by)Fy=da^2-db^2
cx^2-dx^2-2(cx-dx)Fx+cy^2-dy^2-2(cy-dy)Fy=dc^2-dd^2

...you plug in as much information as you can. You said that we know the locations of a, b, c, and d... (x, y) coordinates. You also said we know the distance from a, b, c, and d to point F (da, db, dc, dd).

We can now plug numbers into all of the locations indicated by the X symbol, below. Now your equations look like this:
X^2 - X^2 - 2(X-X)Fx + X^2 - X^2 - 2(X-X)Fy = X^2 - X^2
X^2 - X^2 - 2(X-X)Fx + X^2 - X^2 - 2(X-x)Fy = X^2 - X^2

By simplifying the equations (using the real numbers, not X's), you'd wind up something like this (used random numbers, just to illustrate what you might see):
equation #1) 5 - 6.2*Fx + 1.23 - 7.95*Fy = 43
equation #2) 3 - 3.9*Fx + 2.37 - 1.54*Fy = 12.2
(WARNING: I just pulled these numbers out of thin air, so they probably won't give legitimate values for Fx and Fy)

You can solve a simultaneous set of linear equations in two common ways. 1) Substitution, or 2) subtracting equations (elimination method)

For the substitution method, pick one of the two simultaneous equation and rewrite it such that you have Fx = <some equation with Fy>, or Fy = <some equation with Fx>. Next, you pick one of the rewritten equations (say Fx = <some equation with Fy> ) and plug it into the second simultaneous equation, in place of Fx. Now you have a really long equation with only Fy and some numbers. You do some algebra and work your way down to find Fy = <a number>. Once you find the value of Fy, plug it into back into one of the original simultaneous equations. Now you have an equation with Fx and some numbers. A little more algebra and you will find Fx = <a number>. You now have the x and y location of F, so plot (Fx, Fy) on your graph, and you're done.

For the elimination method , take a look at **broken link removed** to see how it's done. Their pictures are a lot better than what I can put in with this text editor.

P.S. The math needed to solve the two equations shown should be easily within your grasp. This is stuff that I recall learning in middle school/junior high school (e.g. Algebra II). It just looks ugly until you realize all of the locations where you can put in a real number.

 

[hshah8970]

The methods shared above would have been completely valid, had the points shown in the diagram been actual points in space; however, they aren't points. Instead, they are just values. The 'distance' between these 'points' should also be treated as values/variables.

If the OP is asking for the "value of F," then I'm guessing F has a fixed value, unlike the other variables. Keeping this in mind, we can come up with a relationship between F and all the variables involved.

Key for understanding equations:

A = value of 'point' a
B = value of 'point' b
C = value of 'point' c
D = value of 'point' d
a = distance of 'point' F from 'point' A
b = distance of 'point' F from 'point' B
c = distance of 'point' F from 'point' C
d = distance of 'point' F from 'point' D

Top left diagram:
F = k₁a + k₂A + k₃b + k₄B + k₅c + k₆C + k₇d + k₈D
F = 6k₁ + 3k₂ + 5k₃ + 9k₄ + k₅ + 7k₆ + 2k₇ + 2k₈

Top right diagram:
F = k₁ + k₂ + 4k₃ + 3k₄ + 2k₅ + 7k₆ + 3k₇ + 5k₈

Bottom left diagram:
F = k₁ + 2k₂ + 2k₃ + 4k₄ + 4k₅ + 5k₆ + 5k₇ + 8k₈

Bottom right diagram:
F = k₁ + 5k₂ + 2k₃ + 3k₄ + 5k₅ + 6k₆ + 4k₇ + k₈

Therefore our system of equations is:

F = 6k₁ + 3k₂ + 5k₃ + 9k₄ + k₅ + 7k₆ + 2k₇ + 2k₈
F = k₁ + k₂ + 4k₃ + 3k₄ + 2k₅ + 7k₆ + 3k₇ + 5k₈
F = k₁ + 2k₂ + 2k₃ + 4k₄ + 4k₅ + 5k₆ + 5k₇ + 8k₈
F = k₁ + 5k₂ + 2k₃ + 3k₄ + 5k₅ + 6k₆ + 4k₇ + k₈

You can see now that it is not possible to find out F through common simultaneous equation techniques because we have 8 variables and only 4 equations. If this is an assignment, you should point this out to your teacher. If this is something you are exploring on your own, look into Gauss elimination and matrix row reduction. You will need to assume the values of 3 variables, I think, and proceed with the solution.