Radiodrome

The radiodrome ("Leitstrahlkurve", from Latin radius "ray" and Greek dromos "run, run"), or chase curve is a special flat curve . It describes the movement of a point following another point. Both points move at a constant, but not necessarily the same speed.

Construction principle of the straight radiodromes, x and y positive

The “straight radiodrome” describes the simple case in which the victim moves on a straight line. Pierre Bouguer first described it in 1732. It is one of the curves that are referred to with the common name " dog curve ", after a formulation of the question with a dog chasing its master. Pierre-Louis Moreau de Maupertuis soon expanded the problem to include any guide curves. This led to the definition of “general radiodromes”.

The curve typically occurs in tracking in the -Problemen robotics and dynamic simulations on ( tracking problems ).

General equation

Let be the movement of the tracked point and the follower curve. Then you have the equation ${\displaystyle A(t)=(A_{1}(t),A_{2}(t),A_{3}(t))}$ ${\displaystyle P(t)=(P_{1}(t),P_{2}(t),P_{3}(t))}$

${\displaystyle {\frac {(P(t)-A(t)).{\dot {P}}(t)}{\parallel P(t)-A(t)\parallel }}=1}$

for all points in time , where the scalar product means. This equation results from the equation ${\displaystyle t}$ ${\displaystyle .}$

${\displaystyle {\frac {(A(t)-P(t))}{\parallel A(t)-P(t)\parallel }}.{\frac {{\dot {P}}(t)}{\parallel {\dot {P}}(t)\parallel }}=1}$,

which describes that the tangent is parallel to the straight line through and (the scalar product is the product of the lengths of the vectors) and the condition . ${\displaystyle P(t)}$ ${\displaystyle A(t)}$ ${\displaystyle P(t)}$ ${\displaystyle \parallel {\dot {P}}(t)\parallel =1}$

Special radiodromes

Straight radiodromes

Education Act

Be the starting point of a "pursued" and the starting point of a "pursuer". ${\displaystyle A_{0}}$ ${\displaystyle P_{0}}$

If the point moves with the speed on a straight line, and the point always moves with the speed in the direction of the point , then a radiodrome runs through . ${\displaystyle A}$ ${\displaystyle v={\text{const.}}}$ ${\displaystyle P}$ ${\displaystyle w={\text{const.}}}$ ${\displaystyle A}$ ${\displaystyle P}$

Function equation in Cartesian coordinates

Furthermore let the speed ratio . ${\displaystyle k={\frac {v}{w}}}$

${\displaystyle A_{0}(0|0)}$at the origin , on the x-axis , A move along the y-axis . Then moves on the curve ${\displaystyle P_{0}(1|0)}$ ${\displaystyle P}$

${\displaystyle y(x)={\frac {1}{2}}\left({{1-x^{(1-k)}} \over (1-k)}-{{1-x^{(1+k)}} \over {(1+k)}}\right)\quad {\text{ für }}k\neq 1}$

${\displaystyle y(x)={\frac {1}{4}}\cdot \left({x^{2}}-\ln {x^{2}}-1\right)\quad {\text{ für }}k=1}$

The second case is called actual radiodromes . It represents the simplest special case.

Derivation

1. The following applies to the movement of a point with the speed on a function graph: Since the movement should run to the left here, i.e. it should decrease, it is negative. If w is to be represented by a positive value, then constant is used here . ${\displaystyle P}$ ${\displaystyle w}$ ${\displaystyle w={\frac {\mathrm {d} s}{\mathrm {d} t}}={\frac {\sqrt {\mathrm {d} x^{2}+\mathrm {d} y^{2}}}{\mathrm {d} t}}={\frac {{\sqrt {1+{\frac {\mathrm {d} y^{2}}{\mathrm {d} x^{2}}}}}\cdot \mathrm {d} x}{\mathrm {d} t}}={\sqrt {1+y'^{2}}}\cdot {\dot {x}}.}$ ${\displaystyle x}$ ${\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} t}}}$ ${\displaystyle {\dot {x}}=-{\frac {w}{\sqrt {1+y'^{2}}}},w=}$
2. The following also applies in principle: as well as . ${\displaystyle {\dot {y}}={\frac {\mathrm {d} y}{\mathrm {d} t}}={\frac {\mathrm {d} y}{\mathrm {d} x}}{\frac {\mathrm {d} x}{\mathrm {d} t}}=y'\cdot {\dot {x}}}$ ${\displaystyle (y'{\dot {)}}={\frac {\mathrm {d} y'}{\mathrm {d} t}}={\frac {\mathrm {d} y'}{\mathrm {d} x}}{\frac {\mathrm {d} x}{\mathrm {d} t}}=y''\cdot {\dot {x}}}$
3. Now travels up the axis at constant speed , so it has the value at the time . Then the tangent points from P to A on the graph we are looking for, and the tangent condition is obtained . This gives the differential equation : . ${\displaystyle A}$ ${\displaystyle v}$ ${\displaystyle y}$ ${\displaystyle t}$ ${\displaystyle A(0|v\cdot t)}$ ${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {y-vt}{x}}}$ ${\displaystyle y'\cdot x+v\cdot t=y}$
4. Differentiation according to supplies . With what was said under 2. it follows what is too simplified. ${\displaystyle t}$ ${\displaystyle (y'{\dot {)}}\cdot x+y'\cdot {\dot {x}}+v={\dot {y}}}$ ${\displaystyle y''\cdot {\dot {x}}\cdot x+y'\cdot {\dot {x}}+v=y'\cdot {\dot {x}}}$ ${\displaystyle y''\cdot {\dot {x}}\cdot x+v=0}$
5. If you replace after 1., you get ${\displaystyle {\dot {x}}}$ ${\displaystyle -y''\cdot {\frac {w}{\sqrt {1+y'^{2}}}}\cdot x+v=0}$
6. The solution is achieved with integration by the substitution therefore . It follows from this and by separating the variables to with . ${\displaystyle u=y',}$ ${\displaystyle u'=y''}$ ${\displaystyle u'\cdot {\frac {w}{\sqrt {1+u^{2}}}}\cdot x=v}$ ${\displaystyle {\frac {\mathrm {d} u}{\sqrt {1+u^{2}}}}={\frac {k}{x}}\cdot \mathrm {d} x}$ ${\displaystyle k={\frac {v}{w}}}$
7. Integrating provides ( see arsinh ), as well as back substitution and application of the definition formula of sinh  x, with C ₁ = e ^C , to: ${\displaystyle \operatorname {arsinh} (u)=k\cdot \ln x+C}$ ${\displaystyle y'={\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {1}{2}}\left[C_{1}\cdot x^{k}-{\frac {1}{C_{1}}}\cdot x^{-k}\right]}$
8. Then reintegration, taking into account C _2, yields: ${\displaystyle y={\frac {1}{2}}\left({\begin{matrix}\quad \\{\frac {C_{1}\cdot x^{(1+k)}}{(1+k)}}\\\quad \end{matrix}}-\left\lbrace {\begin{matrix}{\frac {x^{(1-k)}}{C_{1}\cdot (1-k)}}\\{\frac {\ln {x}}{C_{1}}}\end{matrix}}\right\rbrace \right)+C_{2}{\begin{cases}{k\neq 1}\\{k=1}\end{cases}}}$
9. Insertion of the starting values ​​of or provide the values ​​for C ₁ and C ₂ . ${\displaystyle y'}$ ${\displaystyle P}$

EW Weisstein gives in a closed parametric representation .

Remarks

• ${\displaystyle C_{1}>0}$, there ${\displaystyle C_{1}=e^{C}}$
• If , well , the chaser catches up with the chased person , so the graph has an intersection with the axis there. If , therefore , there is no catching up, the graph thus asymptotically approaches the -axis. ${\displaystyle w>v}$ ${\displaystyle k<1}$ ${\displaystyle P}$ ${\displaystyle A}$ ${\displaystyle y}$ ${\displaystyle w\leq v}$ ${\displaystyle k\geq 1}$ ${\displaystyle A}$ ${\displaystyle y}$
• If the starting direction is not normal on the guide line, other boundary conditions are obtained. The low point is calculated from . ${\displaystyle y'=0}$
• A suitable coordinate transformation must be carried out for a general position of the guide line .

example

Example radiodrome

${\displaystyle A}$being chased by at double speed, so . If you create a coordinate system with the origin and -axis in the direction of movement of , i.e. perpendicular to it through the -axis, then it may be in . is now moving towards the origin, the tangent of the radiodrome therefore has the slope . Inserting this into the equation from 7 gives with : Which leads to the quadratic equation with the solutions or , where only the positive solution can be used (see 1st remark). Inserted into the equation for from 8. one obtains: Inserting P (9 | 3.75) yields C ₂ = 5.25. This yields with In and the graph has a low point at and bring followers to the persecuted one. The length of the distance covered can also be easily calculated: with the antiderivative . The distance covered from from to the lowest point at is then . The horizontal tangent there shows and has the height (see above), so it has covered the distance , exactly half of , because it is half as fast as . From up sets off back half, so why in of is made. ${\displaystyle P}$ ${\displaystyle k=v/w=0{,}5}$ ${\displaystyle A}$ ${\displaystyle y}$ ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle x}$ ${\displaystyle P}$ ${\displaystyle P(9|3{,}75)}$ ${\displaystyle P}$ ${\displaystyle P}$ ${\displaystyle 3{,}75/9=5/12}$ ${\displaystyle x=9}$ ${\displaystyle {\frac {5}{12}}={\frac {1}{2}}\left[C_{1}\cdot 9^{0{,}5}-{\frac {1}{C_{1}}}\cdot 9^{-0{,}5}\right]={\frac {1}{2}}\left[3\cdot C_{1}-{\frac {1}{3\cdot C_{1}}}\right]}$ ${\displaystyle C_{1}^{2}-{\frac {5}{18}}\cdot C_{1}-{\frac {1}{9}}}$ ${\displaystyle C_{1}={\frac {1}{2}}}$ ${\displaystyle -{\frac {2}{9}}}$ ${\displaystyle y}$ ${\displaystyle y={\frac {1}{2}}\left({\frac {1}{3}}\cdot x\cdot {\sqrt {x}}-4\cdot {\sqrt {x}}\right)+C_{2}.}$ ${\displaystyle y={\frac {1}{2}}\left({\frac {1}{3}}\cdot x\cdot {\sqrt {x}}-4\cdot {\sqrt {x}}\right)+5{,}25}$ ${\displaystyle y'={\frac {1}{2}}\left[{\frac {1}{2}}\cdot {\sqrt {x}}-{\frac {2}{\sqrt {x}}}\right]={\frac {1}{4\cdot {\sqrt {x}}}}\left(x-4\right).}$ ${\displaystyle x=4}$ ${\displaystyle y=2{\frac {7}{12}}}$ ${\displaystyle x=0}$ ${\displaystyle y=5{,}25}$ ${\displaystyle P}$ ${\displaystyle A}$ ${\displaystyle P}$ ${\displaystyle {\sqrt {1+y'^{2}}}={\sqrt {1+{\frac {x^{2}-8x+16}{16x}}}}={\sqrt {\frac {x^{2}+8x+16}{16x}}}={\frac {1}{4\cdot {\sqrt {x}}}}(x+4)={\frac {\sqrt {x}}{4}}+{\frac {1}{\sqrt {x}}}}$ ${\displaystyle F(x)={\frac {x\cdot {\sqrt {x}}}{6}}+2\cdot {\sqrt {x}}}$ ${\displaystyle P}$ ${\displaystyle x=9}$ ${\displaystyle x=4}$ ${\displaystyle F(9)-F(4)=5{\frac {1}{6}}}$ ${\displaystyle A}$ ${\displaystyle y=2{\frac {7}{12}}}$ ${\displaystyle A}$ ${\displaystyle 2{\frac {7}{12}}}$ ${\displaystyle 5{\frac {1}{6}}}$ ${\displaystyle A}$ ${\displaystyle P}$ ${\displaystyle x=9}$ ${\displaystyle x=0}$ ${\displaystyle P}$ ${\displaystyle F(9)-F(0)=10{,}5}$ ${\displaystyle A}$ ${\displaystyle 5{,}25}$ ${\displaystyle A}$ ${\displaystyle A(0|5{,}25)}$ ${\displaystyle P}$

properties

• The line connecting corresponding and is tangent to the radiodromes. ${\displaystyle A}$ ${\displaystyle P}$
• Obviously, it is not negative for all if the starting point is above the axis. ${\displaystyle y}$ ${\displaystyle x}$ ${\displaystyle x}$

Analysis of the speed parameters ${\displaystyle k}$

${\displaystyle k\geq 1}$:

• At is faster than , the curve asymptotically approaches the -axis: The pursuer is slower and does not reach the pursued nor does he cross his path. ${\displaystyle k>1}$ ${\displaystyle A}$ ${\displaystyle P}$ ${\displaystyle y}$
• At the same speed ( ), the pursuer runs at an increasingly equal distance behind the pursued: The curve shows the limit value behavior of a " Traktrix ". ${\displaystyle k=1}$

${\displaystyle k<1}$:

• There is exactly one end point of the graph on the left edge of the definition set . The persecutor is faster than the persecuted and reaches him in a finite time. We call this point “meeting point” or “snap point”, the curve actually ends at the snap point. ${\displaystyle x=0}$

The case is trivial , namely a straight line. The pursuer is “infinitely” fast, or the pursued stands still. ${\displaystyle k=0}$

For rational the function degenerates to an algebraic curve - for example , this curve is of degree . ${\displaystyle k\neq 1}$ ${\displaystyle v,w\in \mathbb {N} }$ ${\displaystyle {\tfrac {vw+w\max\{v,w\}}{ggt^{2}(v,w)}}\quad {\text{falls}}\ v\neq w}$

Circle Radiodromes

Circle radiodrome (red), in which the pursuer catches up with the pursued person after one turn.

If the “pursued” moves on a circular line and the “pursuer” starts in the center, another version results.

If the pursued and the pursued have the same speed, the pursued will be overtaken "after an infinite amount of time", i.e. H. the distance between the pursuer and the pursued converges to 0.

If the following curve has a higher speed than the following curve, it will catch up with it in a finite time.

If the follower curve is slower than the tracked curve, it will approach a circle of smaller diameter.

See also

Commons : Curve of pursuit  - album with pictures, videos and audio files

Web links

• Simulating a chase with Java and tracking problems, menu: Dog Curves - Various Java applets
• Dog curves - detailed derivation
• Pursuit Curve - From MathWorld - Bibliography

Individual evidence

1. ↑ The follower curve should have constant speed and after a suitable choice of the units one can then assume. ${\displaystyle \parallel {\dot {P}}(t)\parallel \equiv 1}$
2. ^ MathWorld, op. Cit.
3. ↑ Michael Lloyd: Pursuit Curves , Academic Forum 24, 2006-07