How to find a parametric equation of uniform motion?

Question

requirement:

1.The trajectory of the parametric equation is a closed polygon. 2.Try not to use trigonometric functions and Mod.Only Abs is allowed.

For example, the three vertices of a triangle are:{0, 0}, {200, 0}, {0, 200}. I tried to solve the triangle trajectory parameter equation, but I failed.Here's the code:

f[t_] := a* Abs[b*t + c] - d*Abs[t + e] + ff*t + g
f2[t_] := a2*Abs[t + b2] - c2*Abs[d2*t + e2] + ff2*t + g2
sol = Solve[
  f[0] == 0 && f2[0] == 0 && f[600/(400 + 200 \[Sqrt]2)] == 200 && 
   f2[600/(400 + 200 \[Sqrt]2)] == 0 && 
   f[(3 (200 + 200 \[Sqrt]2))/(400 + 200 \[Sqrt]2)] == 0 && 
   f2[(3 (200 + 200 \[Sqrt]2))/(400 + 200 \[Sqrt]2)] == 200 && 
   f[3] == 0 && f2[3] == 0 && f[300/(400 + 200 \[Sqrt]2)] == 100 && 
   f2[300/(400 + 200 \[Sqrt]2)] == 0 && 
   f[(3 (200 + 100 \[Sqrt]2))/(400 + 200 \[Sqrt]2)] == 100 && 
   f2[(3 (200 + 100 \[Sqrt]2))/(400 + 200 \[Sqrt]2)] == 100 && 
   f[(3 (300 + 200 \[Sqrt]2))/(400 + 200 \[Sqrt]2)] == 0 && 
   f2[(3 (300 + 200 \[Sqrt]2))/(400 + 200 \[Sqrt]2)] == 100, {a, b, c,
    d, e, ff, g, a2, b2, c2, d2, e2, ff2, g2}, Reals](*0<=t<=3*)

I think there may be a mistake in the equations.A possible correct answer:

{100 (((Sqrt[2] + 1) Abs[Sqrt[2] - (Sqrt[2] + 2) t + 1])/(Sqrt[2] + 
       2) - (2 Sqrt[2] + 3) Abs[t + 1/Sqrt[2] - 1] + (Sqrt[2] + 2) t),
  100 ((3 Sqrt[2] + 4) Abs[
       t + 1/Sqrt[2] - 1] - (2 Sqrt[2] + 
        3) (Abs[Sqrt[2] - (Sqrt[2] + 2) t + 1] + 2 t - 2))/(Sqrt[2] + 
     2)}(*0<=t<=1*)

Related post.

Answer 1

First define triangle as an interpolation function depending on the arclength

p = {{0, 0}, {200, 0}, {0, 200},{0,0}} ; (* points*)
sp = Join[{0}, Accumulate@Map[Norm, Rest[p] - Most[p]]]
ip = Interpolation[MapThread[{#2, #1 } &, {p, sp} ],InterpolationOrder -> 1]

the parametric solution you are looking for is ip[v t]!

ParametricPlot[ip[(400 + 200 Sqrt[2]) t], {t, 0, 1}]
 

addendum

Piecewise gives a explicit solution

pcw[s_] := Piecewise[{
{p[[1]] (200 - s)/200 + p[[2]] s /200, 0 <= s < 200},
{p[[2]] (200 + 200 Sqrt[2] - s )/( 200 Sqrt[2]) +p[[3]] (s - 200   )/(200 Sqrt[2]) , 200 <= s < 200 + 200 Sqrt[2]}, 
{p[[3]] (400 + 200 Sqrt[2] - s )/200  +p[[4]] (s - (200 + 200 Sqrt[2]) )/200  , 200 + 200 Sqrt[2] <= s <= 400 + 200 Sqrt[2]} 
}]

pcw[(400 + 200 Sqrt[2]) t]

$\begin{cases} \left\{\left(400+200 \sqrt{2}\right) t,0\right\} & 0\leq \left(400+200 \sqrt{2}\right) t<200 \\ \left\{\frac{-\left(\left(400+200 \sqrt{2}\right) t\right)+200 \sqrt{2}+200}{\sqrt{2}},\frac{\left(400+200 \sqrt{2}\right) t-200}{\sqrt{2}}\right\} & 200\leq \left(400+200 \sqrt{2}\right) t<200+200 \sqrt{2} \\ \left\{0,-\left(\left(400+200 \sqrt{2}\right) t\right)+200 \sqrt{2}+400\right\} & 200+200 \sqrt{2}\leq \left(400+200 \sqrt{2}\right) t\leq 400+200 \sqrt{2} \end{cases}$

Answer 2

The following code was originally written by @flc:

pts = Transpose[{{0, 0}, {200, 0}, {0, 200}, {0, 0}}];
c = Prepend[#/Last@# &@Accumulate@Abs@Differences@# &[{1, I} . pts], 
   0];
f = FullSimplify[
  RealAbs[t - c] . 
   LinearSolve[Abs@Outer[Subtract, #, #] &@c, Transpose@pts]]
ParametricPlot[f, {t, 0, 1}]
Plot[Evaluate@Norm@D[f, t], {t, 0, 1}]

We obtained the second different correct answer:

{100 ((2 + Sqrt[2]) RealAbs[
      t] - (3 + 2 Sqrt[2]) RealAbs[-1 + 1/Sqrt[2] + t] + 
    RealAbs[1 + Sqrt[2] - (2 + Sqrt[2]) t]/Sqrt[2]), 
 100 ((2 + Sqrt[2]) RealAbs[-1 + t] + (1 + Sqrt[2]) RealAbs[-1 + 1/
       Sqrt[2] + t] - 
    1/2 (2 + Sqrt[2]) RealAbs[1 + Sqrt[2] - (2 + Sqrt[2]) t])}

The 2 function images is congruent within the definition domain [0,1], but not identical outside the definition domain [0,1].