Fitting a curve provided from a table

Question

Code is as follows.

aSet1 = {0.05447969195001433`, {1.62844, 1.79419}};
aSet2 = {0.05448351166864853`, {1.63506, 1.78729}};
aSet3 = {0.054488286316941295`, {1.64428, 1.77771}};
aSet4 = {0.05449306096523404`, {1.65507, 1.76657}};
aSet5 = {0.05449783561352681`, {1.66863, 1.75266}};
aSet6 = {0.054502610261819566`, {1.68992, 1.73102}};
aSet7 = {0.05450356519147811`, {1.69803, 1.72284}};
aSet8 = {0.052521131220325465`, {1.03786, 2.5299}};
aSet9 = {0.042971834634811745`, {0.353918, 3.94334}};
aSet10 = {0.033422538049298026`, {0.102792, 4.96211}};
aSet11 = {0.023873241463784306`, {5.87487, 5.87487}};
aSet12 = {0.014323944878270583`, {6.74069, 6.74069}};
aSet13 = {0.00954929658551372`, {7.16397, 7.16397}};

Tck400 = {aSet1, aSet2, aSet3, aSet4, aSet5, aSet6, aSet7, aSet8, 
aSet9, aSet10, aSet11, aSet12, aSet13};

c400 = ListPlot[{Table[{Tck400[[ii, 2, 2]], Tck400[[ii, 1]]}, {ii, 1, 
 Length[Tck400]}], Table[{Tck400[[ii, 2, 1]], Tck400[[ii, 1]]}, {ii, 1, 
 Length[Tck400]}]}, Frame -> True, RotateLabel -> False, Axes -> False, PlotRange -> All}]

With the following output (plot label wasn't included above):

I would like to ditch the points and plot a line or somehow fit a line to this data. My only thought was changing the format of the data to simple (x,y) points. Anything yall think would be nicer? Thanks!

Here is my solution based on rhermans solution

Tck230 = {fSet1, fSet2, fSet3, fSet4, fSet5, fSet6, fSet7, fSet8, fSet9, fSet10, fSet11, fSet12, fSet13, fSet14, fSet15, fSet16};
line230 = SortBy[First]@Flatten[Distribute[Reverse@#, List] & /@ Tck230, 1];

Here is my solution based on user1066 solution

data230 = DeleteDuplicates[Catenate[{Table[{Tck230[[ii, 2, 2]], Tck230[[ii, 1]]}, {ii, 1, Length[Tck230]}], Table[{Tck230[[ii, 2, 1]], Tck230[[ii, 1]]}, {ii, 1, Length[Tck230]}]}]];
f230 = Interpolation@data230;
data230 // Plot[f230[x], {x, Min@(#[[All, 1]]), Max@(#[[All, 1]])}, PlotRange -> {{0, 7.5}, {0, 0.12}}, PlotStyle -> Red, Frame -> True, Axes -> False] &

This was done for all curves, and the following was done for the overlay as done above:

Answer 1

1. Joined -> True

If you just want to join the points:

With[
 {
  data = SortBy[First]@
    Flatten[Distribute[Reverse@#, List] & /@ Tck400, 1]
  },
 Show[
  ListPlot[
   data
   , PlotStyle -> Blue
   , InterpolationOrder -> 2
   , Joined -> True
   , PlotTheme -> "Scientific"
   , PlotRange -> All
   ]
  ]
 ]

2. FindFormula

If you don't have a model, you can try to guess one using FindFormula

With[
 {
  data = SortBy[First]@
    Flatten[Distribute[Reverse@#, List] & /@ Tck400, 1]
  },
 Plot[
  Evaluate@FindFormula[data, x, SpecificityGoal -> "High"]
  , {x, Sequence @@ MinMax[data[[All, 1]]]}
  , PlotStyle -> Red
  , PlotRange -> {0, All}
  , PlotTheme -> "Scientific"
  ]
 ]

Answer 2

f = Interpolation@data;
data // Plot[f[x], {x, Min@(#[[All, 1]]), Max@(#[[All, 1]])}, 
PlotRange -> {{0, 7.5}, {0, 0.07}},
PlotStyle -> Red, 
Frame -> True, 
Axes -> False, 
FrameTicks -> {Subdivide[0, 7, 7], Subdivide[0.01, 0.06, 5]}] &

 data = DeleteDuplicates[Catenate[{Table[{Tck400[[ii, 2, 2]], Tck400[[ii, 1]]}, {ii, 1, 
   Length[Tck400]}], Table[{Tck400[[ii, 2, 1]], Tck400[[ii, 1]]}, {ii, 1, 
   Length[Tck400]}]}]];

Edit

In response to the OP comment the curves thus far are not smooth, thus providing ammunition for his advisor, it might be better to fit the data to a specific model rather than use interpolation. Nevertheless, increasing the interpolation order to 4 (from the default of 3) does seem to improve things.

f = Interpolation[data, InterpolationOrder -> 4];
data // Plot[f[x], {x, Min@(#[[All, 1]]), Max@(#[[All, 1]])}, 
PlotRange -> {{0, 7.5}, {0, 0.07}}, PlotStyle -> Red, 
Frame -> True, Axes -> False, 
FrameTicks -> {Subdivide[0, 7, 7], Subdivide[0.01, 0.06, 5]}] &

Note that the interpolation function may be used to calculate y-values, provided that the x-values are within the interpolation range. (Useful for a 'standard curve' situation?)

f@3.5

0.0464684

f /@ Subdivide[Min@data[[All, 1]], 7, 10]

{0.0334225, 0.0510192, 0.0543068, 0.0538228, 0.0507972, 0.0460776, 0.0403554, 0.0337375, 0.0266037, 0.0191165, 0.0114045}