Construction2
The older rendezvous page, Construction1, started from a launch time and an exact associated longitude; calculations were complicated and the radial velocity at capture was large.
A better approach is to start with the capture angle at the construction orbit, which defines an exact radius and radial velocity. From there, find a launch orbit with an apogee, launch angle, and launch time that arrives at the construction orbit with the same radius and arrival velocity. The tangential velocity will be different, on the order of 100 m/s, but we can accommodate that with a "passive net capture system".
So, given the construction orbit, choose an angle from apogee \theta . Construction orbits have low perigees and very high apogees, far beyond GEO, so the apogee velocity is small and the angular velocity very small near apogee. Given a desired arrival time t_{ac} referenced from t_{ac} = 0 at apogee, we can estimate \theta given the apogee radius r_{ac} and apogee velocity v_{ac}
The construction orbit perigee should be well above LEO; assume an altitude of 2000 km above the equatorial radius 6378 km, thus r_{pc} = 8378 km. The construction orbit semimajor axis is defined from the orbit period, which should be a multiple N of an 86164.0905 second sidereal day. From that, we can compute the apogee radius, velocity, and angular velocity \omega_{ac} , then iterate from the desired arrival time to the exact capture angle.

period 
semimajor 
apogee 
apogee V 
ang.freq. 
1 hr angle 
Intercept 

N 
P_c 
a_c 
r_{ac} 
v_{ac} 
\omega_{ac} 
radians 
r_i 
v_i 

sdays 
seconds 
km 
km 
km/s 
radians/sec 
est. 
exact 
km 
km/s 
1 
86164.1 
42164.17 
75950.34 
1.02118 
1.3445E5 
0.048403 
0.048557 
75591.05 
0.19988 
2 
172328.2 
66931.45 
125484.89 
0.63056 
5.0250E6 
0.018090 
0.018104 
125341.34 
0.07978 
3 
258492.3 
87705.01 
167032.01 
0.47745 
2.8584E6 
0.010290 
0.010294 
166948.27 
0.04653 
4 
344656.4 
106247.05 
204116.10 
0.39241 
1.9225E6 
0.006921 
0.006922 
204058.99 
0.03173 
5 
430820.5 
123288.78 
238199.56 
0.33721 
1.4157E6 
0.005096 
0.005097 
238157.13 
0.02357 
6 
516984.5 
139223.02 
270068.04 
0.29802 
1.1035E6 
0.003973 
0.003973 
270034.76 
0.01849 
7 
603148.6 
154291.59 
300205.17 
0.26851 
8.9442E7 
0.003220 
0.003220 
300178.07 
0.01506 
Note that the estimate is pretty good, off by 0.000154 radians for one hour delay and the 1 sidereal day construction orbit; that corresponds to an earth rotation of 2.11 seconds, or about 1 kilometer of distance. Good for capability estimation, too much error for accurate aiming from the loop, which will need precise (millisecond) timing.
Other tables can be constructed for other vehicle capture times.
Starting from the capture time T seconds after construction orbit apogee, we can compute the capture angle L radians to the west of apogee.
Given the launch orbit perigee ( r_{pl} = 6378 km + 80 km loop altitude), the intercept radius r_i, and the intercept radial velocity v_i, we can algebraically compute the launch orbit perigee r_{al} and the angle between intercept and launch perigee L . From those parameters, we can compute the semimajor axis a_l, the eccentricity e_l, the period P_l, then the launch time and launch angle.
The radius and the radial velocity at angle L are given by:
r_i = { \Large { { ( 1  e_l^2 ) a_l } \over { 1 + e_l \cos( L ) } } } ~~~~~~~~~~~~~~~~~~~~~~~~~ v_i = { \Large \sqrt{ \mu \over { ( 1  e_l^2 ) a_l } } } \sin( L )
Let's define two new known parameters in terms of the known parameters r_{pl}, r_i, and v_i$ ,
\alpha \equiv { \Large { { v_i^2 r_pl } \over \mu } } ~~~~~~~~~~~~~ hence ( e_l \sin( L ) )^2 = \alpha ( 1  e_l )
\beta \equiv { \Large { r_pl \over r_i } } ~~~~~~~~~~~~~ hence e_l \cos( L ) = \beta ( 1  e_l )  1
This is enough to create a quadratic equation:
( e_l \sin( L ) )^2 + ( e_l \cos( L ) )^2 ~=~ e_l^2 ~=~ ( \beta  \beta e_l 1 )^2 + \alpha ( 1  e_l )
0 = ( \beta^2  1 ) e_l^2 + ( 2 \beta  2 \beta^2  \alpha ) e_l + ( \beta^2  2 \beta + \alpha + 1 )
Define the coefficients of the quadratic polynomial:
A \equiv ( \beta^2  1 ) ...... B \equiv ( 2 \beta  2 \beta^2  \alpha ) ....... C \equiv ( \beta^2  2 \beta + \alpha + 1 )
And solve for e_l :
e_l = { { B + \Large { \sqrt{ B^2 ~~ 4 A C } } \over { 2 A } } }
The other root of the equation yields e_l = 1 which is nonphysical .
Here's the result of a spreadsheet
mu km3/s3 
398600.44 

re km 
6378.00 

sday s 
86164.09 


prime 
launch 
constr 
solution 

period s 
83238.21 
83229.63 
86164.09 
alpha 
2.0389E4 

omega rad/s 
7.5484E5 
7.5492E5 
7.2921E5 
beta 
8.5156E2 

altitude km 
80.00 
80.00 
2000.00 
A 
0.9927484 

perigee km 
6458.00 
6458.00 
8378.00 
B 
0.15560584 

apogee km 
75950.34 
75944.68 
75950.34 
C 
0.83714253 

semimajor km 
41204.17 
41201.34 
42164.17 
radical 
3.34850072 

v0 km/s 
5.787 
5.787 
5.139 
good e+ 
0.8432575 

eccentricity e from apogee 
0.84327 
0.84326 
0.80130 
check 
0.0E+0 

v perigee 
10.66631 
10.66628 
9.25746 
junk e 
1.00000 

v apogee 
0.90695 
0.90701 
1.02118 
check 
0.0E+0 

intercept theta radians 
0.00000 
0.02299 
0.02724 

intercept radius km 
75950.34 
75836.85 
75836.85 

E eccentric anomaly 
0.00000 
0.07881 
0.08199 

time(theta) sec 
0.000 
879.447 
900.000 

tangent velocity km/s 
0.90695 
0.90830 
1.02271 

radial velocity km/s 
0.00000 
0.11218 
0.11218 

perigee time 
41619.10 
40735.37 
42182.04 