Mod's startup options—
Technology cost scaling -> increase,
Technology start price -> increase,
Additional technology cost multiplier -> increase,
Technology cost curve form -> (linear, quadratic, exponential),
Expanded crafting GUI -> disable.
Technology cost scaling—
["Very Cheap"] = 500,
["More Cheap"] = 1000,
["Cheap"] = 1500,
["Default"] = 2000,
["Less Expensive"] = 3500,
["Expensive"] = 5000,
["Very Expensive"] = 100000
--FORMULA DERIVATION
-- we plot cost (in y) over n number of technologies (in x) -> (y, n)
-- assume start point: A = (0, C) ...C is the start price
-- assume end point: B = (N, X) ...C is the max cost
-- Linear: Form y = ax + b <-> cost(n) = an + b
-- b: use point A -> cost(0) = a0 + b =!= C <-> b = C
-- a: use point B -> cost(N) = a*N + b =!= X <-> a = (X - b)/N
-- Quadratic: Form y = ax^2 + bx + c <-> cost(n) = an^2 + c | here we wont need b
-- c: use point A -> c = C
-- a: use point B -> cost(N) = aN^2 + C =!= X <-> a = (X - C)/N^2
-- Exp: Form y = exp(bx) + c <-> cost(n) = exp(bn) + c
-- c: use point A -> cost(0) = 1 + c =!= C <-> c = C - 1
-- b: use point B -> cost(N) = exp(b*N) + C - 1 <-> b = ln(X + 1 - C)/N
-- we got: cost_lin(n) = (X - C)/N * n + C
-- cost_qua(n) = (X - C)/N^2 * n^2 + C
-- cost_exp(n) = exp(n/N * ln(X + 1 - C)) + C - 1
-- since we need technology_price_multiplier use Ansatz: cost(n) =!= f(...) * C
-- we get: f_lin(...) = (X - C)/(N*C) * n + 1
-- f_qua(...) = (X - C)/(N^2 * C) * n^2 + 1
-- f_exp(...) = (exp(n/N * ln(X + 1 - C)) - 1)/C + 1
-- we use the following variables for X, N, C, n
-- X .. maxCost
-- N .. techCount
-- C .. startPrice
-- n .. currentTechs
-- NOTE: Linear will increase strongly in early game, but will make late game "easier"
-- Exponential will increase slowly in early game, but will make late game much harder
-- Quadratic will begin to get more difficult earlier in the game then exp but wont be as steep as exp
-- OVERALL USAGE:
-- we can integrate the 3 curves to get their area, which corresponds to the total needed science packs
-- integrate from 0 to N ofc
-- S cost_lin(n) dn = N/2 * (X + C)
-- S cost_qua(n) dn = N/3 * (X + 2C)
-- S cost_exp(n) dn = N(C - 1 + (X + 1 - C)/ln(X + 1 - C))
-- for big X we can assume ln(X + 1 - C) ~ ln(X) f.e. ln(X=10k) ~ 9
-- -> Area(exp) = N(C - 1 + 1/9 * (X + 1 - C)) ~ NX/9
-- -> Area(lin) ~ NX/2
-- -> Area(qua) ~ N*X/3
-- therfore overall exponential is cheapest, then quadratic and linear is most difficult
^Direct PreLeyZero quote.
New cost calculations are done upon finishing next research.
