Research Costs
 

Deleted.

Re: Research Costs
 

yes the system of costs for each level is like this (normal res):
40 for first level, 60 for level 2...
after that each level costs = two previous levels.

so level 3 will be 100, then 160, 260 and etc.

Think it's called a fibernacci series or some such.

Re: Research Costs
 

Fibonacci series.

So, it'd be

• Easy: 20, 40, 60, 100, 160, 260, 420, 680, 1100, aggregate total 2840
• Normal: 40, 60, 100, 160, 260, 420, 680, 1100, 1780, aggregate total 4600
• Hard: 60, 100, 160, 260, 420, 680, 1100, 1780, 2880, aggregate total 7440
• Very Hard: 100, 160, 260, 420, 680, 1100, 1780, 2880, 4660, aggregate total 12040

Somebody had better verify that with a calculator, because I had to do that in my head and I might have made mistakes. Even so, those numbers aren't off by much.

Edi

Re: Research Costs
 

I wish you could either mod the first two numbers of the series or scale the series. It's too easy to get to the end of the research tree on larger maps, even on "very hard."

Re: Research Costs
 

If you want a super-slow magic game, try playing with very low gold and very low magic sites. You'll have a much harder time buying or summoning mages...

Re: Research Costs
 

Ygorl, I always set the magic sites frequency very low, but I hadn't thought much about reducing the gold. I guess to balance things out between magic and armies I should probably reduce resources as well.

Thanks for the suggestion!

The more I think about it a generic Fibonacci series doesn't really seem a good match with the recruiting and research mechanics. Under normal game settings and assuming one does not purely concentrate or one or two paths of magic, the regular buildup of research points through recruiting combined with new fort building and finding mage recruiting provinces/sites make it often actually easier to research higher levels in a path than it was to research the earlier levels. If you added a geometric series component it would help. The cost for each level would be the Nth term in the Fibonacci series times (1+scaler)^(N-1).