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Here you can see some examples:
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Here you can see some examples:
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2 |
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3 |
One of the most extreme but still realistic examples for underdrive is a grid with m_1=1, m_2=3, E=0. Here the smaller mex transfers 3.2 energy to the bigger one. The smaller one only works at 45% while the bigger one works at 134%. Together they produce 4.02+0.45=4.47 metal compared to 4 normally.
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One of the most extreme but still realistic examples for underdrive is a grid with m_1=1, m_2=3, E=0. Here the smaller mex transfers 3.2 energy to the bigger one. The smaller one only works at 45% while the bigger one works at 134%. Together they produce 4.02+0.45=4.47 metal compared to 4 normally.
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Let's look at a grid with m_1=2, m_2=3, E=2:
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Let's look at a grid with m_1=2, m_2=3, E=2:
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Old system: e_1=0.62, e_2=1.38, M_1=2.15, M_2=3.48, M=M_1+M_2=5.63
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Old system: e_1=0.62, e_2=1.38, M_1=2.15, M_2=3.48, M=M_1+M_2=5.63
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Ideal: e_1=0, e_2=2, M_1=2, M_2=3.67, M=5.67
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Ideal: e_1=0, e_2=2, M_1=2, M_2=3.67, M=5.67
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Underdrive: e_1= -0.92, e_2=2.92, M_1=1.75, M_2=3.95, M=5.70
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Underdrive: e_1= -0.92, e_2=2.92, M_1=1.75, M_2=3.95, M=5.70
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Symmetric (b=1/4): e_1=0.62, e_2=1.38 (like old), M_1=2.39, M_2=3.88, M=6.27
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Symmetric (b=1/4): e_1=0.62, e_2=1.38 (like old), M_1=2.39, M_2=3.88, M=6.27
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11 |
Maybe
b
has
to
be
reduced.
Probably
b=0.
2
would
be
better.
This
is
closer
to
the
current
system
for
low
OD
and
improves
the
[url=http://zero-k.
info/Forum/Thread/11225?page=2]balance[/url]
of
singu
spam
in
team
games
a
bit.
(
b=0.
16
would
be
even
closer,
but
maybe
high
OD
becomes
too
inefficient
then.
)
|
11 |
Maybe
b
has
to
be
reduced.
Probably
b=0.
2
would
be
better.
This
is
closer
to
the
current
system
for
low
OD
and
improves
the
[url=http://zero-k.
info/Forum/Thread/11225?page=2]balance[/url]
of
singu
spam
in
team
games
a
bit.
(
b=0.
16
would
be
even
closer,
but
maybe
high
OD
becomes
too
inefficient
then.
)
|
12 |
\n
|
12 |
\n
|
13 |
[url=http://www.wolframalpha.com/input/?i=sqrt%281%2Bx%2F4%29%2C+1%2Bsqrt%28x%29%2F4+from+x%3D-5+to+20]Here[/url] you can see a graph of current gamma(e_i)=sqrt(1+e_i/4) and symmetric gamma_(b=1/4)=1+sqrt(e_i)/4.
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13 |
[url=http://www.wolframalpha.com/input/?i=sqrt%281%2Bx%2F4%29%2C+1%2Bsqrt%28x%29%2F4+from+x%3D-5+to+20]Here[/url] you can see a graph of current gamma(e_i)=sqrt(1+e_i/4) and symmetric gamma_(b=1/4)=1+sqrt(e_i)/4.
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