3 |
It's a trick question! Haha, got you! with careful overdrive grids, the 9 mex scenario makes up that 2 m/s!
|
3 |
It's a trick question! Haha, got you! with careful overdrive grids, the 9 mex scenario makes up that 2 m/s!
|
5 |
cathartes
and
Rhade
did
an
attempt
at
solving
it
in
this
thread,
good
on
you!
But,
cathartes
overlooked
a
key
problem,
if
you
overdrive
all
the
excess
energy
to
metal,
you
have
no
energy
to
SPEND
that
metal.
Going
over
the
scenarios
again
but
only
overdriving
8
energy
each,
you
get
much
cleaner
numbers.
In
10
mex
scenario,
8e
=
4e/mex
*
2mex
then
2(
4e
=
50%OD)
=
2
*
1
=
2m,
and
you
can
spend
that
2m
with
the
left
over
2
energy.
In
the
9
mex
scenario,
split
12
excess
e
into
8
and
4,
8e
=
1e/mex
*8mex
then
we
get
8(
1e
=
25%OD)
=
8
*
.
5
=
4
metal.
the
4
metal
can
be
spent
with
that
4
energy
I
reserved
from
earlier.
the
9
mex
scenario
generates
an
extra
2m/s,
gapping
the
bridge
between
the
mex
difference
generating
the
same
income.
|
5 |
cathartes
and
Rhade
did
an
attempt
at
solving
it
in
this
thread,
good
on
you!
But,
cathartes
overlooked
a
key
problem,
if
you
overdrive
all
the
excess
energy
to
metal,
you
have
no
energy
to
SPEND
that
metal.
Going
over
the
scenarios
again
but
only
overdriving
8
energy
each,
you
get
much
cleaner
numbers.
In
10
mex
scenario,
8e=
4e/mex
*
2mex
then
2(
4e=
50%OD)
=
2
*
1=
2m,
and
you
can
spend
that
2m
with
the
left
over
2
energy.
In
the
9
mex
scenario,
split
12
excess
e
into
8
and
4,
8e=
1e/mex
*8mex
then
we
get
8(
1e=
25%OD)
=
8
*
.
5=
4
metal.
the
4
metal
can
be
spent
with
that
4
energy
I
reserved
from
earlier.
the
9
mex
scenario
generates
an
extra
2m/s,
gapping
the
bridge
between
the
mex
difference
generating
the
same
income.
|