1 |
Here is how it is currently:
|
1 |
Here is how it is currently:
|
2 |
{{{energy for OD = energy rate * energy storage / max storage.}}}Note that "energy for OD" is only the provided energy. What is actually used is also limited by the grid.
|
2 |
{{{energy for OD = energy rate * energy storage / max storage.}}}Note that "energy for OD" is only the provided energy. What is actually used is also limited by the grid.
|
3 |
|
3 |
|
4 |
@aeonios only wants to spend energy on OD if storage is full:
|
4 |
@aeonios only wants to spend energy on OD if storage is full:
|
5 |
{{{energy for OD = energy rate * [energy storage / max storage],}}}where [x] is rounding down.
|
5 |
{{{energy for OD = energy rate * [energy storage / max storage],}}}where [x] is rounding down.
|
6 |
A
compromise
that
considers
metal
storage
as
@Skasi
wants
would
be
either
applying
the
same
as
currently
only
on
that
part
of
stored
energy
that
exceeds
stored
metal{
{
{
energy
for
OD
=
energy
rate
*
abs(
energy
storage
-
metal
storage)
/
max
storage,
}
}
}
where
abs(
x)
=x
if
x>=0,
abs(
x)
=0
if
x<0,
or
using
all
energy
that
exceeds
current
metal
for
OD:
|
6 |
A
compromise
that
considers
metal
storage
as
@Skasi
wants
would
be
either
applying
the
same
as
currently
only
on
that
part
of
stored
energy
that
exceeds
stored
metal{
{
{
energy
for
OD
=
energy
rate
*
abs(
energy
storage
-
metal
storage)
/
(
max
storage
-
metal
storage)
,
}
}
}
where
abs(
x)
=x
if
x>=0,
abs(
x)
=0
if
x<0,
or
using
all
energy
that
exceeds
current
metal
for
OD:
|
7 |
{{{energy for OD = energy rate * theta(energy storage - metal storage),}}} where theta is Heavyside's step function, theta(x)=1 if x>=0, theta(x)=0 if x<0. Those are rather quick fix ideas than ideal solutions, though.
|
7 |
{{{energy for OD = energy rate * theta(energy storage - metal storage),}}} where theta is Heavyside's step function, theta(x)=1 if x>=0, theta(x)=0 if x<0. Those are rather quick fix ideas than ideal solutions, though.
|