Teathimble_Firmware/maths.c

188 wiersze
4.9 KiB
C

#include "maths.h"
/*!
Pre-calculated constant values for axis um <=> steps conversions.
These should be calculated at run-time once in dda_init() if the
STEPS_PER_M_* constants are ever replaced with run-time options.
*/
const axes_uint32_t PROGMEM axis_qn_P = {
(uint32_t)STEPS_PER_M_X / UM_PER_METER,
(uint32_t)STEPS_PER_M_Y / UM_PER_METER
};
const axes_uint32_t PROGMEM axis_qr_P = {
(uint32_t)STEPS_PER_M_X % UM_PER_METER,
(uint32_t)STEPS_PER_M_Y % UM_PER_METER
};
/*!
Integer multiply-divide algorithm. Returns the same as muldiv(multiplicand, multiplier, divisor), but also allowing to use precalculated quotients and remainders.
\param multiplicand
\param qn ( = multiplier / divisor )
\param rn ( = multiplier % divisor )
\param divisor
\return rounded result of multiplicand * multiplier / divisor
Calculate a * b / c, without overflowing and without using 64-bit integers.
Doing this the standard way, a * b could easily overflow, even if the correct
overall result fits into 32 bits. This algorithm avoids this intermediate
overflow and delivers valid results for all cases where each of the three
operators as well as the result fits into 32 bits.
Found on http://stackoverflow.com/questions/4144232/
how-to-calculate-a-times-b-divided-by-c-only-using-32-bit-integer-types-even-i
*/
int32_t muldivQR(int32_t multiplicand, uint32_t qn, uint32_t rn,
uint32_t divisor) {
uint32_t quotient = 0;
uint32_t remainder = 0;
uint8_t negative_flag = 0;
if (multiplicand < 0) {
negative_flag = 1;
multiplicand = -multiplicand;
}
while(multiplicand) {
if (multiplicand & 1) {
quotient += qn;
remainder += rn;
if (remainder >= divisor) {
quotient++;
remainder -= divisor;
}
}
multiplicand >>= 1;
qn <<= 1;
rn <<= 1;
if (rn >= divisor) {
qn++;
rn -= divisor;
}
}
// rounding
if (remainder > divisor / 2)
quotient++;
// remainder is valid here, but not returned
return negative_flag ? -((int32_t)quotient) : (int32_t)quotient;
}
/*!
integer inverse square root algorithm
\param a find the inverse of the square root of this number
\return 0x1000 / sqrt(a) - 1 < returnvalue <= 0x1000 / sqrt(a)
This is a binary search but it uses only the minimum required bits for each step.
*/
// courtesy of http://www.flipcode.com/archives/Fast_Approximate_Distance_Functions.shtml
/*! linear approximation 2d distance formula
\param dx distance in X plane
\param dy distance in Y plane
\return 3-part linear approximation of \f$\sqrt{\Delta x^2 + \Delta y^2}\f$
see http://www.flipcode.com/archives/Fast_Approximate_Distance_Functions.shtml
*/
uint32_t approx_distance(uint32_t dx, uint32_t dy) {
uint32_t min, max, approx;
// If either axis is zero, return the other one.
if (dx == 0 || dy == 0) return dx + dy;
if ( dx < dy ) {
min = dx;
max = dy;
} else {
min = dy;
max = dx;
}
approx = ( max * 1007 ) + ( min * 441 );
if ( max < ( min << 4 ))
approx -= ( max * 40 );
// add 512 for proper rounding
return (( approx + 512 ) >> 10 );
}
// courtesy of http://www.oroboro.com/rafael/docserv.php/index/programming/article/distance
/*! linear approximation 3d distance formula
\param dx distance in X plane
\param dy distance in Y plane
\param dz distance in Z plane
\return 3-part linear approximation of \f$\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}\f$
see http://www.oroboro.com/rafael/docserv.php/index/programming/article/distance
*/
uint32_t approx_distance_3(uint32_t dx, uint32_t dy, uint32_t dz) {
uint32_t min, med, max, approx;
if ( dx < dy ) {
min = dy;
med = dx;
} else {
min = dx;
med = dy;
}
if ( dz < min ) {
max = med;
med = min;
min = dz;
} else if ( dz < med ) {
max = med;
med = dz;
} else {
max = dz;
}
approx = ( max * 860 ) + ( med * 851 ) + ( min * 520 );
if ( max < ( med << 1 )) approx -= ( max * 294 );
if ( max < ( min << 2 )) approx -= ( max * 113 );
if ( med < ( min << 2 )) approx -= ( med * 40 );
// add 512 for proper rounding
return (( approx + 512 ) >> 10 );
}
uint16_t int_inv_sqrt(uint16_t a) {
/// 16bits inverse (much faster than doing a full 32bits inverse)
/// the 0xFFFFU instead of 0x10000UL hack allows using 16bits and 8bits
/// variable for the first 8 steps without overflowing and it seems to
/// give better results for the ramping equation too :)
uint8_t z = 0, i;
uint16_t x, j;
uint32_t q = ((uint32_t)(0xFFFFU / a)) << 8;
for (i = 0x80; i; i >>= 1) {
uint16_t y;
z |= i;
y = (uint16_t)z * z;
if (y > (q >> 8))
z ^= i;
}
x = z << 4;
for (j = 0x8; j; j >>= 1) {
uint32_t y;
x |= j;
y = (uint32_t)x * x;
if (y > q)
x ^= j;
}
return x;
}
uint32_t acc_ramp_len(uint32_t feedrate, uint32_t steps_per_m) {
return (feedrate * feedrate) /
(((uint32_t)7200000UL * ACCELERATION) / steps_per_m);
}