# Feedrate planning at zero speed

**URL:** <http://discourse.heig-vd.ch/t/feedrate-planning-at-zero-speed/260>\
**Category:** OpenCN Framework\
**Created:** [January 9, 2022, 2:43am UTC](http://discourse.heig-vd.ch/t/feedrate-planning-at-zero-speed/260 "2022-01-09T02:43:45Z")\
**Posts on this page:** 4\
**Page:** 1

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**Author:** ![yakunix](http://discourse.heig-vd.ch/letter_avatar_proxy/v4/letter/y/bb73d2/32.png) [@yakunix](http://discourse.heig-vd.ch/u/yakunix)\
**Post date:** [January 9, 2022, 2:43am UTC](http://discourse.heig-vd.ch/t/feedrate-planning-at-zero-speed/260/1 "2022-01-09T02:43:45Z")

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Hi Raoul,  
In the HPC paper, the remaining time is approximated by the following equation:

![image](https://discourse.heig-vd.ch/uploads/default/original/1X/7fadd3b3a2b6f21f0919ba4c482c12a7123b78ae.png)

I’m wondering if this strategy could also be used for singularity handling. At the start of zero velocity, we can assume the q(u) to be linear, so we have the relation:

![](https://latex.codecogs.com/svg.image?%5Ccolor%7BYellow%7D(u_%7Bk%7D&space;-&space;0)/%5Cleft&space;%5B&space;(%5Csqrt%7Bq(u_%7Bk%7D)%7D&space;-&space;0)/2&space;%5Cright&space;%5D=t_%7Bk%7D "(u\_{k} - 0)/\left [(\sqrt{q(u\_{k})} - 0)/2 \right]=t\_{k}")

Then we have:  
 ![](https://latex.codecogs.com/svg.image?%5Ccolor%7BYellow%7D%5Cfrac%7B4*%7Bu_%7Bk%7D%7D%5E%7B2%7D%7D%7Bq(u_%7Bk%7D)%7D=%7Bt_%7Bk%7D%7D%5E%7B2%7D "\frac{4\*{u\_{k}}^{2}}{q(u\_{k})}={t\_{k}}^{2}")  
As q(uk) is a cubic spline and because q(0) is zero at start, we have q(u) to be the following format:  
 ![](https://latex.codecogs.com/svg.image?%5Ccolor%7BYellow%7Da*%7Bu_%7Bk%7D%7D%5E3+b*%7Bu_%7Bk%7D%7D%5E2+c*%7Bu_%7Bk%7D%7D "a\*{u\_{k}}^3+b\*{u\_{k}}^2+c\*{u\_{k}}")  
Substitute it into the denominator, we have:

![](https://latex.codecogs.com/svg.image?%5Ccolor%7BYellow%7D%5Cfrac%7B4*%7Bu_%7Bk%7D%7D%5E2%7D%7Ba*%7Bu_%7Bk%7D%7D%5E3+b*%7Bu_%7Bk%7D%7D%5E2+c*%7Bu_%7Bk%7D%7D%7D=%7Bt_%7Bk%7D%7D%5E2 "\frac{4\*{u\_{k}}^2}{a\*{u\_{k}}^3+b\*{u\_{k}}^2+c\*{u\_{k}}}={t\_{k}}^2")

This uk has a closed form of solution to be:

 ![image](https://discourse.heig-vd.ch/uploads/default/original/1X/8b673b1d352a6c7cd02bf3f48ea0e3046e52a0ca.png)

And we can choose the minimum positive solution.

We can use very small tk like 1e-8 to solve the uk, and the spline q(u) will be almost linear at the infinite-small range . Once we have the first solution, then we are safe to use Taylor series expansion to get the remaining uk.

Do you think the derivation is correct or did I mistake something?

Thanks!

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**Author:** ![Raoul](http://discourse.heig-vd.ch/user_avatar/discourse.heig-vd.ch/raoul/32/21_2.png) [@Raoul](http://discourse.heig-vd.ch/u/Raoul)\
**Post date:** [January 9, 2022, 1:01pm UTC](http://discourse.heig-vd.ch/t/feedrate-planning-at-zero-speed/260/2 "2022-01-09T13:01:52Z")

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dear Yakunix,

please find the attached pdf document prepared by my mathematician friend Philippe Blanc.  
[ZeroSpeedProblem.pdf](http://discourse.heig-vd.ch/uploads/short-url/heGZUALBu03NFMP2iBslxm6S8uZ.pdf) (85.5 KB)

I hope this makes things clearer.

best regards

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**Author:** ![yakunix](http://discourse.heig-vd.ch/letter_avatar_proxy/v4/letter/y/bb73d2/32.png) [@yakunix](http://discourse.heig-vd.ch/u/yakunix)\
**Post date:** [January 10, 2022, 5:04am UTC](http://discourse.heig-vd.ch/t/feedrate-planning-at-zero-speed/260/3 "2022-01-10T05:04:32Z")

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Thank you Raoul.

This explanation is cool!

In essence, the initial stage could not be approximated by the spline.

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**Author:** ![Raoul](http://discourse.heig-vd.ch/user_avatar/discourse.heig-vd.ch/raoul/32/21_2.png) [@Raoul](http://discourse.heig-vd.ch/u/Raoul)\
**Post date:** [January 10, 2022, 5:41am UTC](http://discourse.heig-vd.ch/t/feedrate-planning-at-zero-speed/260/4 "2022-01-10T05:41:17Z")

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Hi Yakunix,  
100% correct