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malawi_glenn
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That is what you would get as suggested in #21kuruman said:While playing with possible trajectory equations
That is what you would get as suggested in #21kuruman said:While playing with possible trajectory equations
Perhaps. This is more specific than than the suggestions in #21 which include an exponential and a piecewise constant acceleration.malawi_glenn said:That is what you would get as suggested in #21
I thoughtkuruman said:Perhaps. This is more specific than than the suggestions in #21
was pretty specific?malawi_glenn said:or power function a(t)=k⋅tα−2,
The program I use doesn't miss a beat!malawi_glenn said:Integrating logarithm functions
This I did not check, but I wonder how you recognize so quickly that it won't be able to produce a solution?malawi_glenn said:But, such function will not work.
malawi_glenn said:Because of its convexity/concavity
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we already know that linear velocity won't work
erobz said:I don't know if I helped much (If I didn't confuse you more), but I don't believe you ever stated how the circular part of the animation works? What calculations does it use to generate the pixels that will light over time? This is basically, for my own curiosity (since you say it works fine)...but I'm assuming it is some method different from what I asked in post #52 ?
Is there an easy way of determining what the v0 should be for a particular curve, such that the area actually is 10? In your case above it's not actually possible given the shape.malawi_glenn said:View attachment 304860
now the only way to get that area to become 10 is if v0 is negative, because the picture you posted shows the minimal area with v0≥0
You can come up with any kind of v(t) for your problem as long as the graph is above the horisontal axis between 0 < t < 4, it has v(4) = 20, and the area is equal to 10.
Depends on the shape of the curve. But you can see that for a v0>0 you must have an even larger acceleration (slope) in the v-t diagram to make sure the final speed is 20 m/s and only covers 10 m (area)Jack7 said:Is there an easy way of determining what the v0 should be for a particular curve, such that the area actually is 10? In your case above it's not actually possible given the shape.
This seems a bit like a chicken and egg situation. If we want v0 to always be positive (one direction of travel), then don't the curve and v0 need to be defined together through manual iteration?malawi_glenn said:Depends on the shape of the curve. But you can see that for a v0>0 you must have an even larger acceleration (slope) in the v-t diagram to make sure the final speed is 20 m/s and only covers 10 m (area)
malawi_glenn said:This should illustrate the problem with having a non-zero v0
View attachment 304914
These velocity profiles have the same final speed and the particle will cover the same distance.
But the particle with higher v0 has to experience a much larger acceleration at the end of its trajectory.
Ahh! I see! Thank you so much for explaining it like that, that makes a lot of sense!malawi_glenn said:You could define piece-wise constant accelerations over several time-intervals.
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Since the particle has to cover most of its distance in the end of its trajectory, i would not worry too much about the smoothness. You are dealing with pixels, even if you had a continuous velocity function to start with, it will become discretized anyway in your program.Jack7 said:The key challenge I foresee with doing this automated is keeping the shape of the curve smooth
malawi_glenn said:Since the particle has to cover most of its distance in the end of its trajectory, i would not worry too much about the smoothness. You are dealing with pixels, even if you had a continuous velocity function to start with, it will become discretized anyway in your program.
I think it was determined that it wasn't a homework problem, but instead just a hobbyist game designer question about some unexpected results. @Jack7 mentioned about being succinct to decrease the length of the initial post.bob012345 said:Reading through all this I wonder if the first post had all the information given in this challenge? What was the exact statement of the problem?
This can be nicely generalized.kuruman said:While playing with possible trajectory equations, it occurred to me that one could exploit the fact the factors of 2 are prominent. Ignoring units, the final speed is twice the distance and the time is 22. There is a single power of ##t## that will do the trick here:$$x(t)=\frac{5}{2^{15}}t^8~\implies v(t)=\frac{8\times 5}{2^{15}}t^7.$$With this,$$x(2^2)=\frac{5}{2^{15}}2^{16}=10~;~~v(2^2)=\frac{8\times5}{2^{15}}2^{14}=20.$$Note the effect of the many vanishing derivatives at small values of time.
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I realize that there have been many replies to this thread.Jack7 said:So to summarise:
- A body starts at Point A
- It moves in a straight line to Point B, covering a distance of 10m
- The time taken to travel this distance is 4 seconds
- When it reaches Point B, it has a velocity of 20 m/s
haruspex said:Suppose a low acceleration a1 for time t and a greater acceleration a2 for the rest. That's three unknowns, which is one too many for the given constraints, so you can in principle pick a value for anyone and solve for the other two. But for some choices you will still end up getting illegal moves, so you will have to play about with your choice until it gives a valid solution.