Need help with calculating trajectory for a grenade in a game

[young solstice]

i'm a game developer who's not that good at math and i need to calculate the trajectory of a grenade in various different ways:

first, having an origin, end point and a force, i will have to calculate the angle at which the projectile will have to go starting from the origin to reach the end point, and it's okay if it can't reach it fully, but it will have to respect the force.

after that, I have a "beam" which is actually a bezier curve, and I have to modify what I can (attachments and curve sizes) to match the trajectory of the part which is re-calculated. Check the image for more information on how it works.

[dull shuttle]

Is the force like a certain speed?

The way projectiles work is they will always have the same horizontal speed, while their vertical speed accelerates due to gravity

So using trigonometry, the horizontal speed is ucosA, while the vertical speed is usinA, where u is the starting speed of the object and A is the angle

[quaint parrot]

ImOakley

[dull shuttle]

s is vertical height and a is acceleration (on earth it would be -9.8m/s^2)

Whatever end point youre using is another s variable away from the origin (call it d)

The horizontal speed is ucosA so d=utcosA

You would then find where the times are equal and use that to find the angle you need

[young solstice]

yes, it's the speed applied in the beginning

[young solstice]

yeah but you're only calculating the speed in X and Y, i need X, Y and Z

[young solstice]

how would you do that? i still haven't studied anything about trigonometry, i start next year

[dull shuttle]

Its the same thing, just translate the direction to 2d and then use pythagoras or smthn for the vectors

[dull shuttle]

Ill explain later cant do much rn

[dull shuttle]

So like do you anything about sin cos and tan?

[young solstice]

i know how they work and what they do but i've never used them in maths and i dont know any formulas

[dull shuttle]

Does the end point land at any points of higher elevation?

The way trig works here is that when the grenade has its initial speed, that speed can be defined into the x direction and y direction (and z direction)

Like if you were moving in a diagonal line in a coordinate grid you would be changing y position at a constant and changing x coordinate at another constant

So the rates that those coordinates change can be found using a right triangle, where the hypotenuse is the absolute speed

Thats why the vertical speed is usinA and horizontal is ucosA

gravity only affects the vertical

[dull shuttle]

This has nothing to do with the trigonometry its just a uniform acceleration formula

[young solstice]

like, if the end point is higher than the beginning point?

[dull shuttle]

yeah

[young solstice]

well that's variable

depends on where the user points the grenade

could be higher, lower

[dull shuttle]

ok so basically $h=u\sin(A)t+\frac{1}{2}(-9.8)t^2$

[quaint parrot]

ImOakley

[dull shuttle]

h is the landing altitude, t is time, A is angle, u is launch speed

so $-4.9t^2+u\sin(A)t-h=0$

[quaint parrot]

ImOakley

[dull shuttle]

using quadratic formula $t=\frac{-u\sin{A}+\sqrt{u^2\sin^2A-4(-4.9)(-h)}}{2(-4.9)}$

[quaint parrot]

ImOakley

[dull shuttle]

not plus or minus because it should be the higher of the two values for t

this value should be equal to $\frac{s}{u\cos{A}}$ since the horizontal distance is $s=u\sin{(A)}t$

[quaint parrot]

ImOakley

[dull shuttle]

you would use this to find a value for A

its a lot less complicated if the end altitude is 0

because then $-4.9t^2+u\sin(A)t=0$

[quaint parrot]

ImOakley

[dull shuttle]

so either $t=0$ or $-4.9t+u\sin{A}=0$

[quaint parrot]

ImOakley

[dull shuttle]

obviously the latter would be correct since t cant be zero in this context

so just $t=\frac{u\sin{A}}{4.9}$

[quaint parrot]

ImOakley

[dull shuttle]

then you would have $\frac{s}{u\cos{A}}=\frac{u\sin{A}}{4.9}$

[quaint parrot]

ImOakley

[dull shuttle]

$s=\frac{u^2\sin{A}\cos{A}}{4.9}$

[quaint parrot]

ImOakley

[dull shuttle]

theres a trig equation you dont need to worry about where $2\sin{A}\cos{A}=\sin{2A}$

[quaint parrot]

ImOakley

[dull shuttle]

so $s=\frac{u^2\sin{2A}}{9.8}$

[quaint parrot]

ImOakley

[dull shuttle]

$\sin{2A}=\frac{9.8s}{u^2}$

[quaint parrot]

ImOakley

[dull shuttle]

$2A=\sin^{-1}{(\frac{9.8s}{u^2})}$

[quaint parrot]

ImOakley

[dull shuttle]

so theres the equation you need for the angle

divided by 2

but only if its on ground level

[quaint parrot]

ImOakley

[dull shuttle]

ok so including different altitude its this freaking mess: $$A=\tan^{-1}{\frac{u^2s-u^2\sqrt{s^2-\frac{96.04s^4+19.6s^2u^2h}{u^4}}}{9.8s^2}}$$

[quaint parrot]

ImOakley

[dull shuttle]

simplified to this: $$A=\tan^{-1}{(\frac{u^2-\sqrt{u^4-96.04s^2-19.6u^2h}}{9.8s})}$$

[quaint parrot]

ImOakley

[young solstice]

holy

okay but, if i need 3 coordinates for the angle, do i just replace u with each coordinate of the launch speed?

[dull shuttle]

like whats the input

[young solstice]

origin position (XYZ Vector) and launch direction (XYZ Unit Vector)