|
I worked this out today on how long it takes you to kill a target needed to be killed.I worked it out today during school so it probably will be a little off.Well here it goes.
___________________________________________________________________________
You need to figure out the following:
*Time to click button /CT
*Distance (Up/Down) /D
*Speed (Left/Right) /S
*Target's distance /T
So the equation will be:
CT=D(S)=T
___________________________________________________________________________
So here is the problem I worked out.
Say it takes one second to click the button
1=D(S)=T
Then the distance going up is 5 studs per milisecond and the target is 750 feet across and 100 feet up.It will take 1.5 seconds to get up the target.
1=1.5(S)=(750x1000)
Now the distance going up is 1000 and I calculated it to be 2 seconds.
1=1.5(2)=(750x1000)
I calculated it and I got 1=3=(750x1000)
So the average gun like this will take 3 seconds at this point to hit it.
___________________________________________________________________________
If you guys and girls want to, go test this out and record your conclusion.Reply what you think your hypothesis is and what you got for a conclusion.
Goodbye!
-Gavindude97 |
|
|
My conclusion: That's a load of plop.
For starters, it makes no sense.
You didn't use SI units.
Targeting could mean all sorts.
T is always Time.
The average gun will fire a round at a speed between 300 and 900 m/s.
Size doesn't matter.
That isn't an equation as there are 2 '='s
Go back to school and listen to that person at the front instead of making up "equations".
I'm done. |
|
|
First:
I am not good at scripting and this is just an idea for a gun I am making.
Second:
I am sorry I did not satisfy you but not everybody will be.
Third:
I do listen to the teacher in front.That is why I get all A's. |
|
SamacadoJoin Date: 2008-02-16
Post Count: 25119 |
400 meters per second, I believe is the average speed of a rifled hand gun. If a target is more than say, 100 meters away, accuracy is an issue and speed is negligible at that point. |
|
miloguyJoin Date: 2009-12-19
Post Count: 7702 |
@gavin
What grade are you in? Not to look for personal information...
But you should know that this seems like algebra more than scripting, and algebraic variables must have only one letter, because putting two variables next to eachother would mean to multiply them.
noob = n times o times o times b |
|
|
@miloguy
That's not true... You can use any symbol as a variable as long as it doesn't lead to confusion. If you aren't using "n", "o", and "b" as separate variables, there's nothing wrong with using "noob" as a variable, if you define it as such.
@gavin
That "formula" calculates nothing. Stringing a bunch of arbitrary symbols and words together and hoping that it will make valid predictions usually doesn't work, unless you can explain how you derived the formula. How did you get that formula? Generally, if you can't explain why something should work and why that is the case, it won't. |
|
miloguyJoin Date: 2009-12-19
Post Count: 7702 |
@Fungal
I didn't know that... Well he did use the variable "T" and also "CT" which could cause some confusion |
|
|
@MRBlockson by
CT=D(S)=T
He means you can say that
C=D(S)
or
C=T
Although the formula doesn't make sense as
D=100
S=10
1 != 1000
So the real formula should be....
K = KillTime
T = TimeToGetToPoint
D = Distance
S = Speed
C = Time to click
as 100*10 = 1000 would mean going at higher speeds takes longer so it should be 100/10 = 10
K = D/S + C
or
K = T + C
Where as lets say they are 100 meters away. We go at a speed of 10 meters/sec and the time it took us to click the button or pull the trigger is 0.25. Well lets plug it in.
K = 1000/10 + 0.25
k = 10m/s + 0.25s
Now we can modify or formula to be something like....
K = D/S + C
K = C/(D/S)+(D/S)
So now we can say that...
K = .25/(100/10)+(100/10)
.25/(10) + 10
.025 + 10
10.025m/s(that's including time to shoot without is 10)
How'd you get .025? Well we want it to add up to .25 or 100.25(distance+timetoshoot) so we get times it takes to multiply the number to that and divide the seconds to fire by taht than add that to the time it takes to fire. Not even sure this is accurate either as I through it together. But this should be what it looks like if no gravity existed. This is not tested so can someone test it for me.
|
|
pwnedu46Join Date: 2009-05-23
Post Count: 7534 |
Go to wikipedia and search kinematics. (worst.semester.of.physics.evar. don't get me started on 2 dimensional or rotational motion DX) Don't get caught up in the proofs (calculus st00f), just find the equations;
x=xo+vo*t+(1/2)*a*t^2
v^2 * vo^2+2(x-xo)
v = vo+a*t
x = final displacement
xo = initail displacement
v = final velocity
vo = initial velocity
a = acceleration
t = time
[sarcasm]
so...much...fun!!!!
[\sarcasm] |
|
pwnedu46Join Date: 2009-05-23
Post Count: 7534 |
Sometimes, you can just remove terms that have xo and/or vo because they're 0.
for example:
v = vo+a*t;
vo = 0
a = 5 m/s^2
t = 5 seconds
simplified equation:
(=== in this case means = to by definition of defined vars above)
v = a*t === v = 5*5 === v = 25. |
|
|
"Don't get caught up in the proofs"
That reminds me.
I have 6 proofs I have due Wednesday for Geometry.
Plus another 10 at least from other homework assignments I never did.
FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
I HATE PROOOOOOOOOOOOOOOOOOOOFFFFFFSSSSSS! |
|
|
I worked and reworked it in class at least 20 times until I got to my conclusion.I need to learn more about algebra since my 8th grade teacher quit and was replaced due to here maternity. |
|
pwnedu46Join Date: 2009-05-23
Post Count: 7534 |
"I have 6 proofs I have due Wednesday for Geometry"
*cough*SPARKNOTES FTW*cough*
nosrsly... sparknotes has saved me so many times. <3
What are they? Trig proofs?
tanθ = sinθ/cosθ OR secθ/cscθ
cotθ = (tanθ)^-1 = 1/(tanθ) OR cscθ/secθ
secθ = (cosθ)^-1 = 1/(cosθ)
cscθ = (sinθ)^-1 = 1/(sinθ)
cosθ = -cosθ
sinθ = -sinθ |
|
|
I am going to learn that next year I think.Let's just hope that next year I know some people that can help me make a better equation than this. |
|
pwnedu46Join Date: 2009-05-23
Post Count: 7534 |
@gavin: The equations I posted (not trig ones, the kinematics ones) are the easiest you'll see out of all the kinematics: 1 dimensional, linear. It gets worse when you add the y axis, because you have to apply the trig to get x and y vector components. Rotational motion requires conversion to radians AND trig, and I haven't done any 3 dimensional kinematics. I _SERIOUSLY_ underestimated my Pre AP Physics class when I was signing up for classes. DX |
|