8 replies to this topic

[414de7fe6]

A shot from a Armor mobile (yes gunbound references) is 90% heavier than a boomer shot.

An armor shot will travel 5cm across the screen on half power (50.0). How much power would be deducted to work the shot to the exact spot  the armor spot landed?

If the wind is blowing a power of 19 to the northwest, How much more power would be added naturally to the boomer shot due to the wind?

[Frizzle]

Remind me of what the Armour mobile is. I haven't played in a long while.

[xx_WLT_xx]

Wee, I'd use an aimbot for this just kidding

Anyway, my guess is that there is a very big difference between the two mobiles. So, if the Armor's shot is 90% heavier than that of the Boomer's, I'd say deduct about 0.4 bars (random guess). And for the boomer, no additions but deductions can occurs.

Wee, that was a random calculation

[Noitidart]

Ewwwwwww!!!! Physics!!! I hate physics!!!! >_<

[osh]

My guess = about 23.67~ reduction


and if northwest = wind is against your shot, about 71.2~

If it's towards enemy.. then I'm guessing around 12 ish


//edit

I'm probably wrong, but oh well.  .  I'm only 14 and it was worth a try.  .  I'm not really a shot calculator.

Edited by osh, 05 May 2005 - 10:40 PM.

[Smeags]

The problem here is boomer doesn't follow a linear model.  At certain wind + angle levels it will divert backwards and at other it will skim low to the ground.

The angle + power correlation isn't like other mobiles as you have probably noticed. Probably when graphed it looks something like an absolute value function. With a set variable distance from the mobile.

Lets make an assumption that it is a linear model and all mobiles shoot the same**

D = vt
Velocity x is constant as air resistance is negligable.
Velocity y0 = v0sintheta
Velocity x0 = v0costheta
t = vy0/g
y = vy0t - 1/2gt^2

F = ma (so I would assume 26.31% instead of half)

**At this point you release you are missing the angle. Maybe I will look into it later but that is the first equation that pops into my head. G is some predeterimened number.

Also note as I am a gunbound geek from back in the day there is a chart that tells you the effective number of wind at a certain angle. I just remember the values range from 1-6 and I think northwest against wind was something like 4.

With other mobiles there is a simple calculator program that could estimate with 25% accuracy how much power you needed to get a certain amount of distance at a certain angle and certain wind. I'd give it to you but I don't have it on my calculator anymore.

Don't mind me I am currently studying for the AP Physics Exam.

Edited by Smeags, 06 May 2005 - 04:52 AM.

[Jamie]

you play gb?

add me, I'm BruteForcer

[414de7fe6]

Wind ^ buffers the shot upwards by a angle increase of 3% - meaning you'd have to deduct 18% of power for it to hit the enemy.

Using armor however, the shot is buffered by a angle + of 0.8%, so the shot would travel the same distance as it would normally, but you'd need to adjust the power by 0.4% (48.05) in order for the shot to hit.

That was with normal angle, but with true angle..

True angle on boomer is restricted to 30-50
on armor, 10-20

Firing a true angle shot with wind < is only possible of a slant of 86 degrees.

For boomer however, a hook can be achieved at 43-47 true angle (woflclaw) and will hit the enemy 6 cm away with a adjustment of 1.2% in power.

I've worked majority of this out, boomer is really really easy once you work out the physics.

[Jamie]

Boomer is also very easy when useing an aimbot