30 replies to this topic

[ManDaisy]

OK calm yourself.. I think its apparent that no one is asking for mw2 jump skaters.

[Alois Hammer]

ManDaisy, on 08 March 2013 - 06:38 AM, said:

OK calm yourself.. I think its apparent that no one is asking for mw2 jump skaters.

I think the exact opposite is pretty obvious- more maneuverability, for example changing directions once in-flight, is a move toward more MW2 jump skating and not the direction we need to be going in.

[ManDaisy]

It doesn't have to be one extreme or the other, that just depends on the amount of tweaking done. Adding a bit more maneuverability isn't going to suddenly turn it into gundam style anime flying.

[Phaesphoros]

Oh well, now that we got a poll, I'll say something about the "math" again (it's physics)

First of all: force = mass * acceleration. JJs provide a force, so divide by mass to get the acceleration (neglecting drag).

JJs currently lift a mech, that is, they're able to accelerate it at least 10 m/s^2 (1 G). Now the JJs don't just keep it floating (exactly 1 G), but LIFT the mech. The remaining JJ force/acceleration besides the 1 G provides acceleration - the more, the faster the mech accelerates.
Top speed in RL is limited by drag, as drag is proportional to speed^2 (2*speed => 4*drag). So you accelerate => more speed => more drag. Eventually drag becomes equal to JJ force. Then no acceleration, but you keep the speed until you run out of fuel.
Air resistance / drag becomes negligible at high weight and/or low speed, and for aerodynamic mechs (sharkskin coating whatever).

If you use just the 1 G to keep the mech floating, the rest of the force could be used to de/accelerate the mech horizontally. There's more drag if you move a CTF forward compared to lifting it, so the top speed and acceleration are slightly lower. If you neglect drag (e.g. your standing still and then use your JJ to accelerate to 10 km/h <- low speed, high weight), you can reach the same speed vertically as horizontally.Because your leg movement has to match your speed whilst on ground, you cannot just use all of the force to provide a MASC-like boost. But you can do that in mid-air.

The calculation I've done earlier contained a slight mistake (250 t instead of 25 t d'oh) and the result wasn't presented in a very clever way (I'll fix that in the original thread, too).

• It takes 4 s to slow the 100 km/h COM down to 90 km/h by drag only.
  
• It takes 30 s to slow it down to 50 km/h.
  
• It takes 30 min to slow it down to 2 km/h.

(For a truck, friction between the wheels and the road will become important at low speed.)

Edited by Phaesphoros, 08 March 2013 - 08:19 AM.

[Phaesphoros]

I'd love to see this (change trajectory) using the W and S keys to control thrust vs lift. Currently, you can control your throttle with them, immediately slowing down from 100 km/h mid-air to 0 when landing (silly, should cause leg damage).

I've suggested this somewhere before but cannot find it o.O

[ManDaisy]

Well, mathmatically speaking... what would be the quantity of acceleration that can take place in the air if you assume no thrust was diverted to keeping the mech in the air and all of it was used to add in other additional directional forces?

For example:

1) Deceleration rate in Direct opposition to forward momentum caused by jump,
2) A horizontal force perpendicular to forward momentum, that would result in a diagonal movement.

Hmm my brain is hurting.. as I think you would need velocity to get the speed right. Force example a spider would have X force when traveling at some speed. To get the amount of time needed to slow it to 0 you would only need to get the force equal to zero. But assuming an object is in motions the equaltion force = mass * acceleration does not apply because if would no longer be accelerating but still at 30 ton object traveling at a velocity.

As acceleration is cumulative, oh I dunno anymore...

p=mv would have to used somewhere and I dont want to dig up anymore physics or I'll be destracted all day.

Edited by ManDaisy, 08 March 2013 - 08:51 AM.

[Phaesphoros]

o.O I'm confused, guess that makes 2 then, I'll just note something about force/momentum but this currently bloats the thread a bit.

p = m*v <- momentum = mass*velocity
F = m*a <- force = mass*acceleration
F = dp/dt <- force is how momentum changes over time
a = dv/dt <- acceleration is how velocity changes over time

When there is no force or all the forces cancel out, a mech/body moves on with the speed it currently has. Forever.
When you're in a car on the highway, you only keep the speed when you keep pushing the gas pedal (keeping it at the same level, which is not 0). Why? Because the force (of the motor) has to cancel out drag and friction. Same with JJs or walking mechs.

When you use JJ, you'll accelerate until either you run out of fuel or drag force is equal to JJ force (then you keep the speed until running out of fuel).

When you keep the mech floating (1 G) and use all the remaining JJ propulsion to move it horizontally, the acceleration/top speed is almost the same as it currently is vertically. To get the impact on manoeuvrability when ALSO use this 1 G, you need to know how much force the JJ actually have.

Oh well, I've made a simulation finally Used the same numbers as before, but top area is 2 m^2, 25 m^2 front. The numbers outside the brackets refer to moving forward/backward, the number inside the brackets refer to moving up/down. This is different as air resistance is proportional to area (and top area is different from front area).

• Starting from 0 km/h, if you use 1 G to accelerate, you reach 110 km/h (115), having covered 50 m (50) before your 3.25 s fuel runs out. Sounds somewhat reasonable. If you use 2 G JJ force to lift, the result is the same (2 G JJ - 1 G gravity = 1 G acceleration).
  
• Starting from 0 km/h, if you use 2 G to accelerate forward (e.g. falling down by 1 G), you accelerate to 215 km/h (230) and move 100 m (100) before running out of fuel. Of course, it's possible you hit the ground before.
  
• Starting from 100 km/h (cannot neglect drag any more!), if you use 1 G to accelerate, you reach 135 km/h (140), having covered 195 m (210) before your 3.25 s fuel runs out.
  
• Starting from 100 km/h, if you use 2 G to accelerate, you reach 290 km/h (325), having covered 180 m (190) before your 3.25 s fuel runs out.

In 3.25 s, without JJ/lift you'll fall down 50 m (neglecting drag), reaching 115 km/h.

Edited by Phaesphoros, 08 March 2013 - 10:43 AM.

[Oppresor]

Oh Yes! Give this to the Spider; especially if its carrying an ERPPC, things could get very interesting for the other lights.

[M0rpHeu5]

I voted no but if they change trajectory by very little i am ok with that.

Edited by M0rpHeu5, 08 March 2013 - 05:01 PM.

[Chou Senwan]

So by 'no' do you mean 'yes?' Because that's what we're asking for: the ability to change trajectory a little.

Like, if I'm about to miss a rooftop by 2 or 3 meters, I'd like to be able to adjust sideways enough so that I don't fall like an *****. This would be especially helpful, since the game currently doesn't give you very precise analog turning, so if you go for a long distance jump it might actually be impossible to line up the precise angle you need to stick the landing. Let us tweak in mid-air.

[Alois Hammer]

Chou Senwan, on 08 March 2013 - 06:07 PM, said:

So by 'no' do you mean 'yes?'