[Suggestion] How about Mechs

You countered a post by simply saying 'No you're wrong!'

Literally.

You've said nothing else as to why or how...

That is how a Troll works...
...That is how you work.

Ergo - You are a Troll.

I bet you have no idea why they were OP. And here's why. By your own statements the only evidence you give is "It has legs, its freaking OP as hell."

Explain yourself. Explain how a BFR was OP. Oh and don't bring up weapons or armor. Because both of those could be applied to something with wheels and treads. We don't want you contradicting yourself.

If you can't do this, I'm just going to quote you as supporting BFRs because you have no counter argument.

sigh, I've provided the proof in my other posts that you are wrong, I don't have to repeat everything. Just like you didn't repeat anything of what I supposedly did wrong, you only said "you've been disproven!" without any backup. Unfortunately for you the proof you would be referring to is faulty as ****.

I believe you. I don't think you are trolling.

You're just so stupid it may as well be.

the BFR was OP in the context of Planetside 1 Because it was a 1-Man MBT with Better Armor provided through a recharging vehicle shield and a large diversity of weapon options by having 2 Primary weapons which had a variety of combinations to counter any situation (AA, AI, AV - pick 2... or double up on one).

You have to remember in PS1, the MBT's 2-man operated war machines through Driver/Gunner functions... And their load outs where very much so locked as there was no such thing as AP or HE rounds in PS1.

There is also the point that damage on the BFR was localized. Which means un-like all other vehicles which had a mesurable Hit to Kill ratio - the BFR's HtK was variable depending on where you hit, making it even more resilient to enemy fire.

Also - it took them ~7 patches over 2 years, with the first 2-3 patches being nothing but constant BFR nerfs before they managed to balance it against the MBTs... And considering how long it took them to balance the Cert Shovel lol-pods in PS2 and the Dev's penchant to introduce things in an OP State initially (ala Harasser)... trying to introduce BFR's into this system is a really really bad idea.


The entire discussion about Legs started off of the statement from the fact that Mechs in Reality are impractical as hell - being both expensive and no where near as effective, powerful, or stable as a simple tank with tracks. This is what has enveloped a large portion of the thread...

In otherwords you admit that the differences between PS1 and PS2 warrant the point of "it was OP there, it would be OP here" invalid.

Sounds like MBTs/Lightnings to me. Read my suggestion in the beginning pages. It suggests that Mechs follow the same rules as tanks. Where you hit determines total damage. But a unique location based on the unique physiology. If MBTs and Tanks aren't OP because of hit locations causing different damage, then neither would the mech for the same reason.

Since I've played this game I've seen:

SMGs
PA-Shotguns
New iterations of Carbines, LMGs, Assault Rifles, and ES pistols
AV MANA Turret
New ES Rocket Launchers
New MAX weapons
New MAX abilities
Archer
Emissary
New ES Sidearms
Crossbows
Power Knives
Harassers
Valkyries
Wraith Cloak
New ESF Weapons
New Liberator Weapons
Icarus Jumpjets
ES Secondaries
Construction System
Alerts
Continent Locking
And a few other things I'm pronbably missing

Are you suggesting NONE of that should have been added? I'll admit some of that stuff has been OP at first (though never to the degree of what people claim), and many have been UP. Some was just about right. Changes and additions keep the game fresh.

And that was dumb as hell. Mechs/BFR's exist in the Planetside universe.

But seriously, go back to the beginning of the thread and check out my suggestions. They may help alleviate your fears. I've found a way to add them, give them a role, and make them NOT be OP. And definitively not replace any other vehicle. Except maybe the Lightning, but only as a multi-crew vehicle (which still gives the lightning as niche role due to its deployability).

Really i have to do this?
okay...

You have literally described a MagRider equipped with Rocket Pods.

Omni-directional movement, Heavy Armor immune to Small Arms fire, Non-standard movement allowing it to get over barriers that is equipped with an AP/HE Generic direct fire weapon which has a magazine of +10 rounds... Yeah - Mag Rider + Rocket Pods. And lets not forget about the Rumble Seat so we can have a Tag-Along engi to repair it.

And you think that a MagRider with Rocket Pods which can get into Infantry Only Designed Zones with a Pocket Engi repping it as carves a swatch of death is supposed 'alleviate' fears?

You mind posting a screenshot of a so called infantry designated zone? As far as I'm concerned. Those are inside buildings. Which a mech could not hope to fit into.

Basically those tank barriers in bases and such (think Amp Stations)...
you know the very thing you think they should be able to get over.

A number of bases through out each of the continents have these separating partitions designed to keep heavy vehicles (Tanks) out of open air fighting zones designed predominantly for infantry engagements. Either Bollards on stairways or A series of barriers across an open area. The vehicles which can snake their way into those fights (Aircraft, or savvy drivers of ANTs, Sunderers, ATVs, or Harrassers) either can be directly countered through small arms fire or can be engaged with relative ease by exploiting weaknesses (blind spots, narrow firing fields, occasional reloads) which is why they are allowed in those fights.

A good place to get stuck, even with legs. A good place to get that moneyshot C4 placement. A place you would get rocket podded from ESFs anyway (cept a mech can't make a quick escape like a ESF). The place away from the gens, SCU, and control point.

Come on now.

I'm sorry are we just ignoring the fact that you think a Rocket Pod MagRider is a balanced idea?

I don't think the base lightning weapon is OP, no.

I think mechs should be single pilot walking machines with lighter armor than MBTs, but more firepower and mobility - about lighting level armor.

However, I would give them a primary MBT weapon plus duel sunderer level secondary weapons all that the pilot could fire by switching like ESFs do with their 2 weapons, but with much less maximum ammo capacity unless it eliminated a weapon or two for more capacity.

It would have ANT turbo level climbing ability and 1/2 harrasser turbo speed.

As for size, it would be 50% bigger than an MBT, but upright.

It would cost 600 nanites to deploy & engineers/repair vehicles would only repair it at 50% normal speed.

So... where's that explanation?

That is only because it is on the Lightning...

The Viper's strength comes from it's sizable AoE, where it actually out shines the damage output of most Rocket Pods. Give the thing a little height where you can intentionally try and near hit targets and the Viper absolutely ruins ground targets like Infantry. The Lightning, however, provides no such vantage... even if you prop yourself on an over looking hill, the poor gun-depression makes it difficult to fire down on targets.

This is the other side of the problem...
Besides the fact that the proposed mech does not fill any desired role, the nanite to effectiveness ratio is simply too low...

Oh, I think the mech would be used instead of a lighting in many cases - except when the player lacked the nanites to pull it or when their faction had no tech plant as I think the MBT techplant requirement should also apply to mechs.

Originally I had intended to make a detailed explanation for the Mech foot, but considering how things like shear stress, compaction, air resistance and other things are completely misinterpreted time and again the only way to do it any justice and actually convince any of you would be to make an entire book on each and every subject, and even then the pure want to not have Mech legs function will mean you’ll grab anything to try to wiggle out of it. The way shear stress was handled is a perfect example. One word on a wiki gets you searching, you find something that’s called shear stress and even though this particular shear stress is meant for something completely different people grab on to it. When I point out the mistake, I still get insulted.
So there’s no reason for me to go into such detail. I’ll name the things I need to so I can prove myself right (again), and then leave you guys to it.

First on the agenda: stuff getting stuck in the earth.
To understand this I’ll discount the soils capability to carry a load completely. This to give a representation of the worst-case scenario of soil holding an object in place. This is accurate, since the soil pressing down on the sides of an object will offer just as much resistance for upwards movement as downwards movement if there’s no ground supporting the object:





The Grey arrows (GA) represent the forces on the sides of the object caused by the soil (brown), a Pilon (blue) for instance, the red arrow (RA) the force with which the Pilon is pulled down by it’s own weight. These forces are the pressure of the soil on the object. Since the Pilon isn’t supported by ground underneath, it has to have enough force pressing from the side. The deeper the Pilon goes, the more ground pressure will push against the Pilon and cause it to hang still. To calculate how much pressure we need, we need to use friction formula’s: http://formulas.tutorvista.com/physics/friction-formula.html

Unlike the friction formula in the link above, our force acting on the Pilon increases the deeper it goes. We also don’t use gravity to calculate the force but the pressure (although gravity is the cause of the pressure in the soil).
So the friction force has the following formula: Ff=U*Fn.

Where Ff = Friction force
U = coefficient of the soil with the side of the Pilon
Fn = Normal force of the Pilon (Blue arrows/BA). If the blue arrows are smaller than the GA the Pilon would be crushed and compacted until the Fn equalizes. A normal force cannot become bigger than the force acting on it, otherwise putting anything on a weight-bearing surface would launch you in the air.

Now for the interesting part: The Pilon will sag into the hole until the Ff is equal to the F=M*a of the Pilon (because the ground doesn’t carry weight right now). In this case, F= (Pilon Mass*Gravity acceleration)+external forces (the added weight and acceleration a house pushing down on it).

So the moment the Pilon comes at rest you can use the following formula: Fpu=Ffw+Mp*a
Where Fpu = Force pull up
Ffw = Friction force when weight bearing
Mp = mass of the Pilon without weight on it
a = acceleration, in this case gravity.

Now the question, can we take an estimate of the amount of force pressing on the sides of the Pilon? Yes we can. There’s at least two forces acting on the Pilon. One is a ‘spring’. When you compact something, some objects will exert a spring-like behaviour and try to pull back in it’s original shape. Rubber is a good example of a material that does this, sponges as well. Unfortunately things like soil, rock and other ground you stand on usually doesn’t have a lot of this, so we’ll ignore it for now. The other force is generated by the Angle of Repose: https://en.wikipedia.org/wiki/Angle_of_repose




In the above picture you see a pile of dirt that was thrown down. It has reached it’s angle of repose. The edges aren’t pushing outwards because the pressure required to move it sideways isn’t there. But we had a Pilon sitting in the way. If we removed the Pilon a part of this mount will slide downwards because of the pressure and the angle of repose. This means we now know the volume of earth that is pressing down on the Pilon: Everything above the blue line, because that’s all the dirt that would move if you removed the Pilon.
I’m sure there’s a formula to calculate the exact amount of pressure it exerts, but I haven’t found it beyond a doubt. It probably goes similar to this:
Fp=Fmve*g*sin(aof)
Where Fp=Force pressure
Fmve=Force caused by the mass of the volume of earth
g=gravity acceleration
aof=angle of repose
Fpx=vector force of the Force pressure. This is the exact force pressing against the Pilon pushing it sideways. Formula: Fpx=Fp*cos(alpha). The Fpy calculates the pressure this force exerts downwards.

This gives you an estimate of the pressure found on the bottom of the Pilon. However our Pilon isn’t stuck on just one side, but on all sides and we are digging it in:




Below you can see a comparison of a Pilon (or elephants leg), a human Leg and a Cone:




Although I’ve undoubtedly lost just about everyone by now, this step is still fairly simple.
The Pilon in the first picture simply has an increased pressure the farther it goes. This is what most people thought off when picturing tent stakes and feet: Deeper in the ground, more pressure, harder to get it out.
The foot in the second picture shows why a human foot isn’t very well build to get stuk in the ground. Not only do you have to deal with the added pressure, you also have to deal with the added weight on top of your foot. Human legs aren’t very well build to lift something straight up, which means the added weight is going to give you trouble real quick. You can even see it in the vector force: It either offers a friction force against the leg or a vector force downwards on the foot, keeping it in place.
Then we have a masterpiece of “told you so”.
The first thing to notice with the cone in the earth is that it’s closer to the aof. If the cone has less steep angles (the cone is flatter) than the aof there is even 0 pressure on the cone, and it would function 100% the same as lifting a track off the ground, unfortunately for a track that’s stuck in the earth you are looking again at the foot: A lot of earth gets scooped up on the tracks. If the angle of the cone is steeper than the aof, you still have reduced the amount of pressure on the cone compared to non-sloped objects like cone’s, feet and tracks that sunk the same distance.
The second thing to notice is the way the vector forces act. The vector force Fpx pushes against the angled surface of the cone, which means we can create another two vector forces, namely Fpx-y and Fpx-x. The beautiful thing here is that Fpx-y is facing upwards. This is much better than friction for our purposes. The moment you lift the Cone this Fpx-y force does not generate friction to keep the Cone in place, it’ll even reduce the force required to lift it! As a bonus we don’t have to multiply it to a friction coefficient, and even asphalt doesn’t go above 0,9 friction coefficient so we can use 100% of our Fpx-y force to prevent the Cone from going deeper!

I think that’s enough for today. Do you want me to go on about the amount of force soil can resist tomorrow? It has similar results btw.

You have yet to prove yourself right in any one of your posts, hence why you are discredited and insulted at every turn.... Trying to take the high ground while completely and utterly ignoring the tenets of the Scientific Method does not make you a martyr for learned thought my friend.

How do you come to this conclusion?
How does the increased traversal downward create additional horizontal pressure?

You're trying to hand wave it using a formula for Friction to identify the resistance factors at play, however that isn't how friction works... Friction does not create a pinching pressure which holds something in place - but instead produces a resistance that occurs on active connecting surface areas.

You see - this is where the problem occurs...

the Equation for Friction in this situation should be:
Fr = μN

Fr is the resistant force of Friction
μ is the coefficient of friction that occurs on interacting surfaces
N is the Normal Force which pushes the objects together.



Or more to more accurately display in your Pylon image-


You see - there is one flaw in your equation... What is pushing the Pylon or Soil against against the other to produce a high enough amount of Normal Force so that the Friction can counter act the downward momentum of the object?

http://www.geotechdata.info/parameter/soil-young's-modulus.html

Really?

Well, considering that soft soils have an elasticity of .5-5 Mega Pascals - or between 70 to 700 Pounds per Square Inch of pressure, it doesn't seem like you should ignore it... but that seems like par for the course for you... Why address something when you can just as easily ignore it?

Yes... You've managed to lost me.

Not because you've said something smart mind you... but because you've said something so mind numbingly stupid that it is painful to think how some one could of gotten to this point.

Now - to i have to ask...
Why did you assume the Area of Repose was a constant 45 Degree Angle?
I ask this because you intentionally use it as a convinent fact that the conical shaped foot is not effected by the area of Repose produced by your theoretical soils, and thus is seemingly not effected by the force. However - the only materials which have an Area of Repose of ~45 Degrees are things like Gravel and compact soil.... materials where neither an Elephant foot nor a Human one would be driven into the ground by the measurements you've assumed.
In fact, the one material which would submerge a human foot similar to the one you outlined in an image is something closer to a Non-Newtonian Fluid... To which - the area of Repose for those materials is a lot closer to 10-15 Degrees.
If you plug those values into your Math (which is nonsensical at best as it in no way pertains nor calculates the things you want it to) You would find that due to the higher surface area of the conical surface - It would actually be subjected to MORE pressures then the Human or Elephant foot.
Now - to this section specifically.

The first thing you have to notice is that there is no constant in any of your equations... no like measurements, or things to treat as a base line, not even a 'to scale' representation...

In this - the foot you illustrate is literally buried to the knee, the poor elephant's foot has sunk in almost to his belly, while the conical foot effects only the foot. We have provided more then enough math to prove that the mechanized walking beast you're talking about would more then overwhelm the ground pressures of most soft soils to the point that it would literally sink by a measure of Feet rather then Inches.

Yet it is not dug in by feet in your mysterious soft soil which has an angle of repose of 45 degree's but not the surface pressure to prevent a man from sinking 1 foot into it, or an elephant 1 meter.

It is almost as if you used 'wet-finger physics' simply assuming values, throwing out quasi-math phrases, and ignoring what is inconvenient to you in order to try and make an 'I told you so' point which is supposed to seem like you know what you're talking about...

Just like how you say Asphalt has a Friction coefficient of 0,9.
from that it simply seems like you know what you're talking about...

Except friction coefficients are not uniform values, they are varied based on the roughness of individual materials and are calculated based off of the normal force applied and the amount of energy put into the system...
μ=F/N


Second thing to notice - Despite presenting a lot of faux equations... You don't actually USE any of them. You just simply present them in an attempt to sound smart with out actually providing proofs or even trying to prove that they are viable in the least.

For instance - lets use one of the equations you've presented above using generic values.
Fpu=Ffw+Mp*a
Where Fpu = Force pull up
Ffw = Friction force when weight bearing
Mp = mass of the Pilon without weight on it
a = acceleration, in this case gravity.

Assuming a 'Friction force when weight bearing' of 'asphalt', a Mass of a man (80 kilos), and an acceleration of 9.8 m/s^2

Fpu=0.9 + 80*9.8
Fpu=0.9 + 784 netwons
Fpu=784.9 Newtons?

It's kinda weird that you just find the weight of some one then Add the 'Friction force when weight bearing' (what ever that is) and that becomes the Force Pushed Upward...

The Titans are IMO the weakest part of Titanfall (the pilot vs pilot is about the best COD I've played though). But I did like there being a mech on display in a garage (don't remember if I actually saw this in game or just a screenshot its been so long).

If you can't even see the difference between an object getting wedged into the ground without a supporting surface below it and an object sitting on the ground after I point out the difference there's no reason to go on is there?

Oh my god are you really this dense? This is just comical.

Well let's see, imagine there wasn't enough force to keep it in place, then it wouldn't take much extra force to pull it up. This is as simple as it gets and still you are having trouble. It's very simple: Imagine if the soil was lying on top of a rock surface, deep into the ground. You push the Pilon in and the soil you push away straight down disappears into a hole because we are assuming the soil below does not carry a load to give us a worst-case scenario for getting an object stuck in the ground. As the Pilon goes deeper, the pressure on the sides increase because there's more soil pressing against it.

As always, your wrong assumptions are through the roof. Read your own link, the Young Module isn't there to calculate the amount of pressure it exerts against objects, it's there to calculate the amount of settlement and deformation... when a force is applied. Note the "force is applied".

Where do I say it has a constant 45 degree angle? I never do, I never mentioned any type of angle and the angles of repose I drew are at random angles btw. And everything I say is there to help calculate the effects at any type of angle of repose. And everything I say still stands, regardless of using a 45 degree angle.

And you really didn't understand the drawings, since the angles of repose I drew actually do project a force on the Cone. If the red lines of the aof project inside the cone, the pressure of the ground does not push against the cone. I told you I had lost you already.

The aof I used to draw the foot is a lot steeper than 45 degrees, and you interpret it as 10 to 15 degrees? If it was 10 to 15 degrees the edge of the soil would have piled up almost to the top of the leg. You can't even get the angles right!

Also... Where's your math? Where's your counter-proof? You haven't even touched on this subject, and all you say "but but you are wrong! Really! I have nothing to prove and I have just shown I don't understand this but you are wrong!"

... You really want to show off how bad you are at this don't you?
F=M*a is a perfect formula. It tells you what you need to know regardless of the size of the object you are talking about. You need to know the mass, you need to know the acceleration, a force comes out. Easy.

Now we don't have something as simple as that, but it still functions the same. I have given you a string of formula's, a short string by the way considering the amount I usually have to work with in my freaking work where I do this kind of thing more often, and if you simply plug in some data yourself you have your own "like measurements"/baselines and constants.

Besides, what the hell do you need a baseline for? Do you need a baseline for F=M*a? Oh wait, there isn't really one is there? You might say something along the lines of a single kilo on the ground and it's gravitational pull, but that still doesn't say anything about the difference between fast and slow moving objects of different weights. You can simply use the formula and compare two things. There's no need for a baseline.

It's a drawing, not a 1:1 scale perfect representation. I could have drawn different depths but that would have meant less clear images.

Besides you have only provided math about human feet and elephants feet, you haven't taken into account all the advantages that a cone would have, and that's what I'm pointing out. Ofcourse you haven't understood it thus far so it's a small wonder you still haven't made that simple connection... Despite me saying it over and over again.

Gee, it's almost as if you don't have the capacity to understand that you can plug in any number and calculate what happens. So rather than getting stuck in the notion "but a human feet is far lighter" etc, why don't you scale everything up? It's the obvious thing to do.

We now have 3 Mechs, each with different legs. One with feet shaped like elephants foot (but, because you will grab any chance to misinterpret, the elephants foot is the same size as the Cone-foot), one with a human foot (again it's the same size as the cone foot), one with a cone-shaped Mech foot.
What do we see?

The Cone-shaped Mech foot, if you plug in the numbers, has far lower pressure keeping it in place when you pull it out. Part of the pressure that is pushing on the Mech foot actually helps keep it afloat and doesn't exert friction forces to keep it in place when you pull it out. So that's already a double win for the cone shaped foot. Also if you plugged in the numbers the difference would be pretty damn big.

Hmm, let's see.

Ff=U*Fn was the formula I used for calculating friction. Move the Fn to the other side and you get:
U=Ff/Fn, which is the same as your formula.
Now for the fun part. Ff and Fn aren't constants. And the amount of Fn is determined by the amount of force applied. But U is still a constant, try it out. Place objects of different weights on a slope and calculate when they'll fall (here's a simple site that explains how: https://revisionmaths.com/advanced-level-maths-revision/mechanics/coefficient-friction). What do you get? U is always the same number, because it's a constant. You can change this number by wetting the asphalt, but you would get the same coefficient for wet asphalt every time again.

Failed again, how many times is that now that you were completely and utterly wrong about something?

I don't have to use them. You have to understand the formula. I could have calculated the vector forces for you, but what would it accomplish if you don't understand the formula? It would just be a number, and it's size wouldn't tell you anything. If you do understand the formula, it would be a quick test for you to apply it and see that it's correct.

You don't even understand the formula of friction force? And you are trying to tell me about the coefficient?

Just use the normal friction force formula (which you basically did above) and calculate the friction force of a man on asphalt. You get an answer, then tell me what that answer means. You should have had this just after primary school or something. This is the basic of basics, and if you can't even understand this...