Crrraaaab Peeeoplllle

Greetings Mechwarriors,

As a free gift for members who purchased any tier of the upcoming Resistance Pack for MWO, PGI has given away a free King Crab, KGC-000(L) assault Mech! Luckily, I had pre-ordered the Resistance Pack mere hours before they released it! Considering the other variants of the Crab will cost you around $15, this made a great gift for all the qualifying members. I must admit, I had mixed feelings about the Crab as the release inched closer. The first couple of days after the release might as well have been the invasion of the Crab People!

Now, the defamation.

First of all, I don’t really like the look of it, but thankfully it looks better than it does in the vast majority of Battletech literature. Truly hideous! The looks aren’t helped by the way the default, and unchangeable, camo pattern looks. The pattern doesn’t have clearly defined borders for the primary/secondary colors and both are given a light purple gleam. Purple? No, thank you. There are also decals on the “hands”, legs, and torso that are only half changed by selecting a tertiary color. Not to mention the red of the decal on top that you can’t even get a good look at in the Mechlab! Geez. The cockpit also looks undersized compared to the outward appearance.

Second, boy, is this thing slow. At 100 tons, the Crab is certainly a deadly foe. It was designed to be able to bring down enemy Mechs in a single alpha strike and PGI didn’t skimp on this one. The Crab comes with two AC/20’s that will make you a feared opponent, but it will also make you a prime target. This is where the lack of speed really kills the Crab. Although this is lore-friendly, that doesn’t mean it’s good for gameplay. For people who are used to piloting walking walls this might not be a problem, but I just couldn’t take the lack of maneuverability.

For what it’s worth, I do appreciate the free gift from PGI, but this won’t be a Mech that I drop with. I was able to model my build after the KGC-005, which was at least a little fun. I mean, dual LB-10Xs? Boom, boom, motha trucka.

Good luck and good hunting, pilots.

Sources:

http://www.sarna.net/wiki/Main_Page

A Lengthy Love Affair: IV

Greetings MechWarriors,

Over the past few posts, I have been talking about the "Highlander Burial" and the physics involved. Today, we'll be bringing this subject to its smashing conclusion! We'll look at the forces involved in the impact between a Highlander and a filthy Jenner. This analysis involves the concept of impulse and the conservation of momentum. There are a few simplifying assumptions that we will once again make in order to get some useful numbers.

As a re-cap, let's look at what we know:

Highlander velocity right before impact: 32.85 m/s

Weight of the Highlander: 90 tons

Weight of the Jenner: 35 tons

Height of the Jenner: 6m

Let's start by converting the weights of each Mech into kilograms, the standard measure of mass of the metric system. One metric ton is equal to 1000kg, so the weights become easy to convert.

Weight of the Highlander: 90,000 kg

Weight of the Jenner: 35,000 kg

Now, there are a couple of ways that we can look at this:

1. This is a perfectly inelastic collision, the coefficient of restitution is 0, and the Jenner and Highlander become one after collision. (Conservation of momentum)
2. The collision completely crumples the Jenner and there is very little resistive force exerted on the Highlander. (Impulse)
3. The calculation of the Highlanders kinetic energy directly before impact will tell us how "hard" it hits the Jenner. (Energy)

I think we'll look at number 3 and then compare it to number 2. If the kinetic energy calculated in number 3 is far lower than the impulse required in number 2, we can get an idea of what will happen to the Jenner! Alright, let's do it.

So, the kinetic energy of the Highlander is equal to one half the mass multiplied by the square of the velocity. We could calculate the potential energy the Highlander has at the crest of it's jump, but we already know everything we need!

KE_H = ½ m_HV_H²

KE_H = ½(90000kg)*(32.85m/s)²

KE_H = 4.856E⁷  kg m²/s² or 48560 kJ! (Nowhere near enough to power the DeLorean…)

But, what we have is a unit of energy, not force! How do we get the amount of force from that? Energy is a unit of work per a unit of distance. Because the kinetic energy will be turned into “work” as the Highlander continues to the ground (through the Jenner), we will assume that the energy is dissipated over the 6m height of the Jenner.

F_I = KE_H/6m = 8,093,418.75 kg m/s² or 8.09E⁶ Newtons

Now, that’s a lot of force!

So, let’s calculate the force needed to completely crumple the Jenner. Impulse force is defined as the mass of an object multiplied by its change in velocity divided by the time taken for impact. The time for impact can be found by finding how long it would take the Highlander to travel 6m, the height of the Jenner.

6m / 32.85m/s = .183s

Given that impulse is:

F_i = m DV/Dt = (90000kg)*(32.85 m/s) / .183s = 1.62E⁷ N

Comparing this to the force derived from the Highlanders kinetic energy, we find that the kinetic energy is only half the required value!

So, in conclusion, we find that the Highlander lacks the energy required to completely squash that little Jenner! This, of course, doesn’t mean that the Jenner isn’t in for a world of hurt. The force of impact would put enormous strain on the legs, joints, and cockpit components and would certainly lead to failure. Even without taking into consideration the structural stability of the Jenner, it would be safe to assume that it would not fair very well. Let’s assume that the Jenner’s structure was able to absorb the impact; that still says nothing about the ground supporting the Jenner! The ground could certainly give way before the Jenner.

There we have it, a much much much deeper look into this little fanciful situation than was ever called for!

Over this coming week, I hope to post about some House Steiner technology and the early release of the King Crab in MWO!

Good luck and good hunting, pilots.

WVR-6R on HPG Manifold

Not a bad match, but not my best. Enjoy!

WVR-7K on Canyon Network

This thing is fun!

Patch #35 Comes Tonight!

More information on the patch here!

This patch brings some fixes to Viridian Bog. They have apparently toned down the fog effects on the map. Personally, I think having so much stuff to get stuck on was a bigger problem, but hey, UPDATES!

A Lengthy Love Affair: III

Greetings MechWarriors,

Over the past few posts, I have been talking about my adoration for Highlander pilots and their unique maneuver. This time, I’d like to take a peek into what would happen in real life to a Light Mech caught in the path of a dropping Highlander! There are a number of simplifying assumptions that I will make to ball-park useful numbers. This will include the values for acceleration due to gravity, wind resistance, the reality of three dimensional vectors, etc, etc. Although these assumptions remove us from the “real world” and are somewhat idealized, they will allow us to fill in as many gaps as we can to come to a conclusion. And, let’s be real here, there are some details that we just can’t get from a work of fiction!

So, let’s start by determining, as best we can, what the jump capacity of a Highlander is. A 90m jump capacity is what is listed on sarna.net, but a 20m jump capacity is listed in MWO. We’ll split the difference and assume the jump capacity of the average Highlander is 55m. Next, we need to determine the height of a Light Mech for us to crush. An Atlas is referenced as being 13-15m tall. A Jenner, being an average Light, would stand just under half as tall. So, let’s assume that a Jenner is 6m tall.

So, what this tells us is that a Highlander will be able to produce around a DY of 55m. At the peak of its ascent, the Highlander’s +/-V_y and +a_y will be zero, as it’s no longer traveling upward. Given the height of a Jenner, the Highlander will only have 49m to fall before it strikes the Jenner. Now, we need to determine how fast the Highlander will be traveling when it strikes that puny Jenner. To do this, we will assume that the acceleration due to gravity is what it is at sea-level on Earth, ~9.81m/s². For simplicity, we will assume that the Highlander travels all 49m without experiencing positive acceleration or external forces. Given the change in position and the acceleration during motion, we can determine the amount of time it would take the Highlander to strike the Jenner.

DY = a₀ Dt²/2 + V_Y0Dt

This equation states: The change in position is equal to the initial acceleration multiplied by the (change in time)² all divided by two PLUS the initial velocity multiplied by the change in time. Because our Highlander is not moving when it comes to the peak of its jump, the initial velocity is zero. The initial acceleration is not zero because the Highlander’s acceleration has simply changed from a net positive to a net negative, or it is always experiencing gravity. Simplified, we’re left with:

Dt = sqrt(2DY/a₀)

Plugging in our known values, we are able to determine that the Highlander’s descent will last 3.35 seconds! Now, that’s moving pretty quick! Next, let’s determine the velocity at the time of impact.

DV_y  = a₀Dt + V_Y0

Again, our initial velocity is zero, so, we’re able to simplify the equation once more.

DV_y  = a₀Dt = (9.81m/s²)*(3.35s) = 32.85 m/s

There, we have now determined how fast an average Highlander would be travelling when it strikes a Jenner after a full jump in Earth equivalent gravity. How exciting! But, you may be wondering where we are to go from here if we want to find the amount of force a Highlander would impart on a Jenner. Next time, we’ll continue our look into the details of that very question! Anyone that is interested in reading ahead will find what they’re looking for from the concept of momentum and impulse!

Good luck and good hunting, pilots.

Sources:
http://www.sarna.net
http://mwomercs.com/forums/topic/4688-mwo-battlemech-scale/

An End to the Tournament

Greetings MechWarriors,

This morning marks the end of this weekend's MechWarrior Online Tournament! Congratulations to everyone that completed the challenge and who was handsomely rewarded for their efforts!

This was the first tournament that I have participated in and let me tell you, what a grind this one was. The objective of this tournament was to obtain 20 Tournament Points. These Points could be earned in only one way: get at least one kill, at least one assist, survive the match, and win. Sounds easy right? Well, maybe it was for some, but I thought it was a little grueling and, at times, demoralizing. However, I'd like to personally thank whoever it was at Piranha Games that decided to give everyone three days of Premium Time (normally a paid-for service) starting after their first Point! Thanks, pal and/or gal! That really helped me a lot. With Premium Time you gain an extra 50% to your match C-Bills and Exp. That means that while grinding for the prize, I would gain an 80%  C-Bill bonus while running my Hero Mech and a 50% experience bonus as well! F-to-the-yeah, buddy!

The prize for suffering slightly through this tournament was a brand new, crisp, fresh-out-of-teh-box, VTR-9S, Victor. Competition was stiff, but at least a couple of people in my Unit got theirs! Here's a picture of mine.

Now, after all that has been said, I must also say this: I don't like Ultra AC/5's, of  which, the Victor has two. I'll write a post someday as to why I don't like them, if it wasn't obvious after using one. So, once again I'm looking into different builds. I have since changed a couple of the ones that I posted before, but we'll talk about them another day. I've already gone on long enough, so here's a screenshot of what I was thinking for my VTR-9S.

No matter how much I play with this Mech, it will only ever be my second favorite Victor. This would be my first.

Good luck & good hunting, pilots.