Friday, August 7, 2009

Update to project


There's now a not-so-cute rat that chases after the player. Cut me some slack, I'm not an artist.

Basic AI for chasing the player has been setup and works nicely. Now I just need to add the ability to attack (and more artwork showing this). I've also been completely unable to get this to work on anyone elses computer. I have a feeling it has something to do with the directX SDK.

Thursday, August 6, 2009

New Project


It's a very simple game. The only buttons that do anything right now are WASD. And here's the download. It took me a year to figure out object oriented coding and the directx library. It took me about 5 hours to make this. The program needs directX and .NET 3.5 to run. Most windows installations should already have this.

Tuesday, August 4, 2009

Blog 2.0

This blog has been upgraded to the obligatory 2.0.

Woo.

Yea, I'm even less excited than you are. But I am happy with the nifty links section to the comics I read on the right.

Wednesday, March 18, 2009

An Open Letter

To Mr. Ross:

I haven’t done any debunking in about a year. But imagine my surprise when I visited your website recently and found that you had finally crafted a “response” to my refutation of your “Moment Transfer in WTC1” “paper”. Please note the words that I have in quotations: response and paper. This is because I hold these items in contempt, not because they’re imaginary. I also find your lack of testicular fortitude contemptible. When in a lively internet discussion such as the one that we have had, try to have necessary courage to actually inform your opponent that you’ve said something. Otherwise it would appear (as it does now) that you’re afraid of a real response. Let’s take a look at what you’ve written and I’ll respond to it paragraph by paragraph.

If only you could investigate 911 as thorough as you investigated my name.

A few people have written to me over the past few months regarding an article by Mr. Trevor Self, from Albuquerque I believe, styling himself Newton’s Bit. Rather than continue to answer these individually it will save time and effort if this reply is placed on the web and freely available. I have not previously bothered to answer this article because I did not believe that anyone would be taken in by his rubbish, riddled as it is with very basic errors, but for those who have not studied the subject it may prove beneficial to have some of these errors explained. Firstly I will deal with the arithmetical errors, then I will explain the engineering errors.

You’ve outed me! You have done a wonderful investigation and examined all the various clues I’ve left of myself over the internet and figured out my real name! I fear the massive unwashed hordes of Truthers hounding my every step. Or maybe I would, if I wasn’t sure that the hygiene-challenged Truthers were huddled in their parent’s basements playing the latest Halo game.

First of all the conversion from degrees to radians used by Mr Self is incorrect. There are pi (3.142) radians in 180 degrees, except apparently in New Mexico. This introduces an error of 200%.

You got me, an arithmetic error. I intended to use 30 degree angles (pi/6) but incorrectly used 15 degrees (pi/12). Unfortunately for you, 15 degrees still falls within the bounds of my “fudging”. You see, steel columns have ruptured by 8-12 degrees anyways. Let’s just call this one a wash. I made an error, but it doesn’t matter.

There are four rotations in a three point buckle except in the mind of Mr Self who believes there are only three. A further error of 133%.

Incorrect. This should be self-explanatory to an engineer, but I guess you didn’t have to take any Mechanics of Materials classes. Under an arbitrary amount of work, the top and bottom buckle points will rotate X degrees, however the middle one will rotate an angle of 2*X. Each buckle point absorbs the same amount of energy. Let me know if you need me to explain this further, it’s a tad bit complicated (I’m lying here: I’m trying to protect your feelings, it’s really not complicated at all).

Mr Self uses a slenderness ratio which assumes that the columns in the towers were fitted with hinges on every storey. A casual glance at the towers proves this false, and the very fact that they stood for many years would help to confirm the non existence of these hinges. The error in slenderness ratio is 200%.

Do not put words into my mouth, HVAC designer. I never said that the tower was fitted with hinges. I can only surmise that you are making the same basic mistake that Tony Szamboti made regarding the effective length factor “K”. Please see my response to him, I’m really getting tired of having to correct this insanely basic concept of engineering. Here’s the link.

Educate yourself.

Mr Self chooses to call himself Newton’s bit for some reason but his refusal to accept Newton’s laws would have that famous man turning in his grave. Isaac Newton, or “whirling Isaac” as he is now known told us that each action has an equal and opposite reaction, but Mr Self chooses to ignore this fact conveniently allowing him to understate the energies involved by half. An error of 200%.

Hmm. I’m not sure how my handle has anything to do with what I write. Nor do I see where this “error” occurs as my paper only deals with recalculating things you did incorrectly.

Mr. Self ignores the strengthening and bracing effect of the spandrel plates, core bracing, etc. The error is more difficult to quantify but is clearly significant. Why else would they have been included in the original design?


The spandrel plates do not brace from buckling in a direction orthogonal to their length. They provide stiffness to in-plane forces (thus a moment frame) to deliver shear forces to the bottom of the structure. This is basic engineering mechanics. There is no excuse for not understanding this.

These errors when combined add up to ridiculous. It is easy to see therefore why I have previously dismissed this article without much comment. The only interesting part of this episode has been the manner in which supporters of the official story have latched onto it. There are those without the specialised knowledge to judge, who have betrayed their own unthinking bias by adopting Mr Self’s article without question. More importantly there are those who are or claim to be engineers and who do or should have that specialised knowledge and yet they have allowed the article to stand and allowed

Mr. Self to continue to embarrass himself, even when these most basic errors have been pointed out.
I hope that this clears up a few issues for some people, but if questions continue then please do not hesitate to contact me.

As an aside, I have always thought that the custom on the web of allowing everyone to choose their own nickname is a little bit strange. If this were the case in real life then all the Porkys and Kiffys of this world would be calling themselves Ace or Tiger. Mr Self, or Newton’s Bit, as he appears to prefer, is a definite case in point.

You managed to find an arithmetic error (that's posted on the JREF forums) that doesn't actually change any results.  You also showed how ignorant you are of structural design.  Anyhoo, this has been pretty dang entertaining for me.  When you have more “problems” (this quotation is both for contempt and because it’s imaginary), please actually grow a pair and let me know about them instead of hiding it on your website.

Cheers!
Trevor Self

P.S.  My middle name is Newton.  And my blog is my bit.  Hence: Newton's Bit.  Do you get it?

Saturday, February 23, 2008

Why the columns bowed in

Update 9/20/08: Rephrased a couple of sentences, referenced reduction in modulus, fixed some weird formatting.

Why did the columns bow in? There’s been quite a bit of speculation and misinformed opinion about the mechanics of the structure that caused that, so I hope to do a little bit of enlightenment. The first thing that needs to be looked at is the problem. Figure 1 shows a typical building section through a building such as the WTC towers.

Figure 1

Everything here is fairly straight forward. The gravity load path can be seen very easily. The floor trusses deliver the vertical floor loads to the columns which deliver them to the foundations. But what happens when a core column is severed as in Figure 2?



Figure 2

Things start to get a little bit more complicated. The first thing that should dawn on most people is that the floor is no longer being supported by the middle column which is going to cause some problems. The middle column will drop unless there is a force that can resist it, see Figure 3.



Figure 3

As the column drops, the top chord of the floor truss develops tensile forces (it is quite literally stretched). This tensile force has two parts, a vertical portion that pulls the column up and a horizontal force that pulls the rest of the structure in. This can be seen in Figure 4.


Figure 4

The other things to note here is that the left column is still under its full axial loading (P2, which will be important in the analysis). The left column is already shown pulled in to some degree, however it is not to scale. Further modifying the problem, we know that the fires caused the trusses to sag to some degree. If the fire is hot enough, the truss will become a tension only member. This means that the top chord of the truss will act something like a rope and pull inwards at its connections. This can be seen in Figure 5. There is another condition as well (which I have no illustrated) that will cause the heated floor to expand outwards without losing its bending capacity and thus not sagging. It is a condition that likely proceeded that of the sagging floor trusses.


Figure 5

NIST (1-6D) estimates that the total pull-in force at the exterior columns is roughly 6 kips (6,000lbs) at each column. The truss to column connection consisted of (2) 5/8” diameter bolts. Even non-structural grade bolts of this size will have a shear capacity of over 5kips each, so it is reasonable to assume that the top chord of the truss will not pull off of the columns at the connections due to a 6kip load.

Is this 6 kips enough to pull the column in several feet as seen in the photos of the tower?


Figure 6

Math is needed here. First some assumptions need to be made. For the purpose of this analysis, let the exterior column be HSS14x14x5/16 tubes at 25% of the maximum axial load prior to any damage. The column has the following properties: Similar to HSS 14x14x5/16
A = 15.7in^2
I = 739 in^4

S = 92.3 in ^3
Pn = 557k (from AISC LRFD 3rd, table 4-6 with an unbraced length, KL = 12’-4”)

Pu = ¼* 557k = 139k
Mn = 92.3in^3*46ksi = 4645 kip*in (Mn = maximum bending capacity)

The column itself will bend inwardly until it snaps if the column itself ever becomes inelastic. This can be defined by the ratio (I’ve simplified this a bit): Mu/Mn + Pu/Pn <>

Figure 7

CALCULATION 1: DIAPHRAGM DAMAGE, NO FIRE EFFECTS
Unbraced length = 37’-0”

Pu = 139k
Pn = 465k (from AISC LRFD 3rd, table 4-6 with an unbraced length, KL = 37’-0”)
Mn = 4645 kip*in
Mu = P*a

Mu = 6kip*1/3*37ft
Mu = 74kip*ft or 888 kip*in

Deflection = P*a *(3L^2 – 4*a^2)/(24*E*I) (Formula from AISC LRFD 3rd)
a = 1/3*L
= P * 1/3L *(3L^2 – 4/9*L^2)/(24*E*I)

= 23*P*L^3/(1296*E*I)

= 23*6k*(37ft*12in/ft)^3/(1296*29000ksi*739in) = 0.435in

Additional moment due to P-delta

Mu+ = 0.435in*139k = 61.02 kip*in

Additional Deflection
=Mu+*L^2 / (4*EI)

= 61.02kip*in*(37 * 12ft\in)^2 / (4*29000ksi*739in^3)
= 0.140in

Additional moment due to P-delta2

Mu++ = (0.435+0.140)in*139k = 79.93 kip*in

Additional Deflection
= Mu++*L^2 / (4*EI)
= 79.93kip*in*(37 * 12ft\in)^2 / (4*29000ksi*739in^3) = 0.184in

As seen, the first p-delta iteration results in an increased deflection of 0.140in. The second results in a deflection of only 0.184in. We can thus conclude that p-delta will eventually converge and that no further iterations are necessary. The 6kip pull-in force with no effect of fire will not result in the column becoming unstable. At 600C, the Modulus of Elasticity will have reduced to approximately 0.3 of its original value (see Figure 8), and the yield strength to 0.5 of its original value. The effect of the Modulus of Elasticity being so greatly lowered is of far greater important than the yield strength, however.


Figure 8
From AISC Facts for Steel Buildings

CALCULATION 2: DIAPHRAGM DAMAGE, 600C TEMP

Unbraced length = 37’-0”
E = 0.3*29000ksi = 8700ksi
Pu = 139k
Pn = 465k*0.5 = 233k
Mn = 4645 kip*in *0.5 = 2323kip*in Mu = P*a
Mu = 6kip*1/3*37ft
Mu = 74kip*ft or 888 kip*in

Deflection = P*a *(3L^2 – 4*a^2)/(24*E*I) (Formula from AISC LRFD 3rd)
a = 1/3*L
= P * 1/3L *(3L^2 – 4/9*L^2)/(24*E*I)
= 23*P*L^3/(1296*E*I)
= 23*6k*(37ft*12in/ft)^3/(1296*8700ksi*739in)
= 1.45in


Additional moment due to P-delta

Mu+ = 1.45in*139k = 201.6 kip*in

Additional Deflection
= Mu+*L^2 / (4*EI)

= 201.6kip*in*(37 * 12ft\in)^2 / (4*8700ksi*739in^3) = 1.55in

Additional moment due to P-delta2

Mu++ = (1.45+1.55)in*139k = 417 kip*in

Additional Deflection
= Mu++*L^2 / (4*EI)

= 417kip*in*(37 * 12ft\in)^2 / (4*8700ksi*739in^3) = 3.20in

This results in the column becoming unstable due to p-delta. This can easily be seen in that the deflection due to P-delta2 is double that of P-delta1. It can therefore be concluded that it was necessary for both fire and damage to result in the collapse of the towers.

Saturday, January 5, 2008

The Peer Reviewers at J911

I've never really been generally impressed by the folks at the "Journal" of 911 Studies, but my latest encounter with one of their peer reviewers, Tony Szamboti, on the JREF forums leads me to believe that they're not only suffering from group think, but also egregiously incompetent on issues they proclaim to be experts on. Mr. Szamboti posited this question to a group of non-engineers:

What does the slenderness ratio of a structural steel column need to be to be in the inelastic buckling range?

What were the slenderness ratios of the central core columns at the collapse initiation sites of the 98th floor in the North Tower and 82nd floor in the South Tower?

I saw this question and jumped into the discussion, posting:
Inelastic buckling occurs for all slenderness ratios under the Euler limit. That's 4.71 * SQRT(E/Fy). Of course extremely stout members won't buckle inelastically, however none of the columns in the upper floors of the WTC were that stout.

Do you have any clue as to what you're talking about? I recommend picking up an AISC Manual of Steel Construction and see exactly how steel is designed these days. We're not in the 1940's, we know how steel fails now. Maybe you should update you knowledge to modern information.

Mr. Szamboti's replied:

How did I know you would come on.

You used an effective length factor of 1.0 in your letter to Gordon Ross, which is for a pinned connection, when you should have used .5 to .65 for fixed both ends connections for the tower columns. The 1.0 gave you larger slenderness ratios and they still weren't greater than 40. Now you are going to say the tower columns weren't in the short column category and would have been subject to inelastic buckling. The AISC equations you show here and which you used in your paper are conservative for design.

You want to say the tower columns would fail due to buckling. Well how about a test case were an I beam with a slenderness ratio of 20 or lower failed due to inelastic buckling. Do you have any test cases? I have an AISC manual right here. I am familiar with the equations and monograph. You want to go around asking others if they have a clue and you seem to be the one who should be asked that question Mr. Smarty pants.
This is where Mr. Szamboti shows his lack of knowledge as regards to structural engineering. The slenderness ratio that we are talking about, and what Mr. Szamboti struggles to understand, here is defined as K*L/r.

Where:
k = effective length factor
L = length of the column (in)
r = radius of gyratio of the column (in) - [This is a function of the geometric properties of a column]

For the columns that we are talking about, the variables L and r are very well defined and not argued. The effective length factor 'k' is where he slips up. In the commentary of the AISC Manual of Steel Construction (arguably the Bible of how to design steel in structures) this factor 'k' is defined. The first place is in table C2.2 (shown below).


Table C-C2.2 (click to enlarge)

Under column (a) it defines the theoretical k value of 0.5 and a recommended design value of 0.65. It's fairly easy to see that this is where Mr. Szamboti thinks the factor k is defined, as the tower exterior columns were moment frames, which means that the top and bottom portions of the columns were fixed. Column (a) shows a column element fixed at the top and bottom, so he used it. And he's very wrong. The commentary clearly explains how this table is to be used on the page before the table, "These range from simple idealizations of single columns such as shown in Table C-C2.2 to complex buckling solutions for specific frames and loading conditions". In his rush to prove me wrong, I can only surmise that he went through the commentary to find an answer to his question, and stumbling upon the first table that seemed to show an answer that confirmed his bias, he lept to a conclusion. An incorrect one not supported by the document he was referencing.

The correct way to calculate this effective length factor is with the
nomograph chart, shown below. The nomograph table is for frames which can translate horizontally, this contributes significantly to the stability of the frame.


Figure C-C2.4 (click to enlarge)

This table looks nonsensical, but it's fairly simple to use. First Ga and Gb need to be defined, which are simply a comparison of the stiffnesses of the columns to the girders. Ga is the comparative stiffness of the top point of the column and Gb is of the bottom. Then, to get the effective length factor, one merely needs to draw a straight line between these two points (see the figure below with Ga = 1.0 and Gb slightly stiffer).


Example nomograph

In this example, the k factor of a frame that has a column stiffness that is roughly equal to the girder stiffness is about 1.4.

It is very easy to see with this table that the lowest factor k that a column in a moment frame can have is 1.0, rendering Mr. Szamboti's statement that it should be 0.5 or 0.65 completely without merit. The purpose of this isn't to belittle or attack Mr. Szamboti as being an incompetent engineer. I'm sure he is an excellent mechanical engineer, however he is not an expert in structural engineering, far from it. He is most definitely unqualified to "peer-review" papers of a structural engineering focus for anyone.

Thursday, July 5, 2007

Gordon Ross Shows Collapse Progression

Background:
The paper by Gordon Ross claiming that the WTC would self-arrest collapse
My initial response
Ross replies and implies "fake but accurate"


To Mr. Gordon Ross and the Journal of 9/11 Studies:

Regarding the reply to my criticism of the paper by Gordon Ross entitled, “Momentum Transfer Analysis of the Collapse of the Upper Storeys of WTC1”, I have demonstrated below the actual method in which to calculate the strain energy of the columns at the WTC.

For all compression members with eccentric loads or externally produced bending moments, the bending moment is magnified as the axial force approaches the ultimate compression capacity of the member. This is what is known as P-delta effects. As the column deflects due to the bending moment, the axial force becomes eccentric and creates another bending moment. This bending moment causes the column to deflect further which in turn means that the axial force creates yet another larger bending moment because the column has deflected outwards further. This continues until the column either reaches equilibrium or it fails. P-delta can be explained by the differential equation:



Where:
Mz = Magnified Moment (kip*in)
Mi = initial axial load (kip*in)
P = Axial force (kip)
y = initial deflection (in)
E = Modulus of Elasticity (29000ksi for steel)
I = Moment of Inertia (in4)

---Note, most of the calculations following are in the British Imperial Unit System. The numbers at the end will be converted to SI. For reference, a kip is 1000lb, and a ksi is 1000 psi.

A general solution of this equation will not be attempted here, as it is long and outside the scope of what this letter is intended to do. Following Timoshenko and Gere, the general solution is for a column with constant initial bending moments at the top at bottom is:

Where:

And:
Pu = Axial Demand on the Column (kip)
Pn = Axial Capacity from Buckling (kip)

There is a limit to the interaction of bending and compression in a column. This is shown in the classic equation:

Or:

Where:
Mu = Bending Demand (kip*in)
Mn = Bending Capacity (kip*in)
e = eccentricity of vertical load (in)
Pu2 = Axial Load at an eccentricity e (kip).

In all real-world columns, there is always, at a minimum, a small but non-zero eccentricity between the axial force and the column. This is due to a variety of reasons, such as an unequal live load in two adjacent bays or a beam connection on the side of a column. For the purposes of this paper, let e equal a very small but non-zero number. As Pu/Pn approaches 1, the denominator of the term

approaches zero. The magnifier on the Mu/Mn term becomes infinity and the column stops being a compression member and becomes purely a bending member. In reality, the column would transition gradually at somewhere near 0.9 or 0.8 Pu/Pn depending on how eccentric the axial load was, however it is conservative in favor of collapse prevention, and easier math, to assume that this happens at Pu = Pn.

The strain energy of the column can then be calculated by separating the strain energy into two functions: the energy in compression and the energy in bending. These can be referenced from most engineering Strength of Materials books.





For the strain energy in bending, I am assuming 3 buckle points along the length of the column that rotate 30 degrees on each side. This is vastly conservative in favor of collapse prevention. In reality, columns will only go through an 8-12 degree rotation before failing. This of course varies depending on the column strength. The exterior columns were made of varying yield stress materials of up to 100 ksi. High strength steel such as this is more brittle than lower strength steel, such as the A36 wide-flanges, and will fail at a smaller rotation. Even then, a rotational fudge factor of 2.5 has been used to justify that the calculations are completely in favor of collapse prevention

To determine what the strain energy is, actual real column sizes are required. A guess on the cross-sectional area is not enough. To do this, columns based on the same data that Ross references are used to compute a cross-sectional area. The actual column size is then picked by using this cross-sectional area and the same nominal depth of the columns at WTC. The exterior columns were 14x14 box columns and the interior was made up of both box columns (of varying dimensions) and 14” nominal wide-flange shapes. Since the shapes of the interior box columns are not know, the interior columns are all assumed to be 14” nominal wide flanges.

Where (from Ross):
Weight Above = 59000tonnes
FudgeFactor = 4



There are 240 exterior columns. Assuming that the exterior is 50% of the total cross-section of steel, it follows that the cross-sectional area of steel of each is:

A similar shape is the HSS 14x14x1/2. The relevant properties to be used later in this paper are:
Ag = 24.6in2 r = 5.58in Z = 124in3 I = 743in3

There are 47 interior columns. The cross-sectional area of each is therefore:

A similar shape is a W14x370. The relevant properties to be used later in this paper are:
Ag = 109in2 Ix = 5440in3 Zx = 736in3 rx= 7.07in
Iy = 1990in3 Zy=370in3 ry=4.27in


The difference in strength between the two orthogonal axis of a wide-flange are shown with the x and y notation. In column failure, the column will always buckle in the weak-axis. Thus, the smaller numbers of the weak-axis are the relevant ones.

The calculations for the HSS14x14x1/2 follow, please note that the method for calculating the axial compression capacity is inelastic buckling, as seen in my previous letter:





The calculation for the W14x370 are performed in the same manner, the results are:

The total strain energy is:



Or a 269% increase in the absolute maximum. The real column sizes are required to determine what the actual strain energy in the columns are. This calculation is grossly in favor of higher strain energy and even then it is 2.69 times less than the calculation in the Ross paper.

Ross said in his response, “The conclusion of my article, that the energy would be dissipated in many more areas outwith the uppermost storey has been broadly accepted and no serious challenge to that conclusion has been forthcoming.” The summary of the energy losses from his paper:

Energy available;
Kinetic energy 2105MJ
Potential energy Additional downward movement 95MJ
Compression of impacting section 32MJ
Compression of impacted section 24MJ
Total Energy available 2256MJ
Energy required;
Momentum losses 1389MJ
Plastic strain energy in lower impacted storey 244MJ
Plastic strain energy in upper impacted storey 215MJ
Elastic strain energy in lower storeys 64MJ
Elastic strain energy in upper storeys 126MJ
Pulverisation of concrete on impacting floor 304MJ
Pulverisation of concrete on impacted floor 304MJ
Total Energy required 2646MJ
Minimum Energy Deficit -390MJ

Momentum loss and pulverization of concrete: This is another large mistake in Ross’ paper. When two objects collide, their energy is conserved. The typical method to easily to determine this is to assume a plastic collision, where M1V1 = M2V2. This is what Ross has done. He shows that in a perfectly plastic collision, there is an energy loss of 1389MJ. This is also true. However, there is a very important assumption in this that Ross has completely ignored. This loss in kinetic energy is transferred into the plastic deformation of the two bodies. In other words, the difference in energy from impact IS the “pulverization” of the concrete. Ross has calculated the damage of each concrete plate twice. In contrast, in an elastic impact, where the objects are not damaged by the impact, have zero loss in kinetic energy. Energy cannot be created or destroyed, however Ross has allowed 1389MJ (12*107 kip*in) to vanish. Where do you assume this energy goes Mr. Ross? From this alone, there is no longer an energy deficit in your paper, but rather an energy gain from floor to floor.

Mr. Ross, your conclusions and sums and methods have been proven wrong. In my previous letter I offered you the chance to fix and update your calculations out of professional courtesy. Out of respect for your abilities, I said it would be easy for you to do. I had hoped that you would take a harder look at that issue, and take another look at the rest of your paper, but you have chosen not to do so. Your response was nothing more than, “fake but accurate”. This is a disgusting manner for any engineer to respond. Sir, retract your paper or fix your calculations.