baxtrom

Two minor comments, first, I believe you are referring to Roman numerals, not Greek. Second, 90 would be written XC, or "ten from one hundred". Also, good to know Swenglish memory trick when dealing with Roman numerals: Little Camilla's Dancing Mazurka

The obvious advantage of using the diameter is that it is easier to measure than the radius. Pi was defined independently in different ancient cultures so the diameter seems to be a more natural definition of the size of a circle. Regarding the Swedish bikini team I believe they are only interested in circumferences and don't give a crap about diameters and radii.

Hi there, as far as I can see the solution to that DE is a parabolic, [math]y(x) = -\frac{x^2}{2} + C[/math]. According to mathworld this functional DE defines the normal distribution: [math] \frac{\mathrm{d}y(x)}{\mathrm{d}x} = \frac{y(\mu- x)}{\sigma^2} [/math], where [math]\mu[/math] and [math]\sigma[/math] are the mean and standard deviation, respectively.

Yet the effective stress in pure shear is [math]2 \tau[/math] according to Henri Tresca. Since in engineering we use [math]2 \tau[/math] so often I think we should use a new symbol to avoid confusion. Perhaps [math]\pi[/math]?

[math]\tau[/math] denotes shear stress and nothing else!

One could likewise argue that a better constant than [math]\pi[/math] would be [math]\sigma=\frac{\pi}{4}[/math], so that the constant is the ratio of areas of a unit circle to a unit square. Or, [math]\sigma=\pi^2[/math] so that [math]\int_{-\infty}^{\infty} e^{-x^2} \mathrm{d} x = \sigma[/math]. Et cetera. I think there are natural reasons to why the ratio of circumference to diameter instead of radius was chosen thousands of years ago. It's simpler to measure the diameter. The diameter is, in my opinion, a more natural way of describing the size of a circle than the radius. If a handful mathematicians think it will be useful to adopt the [math]\tau[/math] constant, so be it. It will however most likely be known as two times [math]\pi[/math]..

Sorry, actually you misunderstood but it's my fault Without thinking about it I used the term "shell" not for the projectile, but for the plate which is impacted. That was actually a bit stupid, but I'm used to the term shell in the context of "plate" (i.e. the target). So, the penetration decreases with target strength. The strength of the projectile is not considered in the formulas (assumed rigid if you like). Thanks for pointing it out to me. I'll edit the post if I can. Edit: I can't edit the post!

Come to think of it, steel also undergoes deformation hardening up until the tensile limit which typically is perhaps 50-100% higher than the yield strength for structural steels. Also as you point out, the equation does not consider elastic deformation. I imagine in reality a projectile would cause concentric waves propagating away from the area of impact, transporting energy and decreasing the penetration of the projectile. Also, the deformation of the projectile is completely neglected by my formula. However, the end result looks reasonable in one sense, since penetration increases with kinetic energy and decreases with caliber and shell strength. I guess the formula, if it has any correlation with reality at all, would be correct in the high velocity and thin shell limit, where I imagine shearing is the main mechanism of penetration. Thanks for the link. Looks like the software is not available for direct download, and I don't think I'll bother them by asking for it.

Hi there fellow desktop scientists and others that topic some time ago when a game developer asked about realistic damage formulae inspired me to come up with an equation for projectile penetration of a thin elastic-plastic shell. In the derivation below I neglect losses due to deformation of the projectile, and further simply assume that the projectile cuts through the plate by means of shear stress (i.e. wadcutter vs thin shells). Any comments or suggestions? I use the following symbols: velocity [math]v[/math] time [math]t[/math] initial velocity [math]v_0[/math] deceleration [math]c[/math] shell thickness [math]T[/math] force of resistance [math]F[/math] projectile mass [math]m[/math] initial kinetic energy of projectile [math]K[/math] projectile diameter [math]d[/math] shear stress [math]\tau[/math] shell yield strength [math]S_y[/math] Assume a linear relationship between velocity and deceleration, i.e. constant force of resistance, such that the exit velocity of the projectile is zero. Then, [math]v(t) = v_0 - c t[/math], where [math]c = \frac{v_0^2}{2 T}[/math] (this implies that [math]v = 0[/math] after the projectile has decelerated through a distance [math]T[/math]).The deceleration of the projectile is [math]c[/math] which results in a resistance force equal to [math]F = m c = \frac{m v_0^2}{2 T} = \frac{K}{T}[/math]. Assume further that the projectile cuts a hole in the target shell by shear stress along the circumference of the projectile, [math]\tau = \frac{F}{\pi d T}=\frac{K}{\pi d T^2}[/math]. According to Tresca theory (as preferred in the US!) yield will start when [math]2 \tau[/math] equals the yield strength of the shell material. Thus, as a condition for penetration we have [math]\frac{2 K}{\pi d T^2} = S_y[/math]. The maximum shell thickness that the projectile penetrates becomes [math]T = \sqrt{\frac{2 K}{\pi d S_y}}[/math]. This would imply that a 10 mm wadcutter, with a mass of 0.01 kg (10 gram) and traveling at 300 m/s (kinetic energy equal to 450 J), would be capable of penetrating approx. 9 mm of structural steel (350 MPa). Is that a realistic result?

..or it's one of those "one chance in 634 billion billion billion" events that make life interesting.

Ha ha, great subject for Jerry Seinfeld! I think they tangle because of the fourth law of thermodynamics, i.e. Murphy's law!

Personally I see the FFT as an optimized computer algorithm for obtaining the DFT, or Discrete Fourier Transform.

My list would include Laplace, Lagrange and Lasagne, HA HA HA! Hrrrmm. he. Ok, seriously. What about that guy who proved Fermat's theorem? Wiles? And that russian guy who's apparently sitting in some datcha refusing to accept both money and prestigious positions..

Using the plumber methaphor, you don't want discontinous pipes, especially not sewer..

Just relax and enjoy the squabble. I'd put my money on Dr Rocket, he may be a bit grumpy from time to time but he knows his math..

My guess is that it probably lacks in range. A soda can with a jet engine, not too much space for fuel! I bet the military evaluated an attack version with a minigun and hellfire missiles, though! Rule of lawsuit!

Imagine you're a plumber. Let's say you put up a pipe on a wall, and you want some slope on it so that excess water will collect at some point were you can drain it. Perhaps the pipe is 5 ft from the floor at one end of the wall, and 6 ft at the other end. Say the pipe is 10 ft long. Then, the derivative of the vertical position of the pipe is 0.1 ft per ft. The integral is just as simple. Lets say you know you want to start at 5 ft above the floor, and that the slope (derivative) should be 0.1 ft per ft. If you put the pipe up properly, you'll end up at 6 ft vertical position at the other end of the wall. That's integrating.

When I was young man I did a test using an air pistol (125 m/s) and a hunter's air rifle (>250 m/s). Same pellets, 4.5 mm. When shooting at normal incidence on a block of clay, the interesting effect was that the penetration depth was close to identical. However, the rifle pellet created a large conical "wound channel" in the clay with the pellet at the top of the cone. The volume of the cavity created by the rifle pellet was probably ~5 times that of the pistol pellet (perhaps close to the ratio of KEs). If you look mathematically at penetration, I think you'll find that if the material is viscous and the force decelerating the projectile is close to proportional to its velocity (as perhaps expected from a fluid), the penetration depth will be proportional to the momentum [math]m v[/math] of the projectile. If the decelerating force is constant (independent of velocity), the depth will be proportional to the kinetic energy [math]0.5 m v^2[/math]. A constant force is perhaps a simple representation of a solid material. Perhaps you could do a threshold model implementing both KE and momentum in some smart way? For example a geometric average that will only cause damage above some limit value. It's your universe, you do the rules

I was unclear, that was exactly what I did. The plot looks similar too. Scilab?

I guess there is an unfortunate strong correlation between icing up of one pitot tube and the icing up all of them.. If one does, probably others follow very soon. Then again it could be possible to have different types of pitot tubes from different producers. Still I'm interested to know what would be the simplest, down-to-earth way of determining approximate airspeed and avoiding stall if all other systems fail. A force majeure device.

Correct me if I'm wrong, but in gun ballistics there is a strong correlation between KE and damage. However, as already pointed out it's a complicated question. I don't know your exact application but if considering kinetic projectiles vs humans there are a number of parameters that could be of interest, i.e. volume of cavity produced by projectile, shock from rate of change of momentum, depth of penetration et c. I believe there were some rather nasty tests on goats trying to establish "empirical data" (Strasbourg tests) for incapacitation times. In armor ballistics the penetration is the most important and simple models include - I believe - kinetic energy divided by projectile cross sectional area.

I did a quick FFT of a similar vector from the first 1,000 primes and it looks pritty darn random to me Difference is I didn't do the sliding rectangular window thing. Perhaps your peaks are related to the windowing.

That's more like it, similar to my second suggestion. However, it will probably drive pilots nuts.. Perhaps one could imagine a resonant device tuned so that it will be over-critical at normal speed and produce tones only at close-to-stall speeds (say at horizontal angle of attack to simplify things). Basically a mass-spring system.

When I wrote "robust, down-to-earth stone age", that was what I meant. Doh yourself.