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Author: Subject: Calculus! For beginners, with a ‘no theorems’ approach!
Eosin Y
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[*] posted on 11-5-2016 at 12:13


\[D=AX^{1/2}(1+BP_{o})\]


[Edited on 11-5-2016 by Eosin Y]
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Eosin Y
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[*] posted on 11-5-2016 at 12:17


It works! Thanks be to Aga
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[*] posted on 11-5-2016 at 12:40


Eosin Y was shown by aga was shown by blogfast25 who was shown by .....

The great thing about Information is that you can Give it away, and still keep it as well.




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Eosin Y
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[*] posted on 11-5-2016 at 13:02


Thanks for the help. Is my equation correct?
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[*] posted on 11-5-2016 at 13:31


Quote: Originally posted by Eosin Y  
Is my equation correct?

No idea if it's assumptions are correct or gives any useful answer.

I'd write

D=A(X(squared by 1/2))*(1+BP)

as
$$D = (1+BP)AX^{\frac 12}$$
... just for clarity.

It's equivalent to :
$$D = AX^{\frac 12}(1+BP)$$
$$D = A(1+BP)\sqrt X$$
$$D = A\sqrt X + ABPX^{\frac12}$$
$$D = \sqrt X (A + ABP)$$

etc.


[Edited on 11-5-2016 by aga]




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Eosin Y
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[*] posted on 11-5-2016 at 13:48


No, let's find out.
Take RDX as an example. We can fill in certain constants and other information into this equation.
\[D=(1.01*\delta^{1/2})(1+(1.3*1.82))\]
Now, the only thing left to do is find delta when the following is true:
\[\delta = NM^{1/2}Q^{1/2}\]
RDX is C3H6N6O6 the decomposition of which works as such:
C3H6N6O6=3H2O+3CO+3N2
Therefore, N = 9.
Next, to work out M: H2O=18g/mol; CO=28g/mol; N2=28g/mol - plug in.
\[M=\frac{(3*18)+(6*28)}{3}\]
\[M=74\]
Finally, enthalpy of formation. In RDX, bond enthalpy = 7,149kj/mol
Given that this is true:
\[\Delta H_{reaction}^{o}=\sum \Delta H_{f}^{o}(products)-\sum \Delta H_{f}^{o}(reagents)\]
And the bond enthalpy of the products=8,817:
\[\Delta H_{RDX decomposition}^{o}=7,149-8,817\]
Finally, 7,149-8,817=-1668.
Plug these in, and we get:
\[X=9*74^{1/2}*-1668^{1/2}\]
\[X=-3161.96\]
At last:
\[D=(1.01*(-3161.96)^{1/2})(1+(1.30*1.82))\]
D=-191.163, apparently.
Where did it go wrong?

[Edited on 11-5-2016 by Eosin Y]

[Edited on 11-5-2016 by Eosin Y]
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[*] posted on 11-5-2016 at 14:55




Well, you sure learned how to use LaTex quickly, well done! :)

I can't check your result because I don't know whether the following is true:

Quote: Originally posted by Eosin Y  

Finally, enthalpy of formation. In RDX, bond enthalpy = 7,149kj/mol
Given that this is true:
\[\Delta H_{reaction}^{o}=\sum \Delta H_{f}^{o}(products)-\sum \Delta H_{f}^{o}(reagents)\]
And the bond enthalpy of the products=8,817:
\[\Delta H_{RDX decomposition}^{o}=7,149-8,817\]
Finally, 7,149-8,817=-1668.


The principle is correct but where did you get the actual data? Also, reaction enthalpies based purely on bond enthalpies tend to be approximate only. What are the reaction products you base your calculation on?

$$D=(1.01*(-3161.96)^{1/2})(1+(1.30*1.82))$$

One cannot take the square root of a negative number:

$$(-3161.96)^{1/2}=\sqrt{-3161.96}$$
... has no Real solution (only a Complex one).

[Edited on 12-5-2016 by blogfast25]




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Eosin Y
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[*] posted on 11-5-2016 at 21:27


It's a wrong equation. The answer should be ~8800!! I believe that the power for X might be a variable dependent on enthalpy of formation. If I can plot a Cartesian equation for this, I can plug that into x.

All my data about RDX, its products and its enthalpy came from here:https://www.youtube.com/watch?v=HpkGhuV7RPU

My LaTeX equations have been made here:https://www.codecogs.com/latex/eqneditor.php
You just plug LaTeX into the drop-down at the bottom, and lo and behold!

The actual value for the power of X is more like 0.975.
[Edited on 12-5-2016 by Eosin Y]

[Edited on 12-5-2016 by Eosin Y]
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[*] posted on 12-5-2016 at 05:10


Right, let's get going. To start with:
\[\delta _{RDX}=-3161.25\]
I believe that the power of delta in the following equation:
D=A\delta ^{y}(1+BP_{o})
will always be 0.975.
Prove it. Picric acid first.
We can fill in certain constants and other information into this equation.
\[D=(1.01*\delta^{0.975})(1+(1.3*1.82))\]
Now, the only thing left to do is find delta when the following is true:
\[\delta = NM^{1/2}Q^{1/2}\]
Octonitrocubane is C8N8O16.
Density unpacked=1.98g/cm3
Decomposition goes like this: C8N8O16=8CO2+4N2.
Therefore, N=12.
Next, avg. mol weight:
\[M=\frac{(8*44)+(4*28)}{2}\]
\[M=232\]
Finally, Q:
\[\Delta H_{reaction}^{o}=\sum \Delta H_{f}^{o}(products)-\sum \Delta H_{f}^{o}(reagents)\]
\[\Delta H_{reaction}^{o}=((12*348)+(8*293)+(8*201)+(8*607))-((16*799)+(4*941))\]
\[\Delta H_{reaction}^{o}=-3564
Plug in:
\[\delta =12*\sqrt{232}*\sqrt{-3564}\]
\[\delta =-10912\]
Finally,
\[D=(1.01*-10912^{0.975})(1+(1.3*1.98))\]
Apparently, VoD of octonitrocubane = 37,000m/s or so.
I need to do my own equation.
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[*] posted on 12-5-2016 at 06:40


Quote: Originally posted by Eosin Y  


All my data about RDX, its products and its enthalpy came from here:https://www.youtube.com/watch?v=HpkGhuV7RPU

My LaTeX equations have been made here:https://www.codecogs.com/latex/eqneditor.php
You just plug LaTeX into the drop-down at the bottom, and lo and behold!



You can also test drive LaTex by going to:

https://www.mathjax.org/

Scroll down a bit and click 'Try a live demo'.

The values for:

$$\Delta H_{reaction}^{o}=\sum \Delta H_{f}^{o}(products)-\sum \Delta H_{f}^{o}(reagents)$$

... as practiced in that YouTube, using bond enthalpies are only approximate.

That's because the bond enthalpy of, say C-H, depends on its environment. In CH4 the value already differs a bit from in C2H6. And the values for C-H in C3H8 (n-propane) differ for the methylene group and the methyl groups.

The same is true of all other types of bonds. These differences, when calculating large molecules, can add up to quite a bit of error.




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Eosin Y
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[*] posted on 12-5-2016 at 07:07


I doubt that the error with my bond enthalpy formula would really amount to 300%+?
There's something else I could try...
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[*] posted on 12-5-2016 at 07:26


Quote: Originally posted by Eosin Y  
I doubt that the error with my bond enthalpy formula would really amount to 300%+?
There's something else I could try...


I'm not sure where you're coming from with the "300%+". I'm only saying that reaction enthalpies based on bond enthalpies are approximate at best.

I believe the decomposition enthalpies of most HEs must be tabled somewhere, ready for use.

Now this:

$$12*\sqrt{232}*\sqrt{-3564}$$

... is something you should NEVER write in a math exam!! You can't take the square root of a negative number (unless you're happy to obtain a Complex Number!)

Your formula calls for the modulus (aka 'absolute value') of the reaction enthalpy and that is a positive number.

To indicate multiplication, these are better:

$$12\sqrt{232}\sqrt{3564}=12\times \sqrt{232}\times\sqrt{3564}=12.\sqrt{232}.\sqrt{3564}$$

[Edited on 12-5-2016 by blogfast25]




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[*] posted on 12-5-2016 at 08:44


I got three hundred percent as the reading for total VoD - 37000m/s is the number that the formula that it spits out at the end, which is more than 300% of the intended figure - 10100m/s.

I might try seeing whether I can do a correlation on a 3D Cartesian graph with enthalpy of formation on the x axis, density on the z axis and velocity of detonation on the y axis. This would involve a bit of thought, I assume, somehow combining the y = z... and y = x... readings into one equation.

My problem is that this is meant to be a calculation for experimental untested EMs, not known ones. I am using known ones to get the formula correct.
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[*] posted on 12-5-2016 at 09:03


It's probably easier to test the validity of:

$$D=AX^{1/2}(1+BP_{o})$$

... by calculating D for some known HEs, where D, X, A, B and P0 are known: plug in the X, A, B and P0 values and see if the calculated D values relate to the tabled D values.

[Edited on 12-5-2016 by blogfast25]




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Eosin Y
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[*] posted on 12-5-2016 at 09:45


That's what I tried to do for RDX, though I can't find any proper values for X/delta. That might be my problem. Can you find these values anywhere?
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[*] posted on 12-5-2016 at 11:01


Quote: Originally posted by Eosin Y  
That's what I tried to do for RDX, though I can't find any proper values for X/delta. That might be my problem. Can you find these values anywhere?


That question is better asked at the HE section of the forum. HE isn't really main stream chemistry.




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Eosin Y
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[*] posted on 12-5-2016 at 11:02


Fair enough Blogfast. Once I've drawn a graph, I'll post my results.
Anyone know of any decent graph drawing software?

[Edited on 12-5-2016 by Eosin Y]
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[*] posted on 12-5-2016 at 11:51


Quote: Originally posted by Eosin Y  
Fair enough Blogfast. Once I've drawn a graph, I'll post my results.
Anyone know of any decent graph drawing software?



Excel does a pretty good job for 2D graphs.




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[*] posted on 12-5-2016 at 13:09


My first equation:
\[d_{velocity}=6000d_{density} -3000\]
Where d(velocity)=velocity of detonation and d(density)=unpressed density g/cm3.
This is inaccurate, but it's better than that shit equation above. It was worked with a simple Cartesian scatter graph.
My next equation is a Cartesian graph where:
\[y_{axis}=v_{detonation}\]
\[x_{axis}=\Delta H_{bond}^{o}\]
I've got the bonds all down, but it will take me ~1 hour to work them out.
If Equation 1=x and Equation 2=y, this will be quite an accurate measurement:
\[D_{velocity}=\frac{x+y}{2}\]
Or, alternatively:\[D_{velocity}=\frac{(6000d_{density}-3000)+y}{2}\]
I'll post when I work it out tomorrow.

[Edited on 12-5-2016 by Eosin Y]

[Edited on 12-5-2016 by Eosin Y]
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[*] posted on 12-5-2016 at 16:21


I don't really understand your latest reasoning.

Maybe it will become clear when you post next.




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[*] posted on 12-5-2016 at 22:28


That is just the equation for the graph correlation of VoD and density.
I am also doing an equation for graph correlation of VoD and enthalpy of formation.
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[*] posted on 14-5-2016 at 13:13


Finally back to Maths.

OK. My 6 month-old washing machine's brains blew up.
It is kept Outside, so the Warranty is null and void.

Solution: Spend 2 days designing/building an Arduino Uno 'new brain' for the machine.

Did that, and now there are two problems.

Primarily is the Motor Control.
(the second is the particular detergent and foaming which we can ignore for now).

Imagine (or go look at) a Washing machine Drum.

An AC motor hits it with quite a lot of torque when it contains a lot of wet clothes, mostly distributed at the Bottom of the drum.

Begin a Spin cycle.

The Imbalance in the drum causes a HUGE displacement of the position of the machine when the drum spins at 1000 rpm (it jumps 4" to the left)

The Drum has a big concrete weight on top, and is suspended on Big springs, presumably to absorb/counter these effects.

The part i Can control is When the motor fires, even with tacho feedback from the motor to get it slightly more accurate.

An early thought was to fire the motor at 0 then 0+n degrees to try to counter the imbalance, but that did not work, even with an empty drum.

I feel certain there is mathematical solution, possibly involving spinning kinda integrals.

It Oscillates (as a system) so i guess avoiding those would be an objective.






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[*] posted on 14-5-2016 at 13:23


So the motor spins the drum and the system?
Does the rhythm of the shaking of the drum flow with the turns of the motor, and how many degrees is the angle of shaking from the point E where point E=the point exactly at the centre of the drum?
The motor should probably fire at highest speed when the clothes within it are in the middle of dropping to the bottom, meaning that the weight at the bottom will not cause so much deviation from the prescribed course.
If it is spinning at 1000rpm, that's 60/1000=1 rev per 0.06 secs. Given that the clothes probably stick for 1/3 of the circumference (for 0.02 secs) before dropping down and exerting their full momentum (0.01 secs) and repeat this cycle again, I'd probably say make the motor fire in a sequence like the following (each stage taking 0.01 secs) - low torque - low torque - high torque etc. I'd also reinforce the frame a bit, and perhaps get some bigger springs?

It's only 6 months old, which is a great loss. From my medical experience, the foaming at the mouth and the dizziness/imbalance would say rabies perhaps. It's cruelty to children to keep it outside! Baby B (for Bosch) all over again.

[Edited on 14-5-2016 by Eosin Y again]
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[*] posted on 14-5-2016 at 13:35


Close with Bosch, very close: it's a Spanish 'Balay' model.

If i remember i'll post photos tomorrow (dark here now).

One idea was to 'tumble' the clothes-mass to get it to distribute more evenly.
It doesn't.
The wet clothes stay at the bottom.

Washing machine technology is seriously more involved than i ever would have imagined.




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[*] posted on 14-5-2016 at 14:47


Guess it's too hard a mathematical problem then.

I thought so too.




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