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# Hyperon threshold and stellar radii

## Abstract

In this work we show how the emergence of a new degree of freedom in the nuclear bulk not only softens the EoS but reduces the radii of stars with high central density. If an enough repulsive channel, as the strange vector meson, is added to the scheme, we are able to simulate very massive and compact stars. Indeed we are able to construct an equation of state (EoS) that predicts a 2.05 as maximum mass and a radius of 11.51 km for the canonical 1.4 in a thermodynamical consistent way. We also link the radii of the canonical stars to a soft EoS for densities not much above the nuclear saturation point. Moreover, comparing our EoS with results obtained from heavy ion collisions, we show that the presence of a new degree of freedom allows a better agreement between theory and experiment.

###### pacs:
24.10.Jv, 26.60.Kp
1

## I Introduction

The physics of cold strong interacting matter at extreme densities is still not available in terrestrial laboratories. So far, the only place believed to reach such condition is the interior of neutron stars. The observations of pulsars evolved in the last years and provided us more precise information about the macroscopic characteristic of these objects.

For instance, the observations of two hyper massive pulsars, PSR J1614-2230 Demo () and PSR J0348+0432 Antoniadis (), show us that the EoS of beta-equilibrium nuclear matter should be very stiff, being able to produce a two solar masses neutron star. This indicates that hadron interaction at very small distances is strongly repulsive.

In this work we show that the onset of a new degree of freedom, not only causes the known softening of the EoS but also reduces the radii of the stars whose central density is higher than the density of the hyperon threshold ( what we call here subsequent stars), when compared with the EoS without this new degree of freedom. We use here the hyperon as this new degree of freedom, once it is the lowest mass baryon beyond the nucleons. Nevertheless it is important to bear in mind that the nature of the new degree of freedom is not relevant. Similar results can be obtained using or hyperons, or even resonances, by adjusting the strength of the coupling constant for those particles. Even more exotic particles, as dark matter Bridget (); Goldman (); Li (); Muk () can be used without significantly affect the results.

We show that if the hyperon onset happens at relatively low density, and a channel, repulsive enough, as the strange vector meson is added, we can produce very massive and compact neutron stars in agreement with the works above mentioned. Moreover, if an attractive channel, as the strange scalar meson is also added, we can produce even more compact neutron stars. Indeed, we are able to simulate an EoS that predicts 2.05 as maximum mass with a radius of 11.51 km for the canonical 1.4.

We also compare our results with experiences of heavy-ion collisions (HIC). In ref. Daniel (), the authors determine the pressure of the symmetric nuclear matter up to five times nuclear saturation density. They neither exclude the onset of hyperons nor of more exotic behaviour, as the quark-hadron phase transition. We show that the emergence of a new degree of freedom also allows a better agreement of theory with experience.

This paper is organized as follows: in section II we discuss the QHD formalism and present the parametrization of the model alongside some of the physical quantities they foresee for nuclear matter. In section III we expose the results of the threshold of hyperon in the bulk of beta-equilibrium nuclear matter and symmetric nuclear matter within four different approaches. The conclusions are drawn in section IV.

## Ii Formalism

The theory of the strong interacting matter is the QCD, where quarks interact with each other via the exchange of massless gauge bosons called gluons. Since the QCD has no results for dense cold matter, an effective model is required. In this work we use an extended version of the relativistic QHD Serot (), whose Lagrangian density reads:

 LQHD=∑B¯ψB[γμ(i∂μ−gBωωμ−gBρ12→τ⋅→ρμ)−(mB−gBσσ)]ψB−U(σ)+ +12(∂μσ∂μσ−m2sσ2)−14ΩμνΩμν+12m2vωμωμ+12m2ρ→ρμ⋅→ρμ−14Pμν⋅Pμν, (1)

in natural units. are the baryonic Dirac fields, which can be the nucleons, or a new degree of freedom, in this case, hyperon. The , and are the mesonic fields. The are the Yukawa coupling constants that simulate the strong interaction, is the mass of the baryon and , , and are the masses of the , , and mesons respectively. The antisymmetric mesonic field strength tensors are given by their usual expressions as presented in Glen (). The is the self-interaction term introduced in ref. Boguta () to reproduce some of the saturation properties of the nuclear matter and is given by:

 U(σ)=13!κσ3+14!λσ4. (2)

Finally, are the Pauli matrices. In order to describe a neutral, chemically stable matter, we add leptons as free Fermi gases:

 Llep=∑l¯ψl[iγμ∂μ−ml]ψl, (3)

where the sum runs over the two lightest leptons ( and ).

The mesonic fields are obtained via mean field approximation (MFA) Serot (); Glen (); Lopes2012 () and the EoS by thermodynamic relations Glen (); Greiner ().

To describe the properties of nuclear matter we use a slightly modified version of the well-known GM1 parametrization Glen2 (), a widely accepted parametrization Lopes14 (); Glen (); Lopes2012 (); Benito2 (); Paoli (); Weiss1 (); Weiss2 (); Lopes2013 (); Lopes2015 () that is able to reasonably describe both, nuclear matter and stellar structure, consistent with experimental and astrophysical observations Lopes2013 (). In this work we just reduce the strength of the coupling, reducing the symmetry energy slope from 94 MeV to 87.9 MeV Lopes14 (), a value closer to what is inferred in recent observations Lim (); Steiner (); Steiner2 (). This slope can be reduced even further, Rafa2011 (); alex2012 (), but for the purpose of the present work, the modifications introduced in Lopes14 () suffice.

In Table 1 we show the parameters of the model and its previsions for five nuclear matter properties at saturation density: saturation density point (), incompressibility (), binding energy per baryon (), symmetry energy () and its slope ().

## Iii Results

To study the effects of the onset of a new degree of freedom in detail, we divide this section in four different approaches. We consider the hyperon only because the presence of other strange particles would make this study strongly model and parameter dependent.

### iii.1 Role of the coupling constant

The basic constituents of neutron stars are neutrons and protons in -equilibrium. Since both are fermions, as the baryon density increases, so do the Fermi momentum and the Fermi energy, according to the Pauli principle. Ultimately, the Fermi energy exceeds the masses of the heavier baryons. The influence of the hyperons in the EoS and neutron star mass-radius relation, strongly depend of the coupling constant of the hyperons with the mesonic field. It is well known that the potential depth = - 28 MeV Glen2 (), so the coupling constants for and meson cannot be varied independently in principle. So, in this section we fix the potential depth and vary the values of the coupling constants and see how they affect the hyperon onset and population, the EoS, and consequently the mass-radius relation of the neutron stars. Similar works are found in the literature Glen (); Glen2 (), however, they just analyze the effects on the maximum mass. Let’s study, in more detail, the influence on the radii of the neutron stars as well. In Tab. 2 we show four sets of values for the coupling constants utilized in this section and in Fig. 1 we plot the fraction for the four sets.

We see that the higher the value of , the higher the density of the hyperon threshold, varying from 0.302 for set A to 0.431 , for set D, ie, around 1.97 and 2.82 times the nuclear saturation density. Moreover, in general, the population also increases with the increase of . We can see that the curves of sets A and B cross each other (set A and C cross as well at a density around fm). This is due to the Pauli blocking of the hyperons in set A. Once there are more hyperons in set A at low densities, the onset of new particles is more energetically favorable in set B because the Fermi sea for low Fermi moment in set A is already filled. We can also see that in set D is so high that is very suppressed, never reaching more than 0.3, while in other sets this value could reach more than 0.6.

Now we plot in Fig. 2 the EoS for the four sets presented in Tab. 1. We see that the higher the value of , the stiffer the EoS. Unlike Fig. 1 there is no crossing of the curves. This is due to the fact that the main term of the pressure is the vector meson and it is proportional to the density . So, indeed, the curves deviate from each other as the density increases.

To conclude this section we solve the TOV structural equations TOV () and plot the mass-radius relation in Fig. 3. Here, and in the rest of this work we use the BPS BPS () equation to simulate the neutron star crust. The most massive pulsar yet known is the PSR J0348+0432 Antoniadis () with a mass of . So, for an EoS to be valid, it needs to explain this massive neutron star. Another constraint is the radius of the canonical 1.4 pulsar. According to ref. Hebeler (); Lim (); Ozel (); Steiner (); Steiner2 () the maximum radius for these stars lie between 12.45 to 13.9 km. We plot the main results for the TOV solutions in Tab. 3.

By looking at Fig. 3, we can see that according to ref. Antoniadis (), only sets C and D are valid as EoS of dense matter. All the radii are very close, between 13.74 km to 13.76 km, except for set A. This is due to the emergence of a new degree o freedom. A new degree of freedom as the hyperon not only reduces the maximum mass, but compresses the neutron star. When the hyperon fraction becomes relevant, there is a “turn to the left” in the mass-radius relation, compressing the subsequent neutron stars, reducing their radii. For a hyperon potential depth of -28 MeV, the threshold of hyperons appears too late for this “turn to the left” affect the canonical 1.4 to values of radii that agree with the ref. Hebeler (); Lim (); Ozel (); Steiner (); Steiner2 (). But for more massive neutron stars, the large amount of hyperons compresses significantly the star. For instance, if someone wonders about the radius of the 2.01, the PSR J0348+0432, the answer is 12.91 km, if we assume set C as the best parametrization, or 13.40 km if we choose set D, because the “turn to the left” happens earlier in set C than set D. If we want to compress the canonical 1.4 neutron star, the threshold needs to take place at earlier densities. To accomplish that, the potential depth needs to be lower.

### iii.2 Role of the potential depth

Now we study how a more attractive potential depth affects the neutron stars with hyperons. As mentioned earlier in this work, although the is well established around around -28 MeV, the nature of the new degree of freedom is not relevant. Any other particle could be used, just adjusting the values of the coupling constants. Moreover, here we are dealing with highly asymmetric matter, instead of the symmetric matter where the potential was inferred, and the hyperon threshold happens at higher densities than . Because we want the hyperon onset at lower densities we fix the coupling to be very week, as in set A of Tab. 2. Now we vary only the coupling. The parametrization utilized are presented in Tab. 4, and we plot the fraction in Fig. 4.

We can see that the more attractive the , the earlier the hyperon threshold. Also, the more attractive the , the higher the at high densities. Indeed, the lambda fraction can reache values higher than 0.7 for sets G and H. The potential depth can be as low as -126 MeV. This value was not randomly chosen. As we will see in the next topic, this value is the highest value that is able to predict a massive neutron star with the radius agreeing with ref. Hebeler (); Lim (); Ozel (); Steiner (); Steiner2 ().

Now in Fig. 5 we plot the EoS for different values of . Due to the extremely attractive potential, and the great amount of hyperons, we expect very soft EoS. Indeed, as one can see, for sets G and H, the could even be negative. That could indicate a phase transition. These are deeper waters where we will not navigate into serot1995 (). Moreover, since a negative value of , produces unstable neutron stars Glen (), we concentrate this work only in EoS for which is greater than zero. Now we plot the TOV solution for sets E and F in Fig. 6.

Due to the very attractive hyperon potential depth, and once we have a very weak repulsive channel, these soft EoS produces very low maximum mass neutron star. Also, only for sets E and F it is possible to solve the TOV equations, since it requires greater than zero. Indeed, all EoS in this section are ruled out, since none of them is able to explain the massive PSR J0348+0432. Nevertheless, the key point in the section is the “turn to the left”. Looking at Fig. 6, we can see that for = -80 MeV, the “turn to the left” occurs for masses below 0.8. If an even more attractive potential is used, earlier would be this behavior. The star masses are so low due to the very weak repulsive channel. In next section we see how to overcome this issue.

### iii.3 Role of the strange vector meson ϕ

We see that the emergence of a new degree of freedom causes a “turn to the left” in the mass-radius relation, compressing the subsequent stars. The earlier the onset of this new particle, the lower the mass that is affected by this “turn to the left”. But we also see that the threshold of a new particle softens the EoS. To stiffen the EoS, we need to increase the repulsive channel. The term responsible to this in eq. (1) is the coupling constant. However, the increase of the channel will affect the potential, and the “turn to the left” will not affect the canonical 1.4. One way to overcome this difficulty is to add in the Lagrangian a new repulsive channel. We use here here the strange vector meson Weiss1 (); Weiss2 (); Lopes2013 ():

 LYYϕ=−gY,ϕ¯ψY(γμϕμ)ψY+12m2ϕϕμϕμ−14ΦμνΦμν. (4)

The field is analogous to the field and its expected value is also obtained via MFA Serot (); Glen (); Lopes2012 (). Adding a new mesonic field has two advantages. First, since the does not couple to the nucleon, it does not affect the properties of the nuclear matter. Second, since the field is zero below the hyperon threshold density, it does not affect the potential depth, and has little influence on the point of the “turn to the left”. Now we fix the MeV as in set H. Indeed, all sets in this section are derived from set H of Tab. 4, only varying the coupling constant. So, we numbered the sets from H-I to H-IV. In the figures the H is omitted to not burden the notation. We choose that set because, as we will see, this potential depth is the less attractive potential able to produce a 2.05 neutron star, with a radius of 12.45 km. The values utilized in this calculation are presented in Tab. 5.

Now we plot the fraction in Fig. 7. As we can see, since field is zero before the hyperon threshold, and the potential depth are the same for all sets, the hyperon onset is the same for all values of coupling, around 0.177 . This is a relative low value, corresponding to 16 above the nuclear saturation density. Since is a repulsive channel, the stronger the coupling, the stronger the hyperon suppression at high densities. Indeed in set H-IV, never surpasses 0.3. Also, the field contributes in the pressure, stiffening the EoS. Plotting the EoS of Tab. 5 in Fig. 8, we can see that with the field, is always positive.

Now we plot the mass-radius relation in Fig. 9 with the main properties resumed in Tab. 6. We can see that due to the onset of hyperons at low densities, the “turn to the left” occurs at very low masses. Indeed in set H-I, it happens at mass around 0.5. The higher the coupling, the less pronounced is the compression of the subsequent neutron stars. Also, in set H-I, we see that the canonical 1.4 is only 8.11 km, a very compact neutron star. Nevertheless, its maximum mass of only 1.42 indicates that this EoS must be ruled out.

Increasing the coupling, we also increase the maximum mass. In set H-IV, for a strong coupling, the maximum mass reaches 2.05, the upper limit of the , PSR J0348+0432 pulsar. But, unlike sets C and D of Fig. 3, now we have a much lower radius for the canonical 1.4. With this modified GM1 model, the radius was around 13.76 km. Now we have a radius of only 12.45 km. In other words, an EoS that is able to explain the massive PSR J0348+0432 Antoniadis (), and also in agreement with the inferred measures of the pulsar radii Hebeler (); Lim (); Ozel (); Steiner (); Steiner2 (). We show that the emergence of a new degree of freedom is able to compress the neutron stars and still produce very massive pulsars.

Another good constraint to be assessed is the pressure of symmetric matter for densities up to five times the nuclear saturation density, as inferred in ref. Daniel () via HIC. As discussed in ref. Rafa2011 (); Lopes2013 (), the GM1 is in disagreement with this constraint. However, as pointed in ref. Lopes2013 () this could be due to the fact we are not considering the onset of hyperons in the bulk of nuclear matter. Now we define the symmetric hypernuclear matter as in ref. Lopes2013 (). Consider a symmetric matter. In this regime, by definition, the density of the protons is equal to the density of neutrons, . Due to this, the field is zero and the chemical potential of these particles are also the same, . If we compress this matter, the onset of strange particles, as , becomes energetically favorable. Since ref. Daniel () does not rule out neither hyperons nor even more exotic pictures, such as quark-hadron phase transitions, we assume the possibility of onset in the symmetric matter, imposing:

 μp=μn=μΛ. (5)

This choice implies that only symmetric nuclear matter exists until the density is high enough so the creation of strange particles becomes energetically favorable, softening the EoS. The pressure of GM1 symmetric matter and symmetric hypernuclear matter alongside the inferred pressure up to five times nuclear saturation density (hatched area), obtained in ref. Daniel () are plotted in Fig. 10. Since only set H-IV is able to predict a 2.0 we plot only this set.

The hyperon threshold softens the EoS making experiment results from HIC and theory reach an agreement again. While in beta-equilibrium matter the onset of the new degree of freedom is at densities about above the nuclear saturation density, in symmetric matter this onset happens latter, at densities about above the saturation point. This is due to the fact that the field, that is a repulsive channel is zero, and the nuclear matter has a lower chemical potential compared with the stellar one. Indeed, the nuclear matter is bounded for densities below the saturation density. We see that, the emergence of a new degree of freedom, besides being able to explain the astrophysical observations of a very massive pulsar and the inferred low radius for the canonical mass, reconciles these results with those obtained in laboratory.

### iii.4 Role of the strange scalar meson σ∗

Now we give one more step and add a new strange meson, the scalar . As the vector meson, the scalar one just couples to the particles, and does not affect any of the properties of nuclear matter presented in Tab. 1. Also, the field is zero in the absence of particles, therefore it would not affect the hyperon threshold. The Lagrangian of is analogous to the and reads:

 LYYσ∗=gY,σ∗(¯ψYψY)σ∗+12(∂μσ∗∂μσ∗−m2σ∗σ∗2). (6)

The is an attractive scalar meson, while the is a repulsive vector meson. Therefore, the will dominate at low densities, softening the EoS and increasing the amount of hyperons. At high densities, the meson dominates, causing a strong suppression of the hyperons and stiffening the EoS. This behavior is analogous to the competition in the symmetry energy reported in ref. Lopes14 (). In Tab. 7 we show the values of the coupling constants for the and mesons. All these values have been chosen in such a way that the EoS is able to reproduce a 2.05 as a maximum mass. And again, in all cases the set H of Tab. 4 is used to fix the hyperon potential depth at -126 MeV.

Now we plot in Fig. 11 the hyperon fraction for the sets of Tab. 7. We see that the has little influence if the is not high. Sets H-V to H-VII are very similar. For set H-VIII the coupling constant of is two times the one, and the difference is higher. But in all cases, the is low, and again never surpasses 0.3.

The differences in the EoS are even smaller. Since all the EoS are adjusted to reproduce a 2.05 as maximum mass, they are all very similar. Due to this we just plot the set H-V and set H-VIII, that have the weakest and strongest coupling constants. Of course we could increase the repulsive channel and simulate even higher mass neutron stars, however this would imperatively increase the radius of the canonical 1.4. We also plot the results obtained with set C from Tab. 2 since it predicts similar maximum mass. The results are presented in Fig. 12. We see that set H-V produces a lower value of and coupling. Due to this, the EoS is less soft at low density and also less stiffer at high density. In set H-III, and coupling are very high. So meson strongly dominates at low density, softening the EoS, while stiffens it at high densities. Set C has no hyperons at low densities, so it is stiffer in this regime. Also, due to the absence of the meson, this set is softer at high densities. Sets H-VI and H-VII are between set H-V and H-VIII. The stronger the interaction with both mesons, the softer it will be at low density and stiffer at higher ones due to the different nature of the mesons (scalar and vector).

To conclude this section we compare our results for a symmetric hypernuclear matter with the constrain imposed by ref. Daniel (). Again, due to the fact that the EoS are very similar we plot only set H-V and set H-VIII. Since set H-VI and H-VII have intermediate values for the coupling constants, their results will be always between sets H-V and H-VIII. We see in Fig. 14 that the same effects that happen in -equilibrium matter are reproduced in symmetric hypernuclear matter. The stronger the coupling, the softer the EoS. But due to the high value of the coupling needed to reproduce very massive pulsars, this EoS will become stiffer at high densities. We conclude that our model with the onset of a new degree of freedom agrees with astrophysical observations and HIC experiments.

## Iv Final Remarks

In this work we show how the onset of a new degree of freedom is able to reconcile the recent measurements of very massive pulsars Demo (); Antoniadis () with compact ones Hebeler (); Lim (); Ozel (); Steiner (); Steiner2 (), and also satisfy HIC experimental constraints  Daniel (). Although the true nature of this new degree of freedom is an open puzzle, we show that a particle not much heavier than the nucleon and a new repulsive channel is enough to accomplish this goal. Also, we are able to construct a thermodynamical consistent model with no inputs.

We start by reviewing the effects of hyperon threshold in dense nuclear matter. The consequent softening of the EoS and reduction of the maximum mass is a well-known theme in the literature Glen (); Glen2 (); Lopes2012 (), but the “turn to be left” when the hyperon fraction becomes relevant seems to have passed unnoticed. We show that for the well-established potential depth of -28 MeV, there is little influence of the hyperon onset in the radii of stars around the canonical mass of 1.4.

Next we modify the potential depth allowing it to assume more attractive values. Therefore, the onset of the new degree of freedom appears at lower densities and the ’turn to be left” occurss for less massive stars, reaching the canonical mass. Due to the very weak repulsive channel, all resulting neutron stars have very low maximum mass and the corresponding EoS must be ruled out.

To overcome this issue we add a new repulsive channel, through the strange vector meson. This allows us to stiffen the EoS and increase the maximum mass without change any properties of nuclear matter and with no effect on the hyperon threshold. This also allows us to construct a very realistic EoS, that is able to explain the massive PSR J0348+0432 and the measurements of low radii pulsars. The result is even better when compared with the results of HIC. Now we are able to reconcile astrophysical observation and lab constraints.

In the last part of our work we add a new attractive channel, the strange scalar meson. This new attractive field softens the EoS at low densities, making the “turn to be left” more pronounced. This compresses even more the subsequent stars, allowing us to construct an EoS that predicts a maximum mass of 2.05, while the canonical 1.4 has a radius of only 11.51 km. We are also able to link the low radius to a soft EoS at low densities, what seems a more fundamental reason than relating it to the symmetry energy slope. Moreover, the emergence of a new degree of freedom again reconciles this EoS with HIC experiments.

We finish this work by pointing out that the nature of the new degree of freedom needs to be better studied, but it seems that the EoS needs to be soft at low densities to predict very compact neutron stars. However, the radii of canonical neutron stars is not a closed subject. Some studies indicate that their values could be very low, not surpassing 11.1 km Ozel2 (), close but still below the 11.51 km found in this work. Nevertheless, since it is still an open puzzle, a relatively large limit of 10.7 km for the canonical neutron star is presented in ref. Chen () to account for the causality.

Acknowledgments; DPM acknowledges partial suuport from CNPq and LLL from CAPES and CEFET/MG.

### Footnotes

1. preprint: APS/123-QED

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# Diluting Stock Solution with NaOH

So I know that if you want to dilute a stock solution to a smaller concentration you simply use the $M_1V_1 = M_2V_2$ equation where the $M_1$ and $V_1$ are the molar concentrations and volumes of the stock solution and $M_2$ and $V_2$ are the molar concentrations and volumes of the resulting dilute solution. This is assuming if you are diluting with water.

My question is, if you have a stock solution of say para-nitrophenol and you wanted to dilute it with $\pu{0.100 M}$ $\ce{NaOH}$ solution, what modifications would you use to calculate how much of the stock solution you would need to get a dilute para-nitrophenol solution?

What I'm thinking is that the density of the $\ce{NaOH}$ solution would come into play here but I don't know how I would go about it.

• First, you didn't tell us the whole story and what you have done so far. Your sought compound, p-nitrophenol is a solid. Thus, I assume you already have a solution of it (your stock solution). Is it in $\mathrm{0.100M}$ $\ce{NaOH}$ as well? Or, in some other solvent? In any solvent, what is the concentration of your stock solution? You want to the find final concentration of $\ce{NaOH}$ or sodium p-nitrophenolate? (Remember, p-nitrophenol reacts with $\ce{NaOH}$) – Mathew Mahindaratne Apr 11 '18 at 16:18
• Ok, so I really asked this as a general question as in if I were to dilute a solution with something other than water . But I'm doing a lab using a 96-well plate to do spectroscopy to create a standard curve for pNP. I have a 100 micro-molar solution of stock pNP (paranitrophenol, given to us so I assume its dissolved in water). The stock solution is probably colorless and will change after I dilute it because the instructions say that the solution will turn yellow after diluting with NaOH. So the current goal is to create dilute solutions of 10, 20, 30, 40, 50, 60, and 70 micro-molars. – Mayuresh Patel Apr 11 '18 at 17:23
• What is your expected final volume in each well? $\pu{300 \mu L}$? – Mathew Mahindaratne Apr 11 '18 at 20:14
• The final volume for each standard solution should be 1.00 mL. – Mayuresh Patel Apr 12 '18 at 1:59

Since you'd like to use the equation, $\mathrm{M_1V_1 = M_2V_2}$, I'll show you here, how to do dilution for your extreme concentrations. You can follow same method to do the rest.

Suppose, your final volume of each of dilute solutions is $\pu {V \mu L}$. Then, for all calculations, $\mathrm{V_2 = V}$ and $\mathrm{M_1 = 100\space\mu M}$ (concentration of stock solution). Note that, concentration of $\ce{NaOH}$ does not play a role here, since: (1) it is $1000\times$ more than that of the stock solution (pNP) and hence, dialution of itself is minimal; and (2) it acts just as any other solvent, except for reacting with pNP to give the desired color.

Now, let the volume of $\ce{NaOH}$ solution needed to make your highest concentrated pNP solution is $\pu {V_{1} \mu L}$. Then, using the equation $\mathrm{M_1V_1 = M_2V_2}$, you get: $$\mathrm {100\space\mu M \times V_1 = 70\space\mu L \times V\space\mu L}$$ $$\mathrm {V_1 = \frac {70\space\mu M \times V\space\mu L}{100\space\mu M} = \frac {7}{10}V \space\mu L}$$ Thus, you dilute $\pu {0.7V \mu L}$ stock solution with $\pu {0.3V \mu L}$ $\ce{NaOH}$ solution to make $\pu {70 \mu M}$ pNP solution.

To see if there are enough $\ce{NaOH}$ in $\pu {0.3V \mu L}$ portion, calculate $\pu {\mu mol}$ of $\ce{NaOH}$: $$\mathrm {100\times 1000\space\mu M \times 0.3V\space\mu L = 3\times10^{-2}\times V\space\mu mol}$$ Similarly, you can calculate $\pu {\mu mol}$ of pNP in final solution: $$\mathrm {100\space\mu M \times 0.7V\space\mu L = 7\times10^{-5}\times V\space\mu mol}$$ Thus, all p-nitrophenol will be converted to sodium p-nitrophenolate in the highest concentrated solution. Therefore, it is safe to assume that rest of the solutions would undergo the same conversion.

By similar calculations, we can show you must dilute $\pu {0.1V \mu L}$ stock solution with $\pu {0.9V \mu L}$ $\ce{NaOH}$ solution to make $\pu {10 \mu M}$ pNP solution. Since you know value of $\pu {V \mu L}$, you can get numerical values.

Density changes when mixing liquids is not only applicable to NaOH solutions. Even when you dilute a stock solution with water you may find that the volume is not additive. The factors that matter are the starting concentration and the concentration change.
At the concentrations you're using, you can probably safely ignore density changes and assume volumes are additive, unless you need extreme precision (but then you could use volumetric glassware and do the dilutions properly, instead of just adding volumes).

You are correct: when you dilute your $100 \ \mu M$ pNP solution with $0.1 \ M$ NaOH, it will turn yellow, because you form the pNP anion, which is intensely yellow.
However, for the spectroscopy part to work correctly, you must make sure you always have at least the stoichiometric amount of NaOH, i.e. you must convert all pNP to its anion.
This will be the case in all your diluted solutions, simply because the concentration of NaOH is 1000 times larger than the pNP concentration ($0.1 \ M = 100 \ mM = 100000 \ \mu M$), so whenever the volume of NaOH solution you add to the pNP solution is greater than 1/1000 times the volume of the latter, you're already in excess of NaOH.

For the rest, the formula you mentioned works correctly. For instance, to make the $70 \ \mu M$ solution of pNP anion: $V_2 = \frac {100 \cdot V_1} {70} \approx 1.43 \cdot V_1$, so you could add $0.43 \ mL$ of $0.1 \ M$ NaOH to $1 \ mL$ of $100 \ \mu M$ pNP. You can see that you are in huge excess of NaOH, so there is no risk to leave any non-ionised pNP.

You can achieve the same result by a 'quick' subtraction method, which also works when both solutions contain the substance of interest.
In this case, your stock is $100 \ \mu M$ in the substance of interest (pNP). You dilute it with a solution that contains no pNP, so it's $0 \ \mu M$. You want to obtain a $70 \ \mu M$ solution.
Subtract the final desired concentration from the concentration of each solution you're mixing, take the absolute values and swap the results.
So, pNP stock minus final desired solution is $|100 - 70| = 30$. NaOH minus final desired solution is $|0 - 70| = 70$. Swap the two: 30 applies to NaOH and 70 applies to pNP.
This means that, to obtain the desired final solution, you need to mix 30 'parts' of NaOH with 70 'parts' of pNP. You can easily verify that indeed $100/70-1 = 30/70$.

This may sound more complicated than the other formula, but in fact if you write the concentrations of the two 'starting' solutions on the top left and bottom left corner of an ideal square, respectively, and the final concentration in the centre of the square, and then do the subtractions 'across the diagonals', you will get in the right corners of the square the 'parts' to mix for each starting solution.
E.g. if you had a $10 \ \mu M$ solution (A) and a $90 \ \mu M$ solution (B) and wanted to obtain a $40 \ \mu M$ solution by mixing them, this method would quickly tell you that you need to mix 50 'parts' of A and 30 'parts' of B.
Obviously any mixing-based dilution will work only if the final desired concentration is 'bracketed' by the concentrations of the solutions being mixed. In these examples, $0<70<100$ and $10<40<90$.

• Thanks. I actually just asked my professor and she just wanted a percentage of pNP to NaOH. So like 100 uL of pNP and 900 uL NaOH for the 10 uM solution. This was just theoretical and the actual concentrations should be close to these values. I kind of over thought about this :) – Mayuresh Patel Apr 12 '18 at 19:31

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# Work Done and final temperature of 5 mols of Ideal Gas

[SOLVED] Work Done and final temperature of 5 mols of Ideal Gas

## Homework Statement

Five moles of an ideal monatomic gas with an initial temperature of 129 degrees Celsius expand and, in the process, absorb an amount of heat equal to 1160 Joules and do an amount of work equal to 2020 Joules.

What is the final temperature of the gas?

## Homework Equations

$$W = -nR(T_f - T_i)$$

## The Attempt at a Solution

I have used the above formula before, and it worked then but doesn't appear to be working as well now

$$2020 - 1160 = -5*8.31(T_f - 129)$$

$$420 = -41.55(T_f - 129)$$

$$-10.1 = T_f - 129$$

$$T_f = -10.1 + 129$$

This gives me 118.9, which I have entered, and it says that:

" Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures. "

I have tried also 120 as well, with the same response!

I doubt its the formula, because I have used it before? Any Ideas?

TFM

Doc Al
Mentor
How does the internal energy of an ideal monoatomic gas relate to its temperature? (Your formula is not quite right.)

Would that be this:

$$U = \frac{1}{2}nRT$$ per degree of Freedom.

?

TFM

Doc Al
Mentor
Yes! And how many degrees of freedom are there?

There are three degrees of Freedom, since it is a monatomic gas.

TFM

Doc Al
Mentor
Good. Now finish it up.

This would be:

$$U = \frac{3}{2}nRT$$

Can the T part be $$T = (Tf - Ti)$$

?

TFM

Doc Al
Mentor
This would be:

$$U = \frac{3}{2}nRT$$
Yes.

Can the T part be $$T = (Tf - Ti)$$
If you rewrite the above equation in terms of changes in U and T (as you should), yes.

Would the change in U still be 420?

TFM

I tried:

$$U = \frac{3}{2}nR(T_f - T_i)$$

and put in:

$$420 = \frac{3}{2}(5)(8.31)(T_f - 129)$$

IOt gave me an answer of 122, I put it in, And MP gas said itis wrong, and I have used up all my tries. Any ideas where I could have gone wrong?

TFM

Doc Al
Mentor
Arithmetic error: 1160 - 2020 = ?

Drat! I put 2020 - 1600 into my calculator by mistake!!! Putting in 860, I get the final temperature to be 115

Thanks for the help, even if my sillyness cost me ,

TFM


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# [MDX or any DX] State Management.

This topic is 4443 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi I’m kind of new to using Dirext X and so far as I’m sure is common I’m finding the huge number of options to be quite staggering. One thing I’ve noticed is that keeping track of what options are on and which are off seems to be a huge problem for me so far. Take this following code for example:
device.RenderState.SourceBlend = Blend.SourceAlpha
device.RenderState.DestinationBlend = Blend.InvSourceAlpha
device.RenderState.AlphaBlendEnable = True
device.TextureState(0).AlphaOperation = TextureOperation.BlendDiffuseAlpha

device.Transform.World = PosMatrix
obj.DrawObject(device, Drawing.Color.FromArgb(100, obj.OverLayColor))

This seems to work great but when I next draw something I’m using the options I have set for the last thing I draw in my scene. What if I need different options? The question boils down to: Do I have to reset every option every time I change one to draw something?

Dim previousAlpha As Boolean = device.RenderState.AlphaBlendEnable
Dim SourceBlend As Blend = device.RenderState.SourceBlend
Dim DestinationBlend As Blend = device.RenderState.DestinationBlend
Dim AlphaOperation As TextureOperation = device.TextureState(0).AlphaOperation

device.RenderState.SourceBlend = Blend.SourceAlpha
device.RenderState.DestinationBlend = Blend.InvSourceAlpha
device.RenderState.AlphaBlendEnable = True
device.TextureState(0).AlphaOperation = TextureOperation.BlendDiffuseAlpha

device.Transform.World = PosMatrix
obj.DrawObject(device, Drawing.Color.FromArgb(100, obj.OverLayColor))

device.RenderState.AlphaBlendEnable = previousAlpha
device.RenderState.SourceBlend = SourceBlend
device.RenderState.DestinationBlend = DestinationBlend
device.TextureState(0).AlphaOperation = TextureOperation.BlendDiffuseAlpha

Or is there a better way? Looking for any advice. Thank you.

##### Share on other sites
My solution is to develop render state change policy and stick to it. You should decide which render states you'll reset and which ones you'll leave changed, so future rendering will have to set them. For example, you can decide to always turn off blending if you used it, so future rendering can rely on the fact that it's off. Similary, you can decide to leave blending mode changed - as long as blending is off, blending selected operations doesn't matter and when something will need blending, you usually have to set blending mode anyway.

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# Is there anything to do with unlucky 13 ????

Calculus Level pending

Find the area in the first quadrant bounded by [x]+[y]=13 ... Here [.] denotes Greatest Integer Function . Ex : [2.3]=2

×


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# Gary

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#### Archived

This topic is now archived and is closed to further replies.

# How far to take error checking?

This topic is 5109 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hey all. Finally had a bit of time to work towards some basic GDI drawing (busy year), but after designing a class to load and store an external bitmap resource, it''s brought up a question. As the topic says, just how far would you take error handling? Here''s sort of what I''m talking about: BitmapObj.h
// Definition of class, CBitmapObj

#ifndef BITMAPOBJ_H
#define BITMAPOBJ_H
#pragma once

#include <windows.h>
#include <string>

class CBitmapObj
{
public:
CBitmapObj();       // Constructor.

~CBitmapObj();      // Destructor.

// Utility functions.

BOOL LoadBMP(std::string path);  // Load BMP from external file.

BOOL PrepareDC();                // Prepare the memory DC.

BOOL SelectBMP();                // Select the BMP into the DC.

BOOL SaveDimensions();           // Save the BMP attributes.

// Getter/setter functions.

HDC GetMemDC();
int GetWidth();
int GetHeight();

private:
HDC m_hMemDC;         // Memory DC.

HBITMAP m_hBitmap;    // Handle to the bitmap resource.

int m_width;          // Bitmap''s width.

int m_height;         // Bitmap''s height.

};

#endif

BitmapObj.cpp
// Implementation of class, CBitmapObj.

#include "bitmapobj.h"

CBitmapObj::CBitmapObj()
{
m_hMemDC = NULL;
m_hBitmap = NULL;
m_width = m_height = 0;
}

CBitmapObj::~CBitmapObj()
{
m_hBitmap = NULL;

if (m_hMemDC)
DeleteDC(m_hMemDC);
}

{
m_hBitmap = reinterpret_cast< HBITMAP > (LoadImage(
NULL, path.c_str(), IMAGE_BITMAP,
));

return (m_hBitmap == NULL) ? FALSE : TRUE;
}

BOOL CBitmapObj::PrepareDC()
{
m_hMemDC = CreateCompatibleDC(NULL);

return (m_hMemDC == NULL) ? FALSE : TRUE;
}

BOOL CBitmapObj::SelectBMP()
{
// No BMP handle saved, so simply return.

if (m_hBitmap == NULL)
return FALSE;

return (SelectObject(m_hMemDC, m_hBitmap) == NULL) ? FALSE : TRUE;
}

BOOL CBitmapObj::SaveDimensions()
{
// No BMP handle saved, so simply return.

if (m_hBitmap == NULL)
return FALSE;

BITMAP bmpInfo; // Structure to hold BMP attributes.

if (GetObject(m_hBitmap, sizeof(BITMAP), &bmpInfo) == NULL)
return FALSE;

m_width = bmpInfo.bmWidth;
m_height = bmpInfo.bmHeight;

return TRUE;
}

HDC CBitmapObj::GetMemDC() { return m_hMemDC; }
int CBitmapObj::GetWidth() { return m_width; }
int CBitmapObj::GetHeight() { return m_height; }

Apologies for the slight lack of commenting, only just finished that. Anyway, here''s an example of it in use: Extract from WinMain()
	CBitmapObj bmp;
HDC hdc = GetDC(gameWin.m_hwnd);

if (bmp.LoadBMP("Images/red.bmp") == NULL)

if (bmp.PrepareDC() == NULL)
MessageBox(NULL, "PrepareDC", "PrepareDC", MB_OK);

if (bmp.SelectBMP() == NULL)
MessageBox(NULL, "SelectBMP", "SelectBMP", MB_OK);

if (bmp.SaveDimensions() == NULL)
MessageBox(NULL, "SaveDimensions", "SaveDimensions", MB_OK);

if (BitBlt(
hdc, 0, 0, bmp.GetWidth(), bmp.GetHeight(),
bmp.GetMemDC(), 0, 0, SRCCOPY
) == NULL)
MessageBox(NULL, "BitBlt", "BitBlt", MB_OK);

ReleaseDC(gameWin.m_hwnd, hdc);

As you can see, I''ve added several calls to MessageBox(), just to illustrate my point. After seeing that little lot, do you think that would be overly cautious (imagining that calls to MessageBox() wouldn''t be the way any errors would be handled - something like a log file instead)? Thanks for any replies. -hellz

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Mess around with exception handling and see if you like it; I find that it cleans up my code an awful lot, but it took a while for me to convert. I still check function return values too, in case you're wondering, but only for things that wouldn't fail (i.e. checking if a file is open shouldn't throw an exception, but would return a bad value.)

edit: on a side note, i was reading about memory management and heard that you shouldn't check the return values for operator new() anymore; the reasoning is that if it fails, then the computer is so low on memory that it will crash anyway. Unless you're allocating an insane amount of memory, of course.

[edited by - psykr on December 16, 2003 6:07:16 PM]

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Thanks for the reply mate. That''s the sort of thing I was talking about. Funny you should mention about the new operator; someone said something similar to me a few days ago, but I''d not had time to check it out, so thanks for reminding me.

I guess in some ways, error handling is pretty much task specific. I mean, I''d imagine that if Windows had a problem returning a handle to a memory device context, that it''d be unlikely to be able to load a bitmap anyway. On the other hand, it''s easier to think in terms of handling possible errors with situations such as when opening files, because the error is more common and easy to get to grips with.

I should probably check out some code from a reputable engine, or something similar, just to see how other people tackle the problems I''ve been toying with in my mind for a while now.

Thanks again, for the reply.

-hellz

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I think exceptions should only be used if the error is recoverable.

For example, let''s say that your loadBMP fails. What are you going to do... play the game without the bitmap? Is that really acceptable? If you can''t play without the bitmap, then it''s easiest to just display an error and stop the game.

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I just finished my error handling class today, it does alright. I'm going to post the source January 1 if you want it. It's nothing major, but it's pretty nice.

Here's the header for it:
//::|ññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññ|:://::|ññ ---------------------------------------------------------------------------------- ññ|:://::|ññ|						~ A R I E L   P R O D U C T I O N S ~                     |ññ|:://::|ññ|                               ~ ALL RIGHTS RESERVED ~                            |ññ|:://::|ññ ---------------------------------------------------------------------------------- ññ|:://::|ññ| ~ PROGRAM DATA ~													      /-----/ |ññ|:://::|ññ|							            								/-----/ | |ññ|:://::|ññ| PROGRAMMER:: James Dougherty											| |   | | |ññ|:://::|ññ| COPYRIGHT :: ©2003 Ariel Productions									| |   | | |ññ|:://::|ññ| TYPE      :: Utility													| /---|-/ |ññ|:://::|ññ|																		/-----/   |ññ|:://::|ññ ---------------------------------------------------------------------------------- ññ|:://::|ññ| ~ SOURCE DATA ~																  |ññ|:://::|ññ|																				  |ññ|:://::|ññ| FILENAME :: XErrorSystem.h	 													  |ññ|:://::|ññ| VERSION  :: 1.5																  |ññ|:://::|ññ|																				  |ññ|:://::|ññ| HISTORY DATA																	  |ññ|:://::|ññ| | 																				  |ññ|:://::|ññ| |-10.03.03 (JD)																  |ññ|:://::|ññ| |-Initial Build																  |ññ|:://::|ññ| |																				  |ññ|:://::|ññ| |-12.09.03 (JD)																  |ññ|:://::|ññ| |-Fixed clipboard failure														  |ññ|:://::|ññ| |																				  |ññ|:://::|ññ| |-12.16.03 (JD)																  |ññ|:://::|ññ| |-Added a new e-mail feature													  |ññ|:://::|ññ| |																				  |ññ|:://::|ññ|																				  |ññ|:://::|ññ| BUGS																			  |ññ|:://::|ññ| |																				  |ññ|:://::|ññ| |-N/A																			  |ññ|:://::|ññ| |																				  |ññ|:://::|ññ|																				  |ññ|:://::|ññ ---------------------------------------------------------------------------------- ññ|:://::|ññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññññ|::#ifndef XERRORSYSTEM_H#define XERRORSYSTEM_H#include "XDefinition.h"//--//-- This will only enable the e-mail feature in the release build. This will save the//-- programmer a lot of e-mails during the debug/build phase.//--#define ONLY_ENABLE_EMAIL_IN_RELEASE_BUILD 1//--//-- SMTP server address//--#define SMTP_SERVER_ADDRESS ":::::::::[NOBODY NEEDS THAT]::::::::::.net"//--//-- ::NOTE:://-- The e-mail was designed for internal business use. If an error occurs in the program//-- an e-mail can be sent instead of the associates giving you a call. This is a good idea//-- if you have a massive amount of programs to deal with. Where I work at now, we have//-- around 1,200+ programs that we have to deal with. Of course it's not in C++, or we would//-- have 1 :) (Don't you love publisher reports and database programming!)//--#define SYSTEMS_PROGRAMMER_EMAIL_ADDRESS "arielproductions@zoominternet.net"//--//-- This will be the subject on the e-mail when it is sent.//--//-- ::NOTE:://-- Change %Put Program Name Here% to the name of the current project you're working on.//--//-- ::NOTE:://-- Notice there is no "]" at the end of the subject. This is because the error system//-- will put the computers name at the end of the subject so you know what computer it//-- came from. This is helpful when there are 10's to 100's of computers on your network.//-- It becomes usfull because you can find out who was running the computer at the time//-- the error occured so you can talk to them personally to find out what they were doing//-- to cause the error.//-- #define EMAIL_SUBJECT "[ERROR IN PROGRAM %Put Program Name Here% ON SYSTEM "class XERROR{public: //--Error Results	enum XRESULT	{		XNOERROR		= 0x00, //--No errors occured		XFALSE			= 0x01, //--Process failed, but can continue (Warning)		XFATAL			= 0x02, //--Fatal error		XNOMEMORY		= 0x03, //--Not enough memory available to finish the process		XNOTIMPLEMENTED = 0x04, //--Currently not implemented		XFILENOTFOUND   = 0x05  //--File does not exists	};public: //--Error Actions	enum XERRORACTION	{		XEA_NONE   = (1<<0), //--Perform no action		XEA_CLPBRD = (1<<1), //--Copy error message to the clipboard		XEA_MSGBOX = (1<<2), //--Display a message box		XEA_REPORT = (1<<3), //--Save to an error report		XEA_DEBUG  = (1<<4), //--Print out a debug string		XEA_BREAK  = (1<<5), //--Break into the debugger		XEA_EXIT   = (1<<6), //--Exit the application		XEA_EMAIL  = (1<<7), //--Send an e-mail to the systems programmer		XEA_FATAL  = XEA_MSGBOX | XEA_REPORT | XEA_EMAIL  | XEA_EXIT,		XEA_COMMON = XEA_CLPBRD | XEA_MSGBOX | XEA_REPORT | XEA_EMAIL,		XEA_FULL   = XEA_COMMON | XEA_DEBUG  | XEA_BREAK  | XEA_EXIT	};public:	XERROR();	XERROR(Xint);	XERROR(XRESULT);	XERROR(const XERROR&);public:	Xbool   Failed(Xvoid);							//--See if the current error failed	Xbool   Failed(const XERROR&);					//--See if the query error failed	Xbool   Succeeded(Xvoid);						//--See if the current error succeeded	Xbool   Succeeded(const XERROR&);				//--See if the query error succeeded	Xvoid   ProcessError(Xstring ErrorMessage);	    //--Process the error if one occured	Xvoid   ProcessWarning(Xstring WarningMessage); //--Process the warning if one occuredpublic:	Xint    GetFlags(Xvoid);						//--Get the current error action flags	Xvoid   SetFlags(Xint Flags);					//--Set the error action flags	Xvoid   SetResult(XRESULT Result);				//--Set the current error result flag	Xuint   GetErrorID(Xvoid);						//--Get the current error ID	Xuint   GetErrorID(const XERROR&);				//--Get the query error ID	Xstring GetErrorDescription(Xvoid);				//--Get the current error description	Xstring GetErrorDescription(const XERROR&);		//--Get the query error descriptionpublic:	XERROR&	operator= (const XERROR& Error);		//--Copy an error	Xbool	operator==(const XERROR& Error)  const; //--See if two errors are the same	Xbool	operator==(const XRESULT Result) const; //--See if two error results are the same	Xbool	operator!=(const XERROR& Error)  const; //--See if two errors are not the same	Xbool	operator!=(const XRESULT Result) const; //--See if two error results are not the sameprivate:	Xint	m_Flags;		 //--Holds the current error action flags	XRESULT	m_CurrentResult; //--Holds the current error result flag};//------------------------------------------------------------------------------------------------//--//-- Error Handling Macros//-- ---------------------//--//-- POSTERROR(*)//--    ~INTERNAL::PRIVATE~ Called by XERROR()//--//-- POSTWARNING(*)//--    ~INTERNAL::PRIVATE~ Called by XWARNING()//--//-- TRAPERROR(*)//--    If CONDITION is 'true', it wall call POSTERROR() with the ERRORMESSAGE//--    and FLAGS as the parameters.//--//-- TRAPWARNING(*)//--    If CONDITION is 'true', it wall call POSTWARNING() with the ERRORMESSAGE//--    The XEA_EXIT flag is disabled, can't exit on a warning.//--//-- ~ NOTE ~//--	Flags are XERRORACTION flags.//--//------------------------------------------------------------------------------------------------extern Xvoid POSTERROR(Xstring File, Xint Line, Xstring ErrorMessage, Xint Flags);extern Xvoid POSTWARNING(Xstring File, Xint Line, Xstring WarningMessage, Xint Flags);#define TRAPERROR(CONDITION, ERRORMESSAGE, FLAGS)								{																					if((Xint)(CONDITION))															{																					POSTERROR(__FILE__, __LINE__, (Xstring)ERRORMESSAGE, (Xint)(FLAGS));		}																			}#define TRAPWARNING(CONDITION, WARNINGMESSAGE, FLAGS)							{																					if((Xint)(CONDITION))															{																					POSTWARNING(__FILE__, __LINE__, (Xstring)WARNINGMESSAGE, (Xint)(FLAGS));	}																			}				}#endif //--XERRORSYSTEM_H

EDIT:
It put the macros on 1 line!?!

-UltimaX-

"You wished for a white christmas... Now go shovel your wishes!"

[edited by - UltimaX on December 16, 2003 10:04:33 PM]

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Nekhmet: I may have misunderstood you, but I believe you''re saying the exact opposite of what I''m saying. I think that if you have an error that is recoverable (if that bitmap is just an extra feature), then you should return an error instead of throwing an exception. This way, you don''t have to pass the error up levels and levels of functions, instead you can just wait for the appropriate handler, and prevent wasting precious coding time typing those extra function checking routines. (I''ve forgotten the other advantages at the moment..)

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quote:
...heard that you shouldn''t check the return values for operator new() anymore...

A better reason is that the standard says that new throws std::bad_alloc, rather than returning 0 on failure. Checking for 0 only became popular because of compilers that do not directly conform to the standard in that area (like VC++).

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Nekhmet: Not only what psykr said, but how many games do you see shipped with the actual bitmap files. Ususally they use some resource file system anyway. So if your missing 1 image you have a lot more serious problems than just 1 image anyway.

-UltimaX-

"You wished for a white christmas... Now go shovel your wishes!"

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I think what the OP was talking about was the cost of doing all this checking, and is it going to hurt. The answer in short, yes, but not as much as you think. Put all the error checking you want in initalization, but putting TOO much into the main game loop may add a moderate performance hit. One way to avoid this, is to have specific #ifdef debug statements lieing throughout your code, and have it ignore the code when you are not in debug mode. This is tedious, but if you want the fastest code, its a necessity.

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quote:
Original post by merlin9x9
quote:
...heard that you shouldn''t check the return values for operator new() anymore...

A better reason is that the standard says that new throws std::bad_alloc , rather than returning 0 on failure. Checking for 0 only became popular because of compilers that do not directly conform to the standard in that area (like VC++).

Yes. If someone still want to check for 0 though it''s easy to disble the exception by using std::nothrow as in:

FOO* foo = new (std::nothrow) FOO();

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Thanks for the replies guys. Got quite a lot to go on there, and yes PaulCesar, that''s pretty much where I was heading with my question.

I''m thinking about downloading a 3D engine to have a look at; not necessarily for the 3D side of things, but more for the error handling at this point.

There were actually two reasons my question came up:

1. From working on the class I pasted into my original post.
2. Seeing how errors are returned from Unreal Tournament by their calling function(s), as in:

Exception at: MainLoop->GameEngine->Botpack->Warshell->etc.

Some of the error messages I''ve seen from UT have been absolutely huge (message boxes literally taking up an entire screen on 800x600).

Anyway, thanks again. I''ll have a nose around some engine code in a day or so.

-hellz

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UT uses the exception mechanism to build those error messages. They have 2 macros, guard and unguard:

SomeObject::SomeFunc(){   guard(SomeObject::SomeFunc)   // code   unguard;}

The guard macro starts the try block, but also stores the name of the function in a local static variable (and possible some other stuff). The unguard macro closes the try and implements the catch block. When an exception is caught, the function name in the local var is added to a list somewhere and the exception is rethrown. The function names in the list are then used to build a stack trace.

Additionally, two other versions of the macros, guardslow and unguardslow, are used in performance critical areas. These macros can be turned on for testing/debugging purposes, but are turned off in the release version.

The Torque engine takes a somewhat different approach, using a custom assert system. This is used most often to verify function params, but also to catch things like ''divide by zero'' and ''array out of bounds''. There are different macros for fatal and warning conditions, and also macros for release build and those that are turned on/off for testing and debugging. This sort of system can also be implemented in C.

Error checking can take away cycles, but it is important nonetheless. People who knock Java & C# often point at the fact that they are not compiled to native code as a reason for performance issues, when in fact it''s mostly because of all of the error checking going on behind the scenes. Those languages verify that every array access is in bounds, that every divide is safe, and things other than error checking. C and C++ allow you to be as stringent or as loose about error checking as you''d like, which is a great thing when performance is important.

I think a simple rule to follow is, write your code to check for errors everywhere the code can hurt you (verifying function params, return values, array indices where they may go out of bounds, etc...). Remove the checks that hurt performance only after profiling, but see if there are other ways of making the same check without the performance penalty (verifying a pointer once before calling a method multiple times, rather than in the method itself, for example). Also, if the game is moddable, it may be a good idea to look for optimizations elsewhere and leave the error checking code in if it''s something that can be unpredictable at runtime. Macros can greatly help in this, so they can be turned on and off when needed.

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Wow, thanks for the reply, Aldacron. That''s given me quite a nice insight and a lot of food for thought.

Thank you very much.

-hellz

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quote:
Original post by Nekhmet
I think exceptions should only be used if the error is recoverable.

For example, let''s say that your loadBMP fails. What are you going to do... play the game without the bitmap? Is that really acceptable? If you can''t play without the bitmap, then it''s easiest to just display an error and stop the game.

But you can do that with an exception. Just catch fatal exceptions at the top level, cleanup, and bail.

--
Dave Mikesell Software & Consulting

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text:
Thesis Open Access

# Unsteady State Kinetic Modelling of Biomass Gasification in Bubbling Fluidized Bed Gasifier

Tolossa Kebede Tulu

### Dublin Core Export

<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:creator>Tolossa Kebede Tulu</dc:creator>
<dc:date>2021-06-09</dc:date>
<dc:description>From available renewable energy, biomass is the most abundant and promising resource of renewable energy which is environmental friendly. From biomass conversion methods to useful energy, gasification is an adequate technology. The mathematical modeling of biomass gasification in bubbling fluidized bed can be used to predict syngas composition and optimize the operational condition as well as gasifier design, which is a cost-saving option and viable routine. However, the kinetic modeling of biomass gasification is at early phase. With this regard, the aim of this study is to develop kinetic model based on two-phase fluidization theory, which is comprised of reaction kinetics, bed hydrodynamic parameters, diffusion effect, and species transport equations. Furthermore, the thermal cracking of tar was integrated into the model to increase the prediction accuracy of the model and to address the gap in knowledge. In this study, the transient kinetic modeling of biomass gasification in bubbling fluidized bed gasifiers and optimization methods to maximize gasification products were developed. The model was coded in Matlab and simulated. The result depicts good agreement with experimental work in literature. The sensitivity analysis carried out and the effect of temperature from 650oC-850oC and steam to biomass ratio (S/B) from 0.1-2 investigated. The result depicts the increase in temperature and S/B promotes H2 production and reduces CO and CH4. Furthermore, the regression model was carried out for temperature and S/B variation in Design-Expert. The surface response is constructed from the regression model and the mutual effect of temperature and S/B on gasification product and heating value investigated. Besides, the desirability function is employed to optimize gasification product and heating value. The maximum gasification product yield was at 827.866oC and 0.1 S/B in investigated range. The response predicted by desirability function at this optimum operational conditions is 30.096%, 44.07%, 13.20%, 12.90%, 14.035 (MJ/Nm3 ), and 14.536 (MJ/Nm3 ) for H2, CO, CO2, CH4, LHV, and HHV respectively.</dc:description>
<dc:language>eng</dc:language>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:subject>Biomass gasification, Fluidized bed, Kinetic Modeling, Response surface,  Optimization</dc:subject>
<dc:title>Unsteady State Kinetic Modelling of Biomass Gasification in Bubbling Fluidized Bed Gasifier</dc:title>
<dc:type>info:eu-repo/semantics/doctoralThesis</dc:type>
<dc:type>publication-thesis</dc:type>
</oai_dc:dc>

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text:
# Entire solutions of completely coercive quasilinear elliptic equations, II

created by farina on 06 Jul 2010
modified on 16 Sep 2010

[BibTeX]

Accepted Paper

Inserted: 6 jul 2010
Last Updated: 16 sep 2010

Journal: Journal of Differential Equations
Year: 2010

Abstract:

In an earlier paper 3 the authors treated a broad class of quasilinear elliptic equations which have the property that any entire solution must necessarily be constant, a property of course not holding for the simple Laplace equation itself. Here we generalize the earlier class of equations to include cases where the "inhomogeneous terms" depend strongly on the gradient of the solution; see for example the model $p$-Laplace--type equation (2) below, as well as other more general examples discussed later.

Theorems 8 and 9 are particularly interesting in that, in contrast to the earlier conclusions, they require only the most minimal coercive behavior of the inhomogeneous terms when the solution variable lies in some arbitrarily large but bounded set; see especially the model example (15) at the end of the introduction.


text:
# Cunning Complex 6

Algebra Level 4

Find the sum of real values of $$x$$ and $$y$$ for which the following equation is satisfied:

$\frac { \left( 1+i \right) x-2i }{ 3+i } + \frac { \left( 2-3i \right) y+i }{ 3-i } =i.$

×


text:
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 August 26th, 2012, 05:03 AM #1 Newbie   Joined: May 2012 Posts: 6 Thanks: 0 Demonstration help Prove that the 1979 first algarisms after the comma, of the number $(6+\sqrt{35})^{1980}$, are all algarisms 9. *Sorry, maybe there's a grammar error, this is a direct translation from portuguese.
 August 26th, 2012, 05:40 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Demonstration help I don't know the word "algarisms".
 August 26th, 2012, 05:59 AM #3 Newbie   Joined: May 2012 Posts: 6 Thanks: 0 Re: Demonstration help Sorry, edit: Prove that the 1979 first digits after the comma, of the number , are all algarisms 9.
 August 26th, 2012, 06:01 AM #4 Newbie   Joined: May 2012 Posts: 6 Thanks: 0 Re: Demonstration help Prove that the 1979 first digits after the comma, of the number $(6+\sqrt35)^{1980}$, are all algarisms 9. Why can't I edit my posts?
 August 26th, 2012, 06:05 AM #5 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Demonstration help "algarisms" is not an english word.
August 26th, 2012, 06:08 AM   #6
Newbie

Joined: May 2012

Posts: 6
Thanks: 0

Re: Demonstration help

Quote:
 Originally Posted by mathbalarka "algarisms" is not an english word.
Yep, i forgot to change the second one. Sorry, again...

Prove that the 1979 first digits after the comma, of the number $(6+\sqrt{35})^{1980}$ , are all algarisms 9.

 August 26th, 2012, 06:10 AM #7 Newbie   Joined: May 2012 Posts: 6 Thanks: 0 Re: Demonstration help Yep, i forgot to change the second one. Sorry, again... [2] Prove that the 1979 first digits after the comma, of the number $(6+\sqrt{35})^{1980}$ , are all digits 9. I can't edit(or even delete) my old posts.
August 26th, 2012, 06:14 AM   #8
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Re: Demonstration help

Quote:
 Originally Posted by Fingolfin I can't edit(or even delete) my old posts.
You can't edit your posts until you have at least 25 posts.

You can't delete you posts if you are not a moderator.

Balarka

.

August 26th, 2012, 06:24 AM   #9
Math Team

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From: India, West Bengal

Posts: 3,871
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Re: Demonstration help

Quote:
 Originally Posted by Fingolfin Prove that the 1979 first digits after the comma, of the number $(6+\sqrt{35})^{1980}$, are all digits 9.
To be honest, I don't understand what you want to ask.( I observed that $(6+\sqrt{35})^{1980}$ has large numbers of 9's in the later terms of the decimal places of the number).

Balarka

.

 August 26th, 2012, 07:51 AM #10 Newbie   Joined: May 2012 Posts: 6 Thanks: 0 Re: Demonstration help I found the source of the Problem. Sidney Technical highscholl (1996): i) Explain why $(6+\sqrt{35})^{1980} + (6-\sqrt{35})^{1980}$ is a even number. ii) Show that $0 < (6 - \sqrt{35}) < 0.1$ iii) Hence or otherwise prove that the first thousand digits after the decimal point in the value of $(6+\sqrt{35})^{1980}$ are all 9's.

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text:
### Many familiar substances have common, unsystemati…

03:10
Manhattan College
Problem 111

Give the chemical names of each of the following familiar compounds:$(\mathbf{a})$$\mathrm{NaCl}( table salt ) (\mathbf{b})$$\mathrm{NaHCO}_{3}$ (baking soda),$(\mathbf{c}) \mathrm{NaOCl}$ (in many branches)$(\mathbf{d})$ $\mathrm{NaOH}($ caustic soda $)$ $(\mathbf{e})$ $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}($ smelling salts $)$$(\mathbf{f})$$\mathrm{CaSO}_{4}$ (plaster of Paris).

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## Video Transcript

problem 1 11 except to give the chemical name to each of the familiar compounds and a CEO. Oh, we just table salt and h c 03 This is known as baking soda and a O. C. O. Which is used in a block of bleachers. Sorry, Emmett. Many branches I should say Hey! And a ohh. Which is caustic soda and h for to see Oh, three just smelling salts and C A s 04 which is the Glasser of Paris. So remember when we named these things the metal comes first and then the non metal so we have sodium, and then we have chlorine. But remember, it's eye on a compound because it's composed of a metal and non metal. It's a sodium chloride. All right, here we have sodium, and we have a t 03 which knows bicarbonate. Here we have Excuse me. So we have sodium hypo folklore chloride because hypo chlorate is o. C E o. Next year, we have sodium hydroxide, so hydroxide is o H minus. And remember, you have an A plus for our sodium Zagato. So which minds is going to be our non metal or partly atomic. I am here we have and h four, which is gonna be ammonium. Um, when we have to. I have a compound made out of two non metals meadows. We used the positively charged one first and then we use the negative charge. So we have ammonium. And then after that, we have what is known as carbonates. Now, next we have calcium, which is C a and then we are going to have I have so faint. Eso eso four This is very strange folk. So we're just remember the name and convince.


text:
# (Non/)Equlibrium and Newton's Laws of Motion

1. Oct 6, 2007

### Gannon

This is mainly stumping me because of the absence of any force or mass.

1. The problem statement, all variables and given/known data

A crate is sliding down a ramp that is inclined at an angle of 33.6° above the horizontal. The coefficient of kinetic friction between the crate and the ramp surface is 0.485. Find the acceleration of the moving crate.

$$\mu$$$$_{k}$$= .425
$$\theta$$ = 33.6
g = 9.8m/s$$^{2}$$

2. Relevant equations

Well, F$$_{net}$$x = ma , but there is no mass... I don't know where to go from here.

Any help is appreciated. Thanks.

2. Oct 6, 2007

### Bryan52803

If you think about the ideal concepts of this type of motion, you can see that mass does not matter. Start to solve the equation as if you had a mass (just use 'm' in place of the unknown quantity) and you'll see it's not as important as you think. If you get stuck in the actual solving, post again with your progress.

Bryan

3. Oct 7, 2007

### Gannon

Ok... I think I'm on the right track. So the net force will be simply

mgsin$$\theta$$ - $$\mu$$mgcos$$\theta$$ = ma

You can divide everything by mass, leaving

gsin$$\theta$$ - $$\mu$$gcos$$\theta$$ = a.

Is this correct?


text:
# model,report_ge_weight-method

0th

Percentile

##### model method for report_ge_weight' this method uses samples collected over the season to model the variation in weight of glass eel or yellow eels.

model method for report_ge_weight' this method uses samples collected over the season to model the variation in weight of glass eel or yellow eels.

##### Usage
# S4 method for report_ge_weight
model(object, model.type = "seasonal",
silent = FALSE)
##### Arguments
object

An object of class report_ge_weight-class

model.type

default "seasonal", "seasonal1","seasonal2","manual".

silent

Default FALSE, if TRUE the program should no display messages

##### Details

Depending on model.type several models are produced

• model.type="seasonal". The simplest model uses a seasonal variation, it is fitted with a sine wave curve allowing a cyclic variation w ~ a*cos(2*pi*(d'-T)/365)+b with a period T. The modified day d' used is this model is set at 1 the 1st of august doy = d' + d0; d0 = 212, doy=julian days

• model.type="seasonal1". A time component is introduced in the model, which allows for a long term variation along with the seasonal variation. This long term variation is is fitted with a gam, the time variable is set at zero at the beginning of the first day of observed values. The seasonal variation is modeled on the same modified julian time as model.type="seasonal" but here we use a cyclic cubic spline cc, which allows to return at the value of d0=0 at d=365. This model was considered as the best to model size variations by Diaz & Briand in prep. but using a large set of values over years.

• model.type="seasonal2".The seasonal trend in the previous model is now modelled with a sine curve similar to the sine curve used in seasonal. The formula for this is $sin(\omega vt) + cos(\omega vt)$, where vt is the time index variable $\omega$ is a constant that describes how the index variable relates to the full period (here, $2\pi/365=0.0172$). The model is written as following $w~cos(0.0172*doy)+sin(0.0172*doy)+s(time).$

• model.type="manual". The dataset don (the raw data), coe (the coefficients already present in the database, and newcoe the dataset to make the predictions from, are written to the environment envir_stacomi. please see example for further description on how to fit your own model, build the table of coefficients, and write it to the database.

##### Aliases
• model,report_ge_weight-method
• model.report_ge_weight
Documentation reproduced from package stacomiR, version 0.5.3, License: GPL (>= 2)

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text:
Workshop on Prismatic Cohomology

We will study prismatic cohomology, a new cohomology theory developed by Bhargav Bhatt and Peter Scholze. There will be 10 talks, see below for further details.

Organizers: Eike Lau, Torsten Wedhorn, Eva Viehmann / Ulrich Görtz (local organizer)

Date: March 27 – 29, 2019 (We will start on Wednesday at 10am, and the workshop will end on Friday around 2pm)

Place: Department of Mathematics, University of Duisburg-Essen (Thea-Leymann-Str. 9, 45127 Essen)

Program: pdf

Registration

If you are interested in attending, please send an email to Ulrich Görtz, ulrich.goertz@uni-due.de, so that we can keep track of how many people will be around and order enough coffee for the coffee breaks.

Accommodation

Most participants are in B&B Hotel Essen, Helmut-Käutner-Straße 4, 45127 Essen (very close to the math department), Phone: 0201 616960. If you have not yet booked a room, you should contact the hotel as soon as possible. We have reserved a few more rooms until Friday, Feb. 8 (mention “Math Institute/Prof. Görtz”).

Venue

All talks will take place in the math building of the University of Duisburg-Essen, Thea-Leymann-Str. 9/Altendorfer Str. 11, 45127 Essen, seminar room N-U-3.04 (3rd floor, north wing).

To get to the math department from the main station, take the U-Bahn to Berliner Platz. From there, walk down Altendorfer Str. In the ground floor of the building, there is a large Conrad Electronics store, so the building is easy to find.

Talks and schedule

 1 $\delta$-rings Jakob Stix (Frankfurt) Wednesday 10:00 – 11:15 2 Derived completion Jens Hesse (Darmstadt) 11:45 – 12:45 3 Prisms Maria Beckemeyer (Münster) + Karsten Schrödter (Bielefeld) 14:45 – 16:00 4 The prismatic site and the Hodge-Tate comparison map Matteo Tamiozzo (Essen) 16:30 – 17:30 5 Crystalline and Hodge-Tate comparison maps Lennart Gehrmann (Essen) Thursday 9:30 – 10:45 6 Non-abelian derived functors Fritz Hörmann (Freiburg) 11:15 – 12:45 7 Derived prismatic cohomology Kay Rülling (FU Berlin) 15:00 – 16:00 8 Application to perfectoid rings Paul Hamacher (TU München) 16:30 – 17:30 9 Étale comparison Johannes Anschütz (Bonn) Friday 9:30 – 10:45 10 Application: $q$-de Rham cohomology Andreas Pieper (Ulm) 11:15 – 12:30