text:
# Adaptive equalizer for BPSK signals with frequency offset

I'm trying to equalize a BPSK signal which has a residual frequency offset on it. This frequency offset is dealt with at a later stage in the DSP chain, along with phase estimation.

I have implemented a FIR filter which updates its filter taps using gradient descent CMA:

The equalized signal (y) is a discrete-time convolution of the filter taps (W) and the input signal (X):

$$y = W^HX$$

The update error is: $$\epsilon = |y|^2-1$$

Finally I compute the update as: $$W_{i+1} = W_i - \mu \epsilon y^*X$$ Where $\mu$ is the step size = 0.001.

My problem is that due to the frequency offset, my input signal X is complex, which is resulting in complex filter taps in W. This in turn introduces a huge phase error into my signal. One possible solution I've looked at so far is putting the equalizer behind the frequency offset compensation stage. However, since the signal still contains some phase error, it's not entirely constrained to the real axis in the constellation diagram and I have the same problem. My workaround so far has been to update the filter taps as: $$W_{i+1} = W_i - real(\mu \epsilon y^*X)$$ My reasoning being that after frequency offset compensation and phase estimation, there should only be noise in the quadrature part of the signal, and I can safely discard it.

I'm unhappy with this solution for 2 reasons:

1. I'm not entirely sure it's theoretically sound, and at the very least it's rather unorthodox.

2. I would prefer to equalize the signal upstream in the DSP chain.

I would appreciate any help or insights!

• Even if you did not have a frequency offset, it is likely that due to channel distortion effects that you would still need an equalizer with complex filter taps given distortion effects that are not conjugate symmetric centered about the carrier. Are you dealing with a channel where you are sure that is not the case? May 10 '17 at 9:47
• Can you provide more details on the distortion sources you are trying to equalize? My thoughts are constrained to wireless channels where you would certainly want a complex equalizer, and to my understanding this is best and commonly done prior to timing and carrier recovery. May 10 '17 at 10:28
• An approach that I've used before is to use a fractionally-spaced blind equalizer (e.g. using CMA), followed by timing and carrier sync, then followed again by a symbol-spaced equalizer that is decision-directed (or possibly taking advantage of a training sequence if there is one). The carrier synchronization step will eliminate the frequency error that you're seeing. You'll still often have residual phase offset due to ambiguities in the carrier sync algorithm, but your second equalizer should be able to handle a constant to slowly-varying phase offset. May 10 '17 at 13:40
• The receiver is bandwidth limited, which is introducing a large amount of ISI into the signal. That is the main effect I would like to equalize. I should clarify that I'm not trying to avoid using a complex taps, rather, I simply can't get the filter to work if I don't include the real(.) operation in the update algorithm. May 10 '17 at 14:24
• To echo @DanBoschen, if you're receiving BPSK over a wireless channel, then the received signal will be spread over both the in-phase and the quadrature branches, and you may want to have a complex CMA. In my experience, I've used a phase-independent symbol timing estimator first, followed by CMA, and then carrier tracking. This works fairly well for BPSK and QPSK, and probably for M-PSK in general.
– MBaz
May 11 '17 at 2:42

It sounds like your biggest issue is dealing with a phase offset and otherwise you have no other concerns with your equalization approach? In that case, the issue is resolved by simply having a phase correction prior to making your decisions on the real signal. If done correctly your carrier recovery implementation should remove any residual long term phase offset. This post may help give you further insight: Phase synchronization in BPSK. I duplicated the final block diagram below where in your case "I data demod" would be your BPSK output. I want to point out using this diagram, that any residual phase error would accumulate in the PI Loop Filter which would then adjust the NCO to rotate the signal to have zero error (on average). Note that the Loop as shown if completely blind will lock in quadrature or 180° phase positions - if that is your remaining issue then another question on carrier recovery implementations would be more appropriate specific to any issue you have there, but a simple solution is to derive the phase error itself from $\phi \propto Q \text{sign}(I)$ since the objective in this case is to drive any signal on Q down to the noise. The assumption is the signal has been AGC'd prior to this point so such an approach to phase detection would result in a consistent loop gain. Here is another related post with regards to what gain constants you should use (what loop BW should be used): Loop bandwidth for symbol timing recovery

Also notice the similarity of the above diagram to a classical Costas Loop with the following figure from http://michaelgellis.tripod.com/mixerscom.html. This is because the Decision Directed CR Loop is a Costas Loop. The referenced link also diagrams the classical Costas Loop for BPSK specifically, which also shows how the digital implementation such as shown in the diagram above can be modified for the BPSK only case (which is modified the phase error metric as I described earlier). This solution will still have a 0/180° phase ambiguity which is typically resolved with minimum knowledge of the message (header or some other recognized feature in the data message or encoding in the data message).

• Hi Dan, thanks for your help so far. Perhaps I should have clarified that I already have a working CPR system in place. To verify that it's not the CPR or FO compensation causing the problems, I tried using some different adaptive filter implementations, which do actually give me nice results! I used the blind adaptive algorithms described in this paper: spiral.imperial.ac.uk/bitstream/10044/1/589/1/…. So this leads me to believe something is wrong with how I implemented the CMA algorithm. May 11 '17 at 15:50
• @Philip I was keying specifically on your comment of concern with a phase offset. This implies your carrier recovery isn't completely working or you simply need an additional phase rotator since you can easily measure (and rotate) the phase to be completely on the real axis. May 11 '17 at 18:54


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# scipy.special.h2vp¶

scipy.special.h2vp(v, z, n=1)[source]

Compute nth derivative of Hankel function H2v(z) with respect to z.

Parameters: v : float Order of Hankel function z : complex Argument at which to evaluate the derivative n : int, default 1 Order of derivative

Notes

The derivative is computed using the relation DLFM 10.6.7 [R489].

References

 [R488] Zhang, Shanjie and Jin, Jianming. “Computation of Special Functions”, John Wiley and Sons, 1996, chapter 5. https://people.sc.fsu.edu/~jburkardt/f_src/special_functions/special_functions.html
 [R489] (1, 2) NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov/10.6.E7

#### Previous topic

scipy.special.h1vp

#### Next topic

scipy.special.spherical_jn

format: plain


text:
# Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 117: 3.67c

354g nitrogen gas

#### Work Step by Step

Using conversion factors, we have: Volume = $10.0ft^3 * 28.31L/ft^3 = 283L$ Mass = Density * Volume = $1.25g/L * 283L = 354g$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
### State the following statements are True or False: 1. Utility depends upon intensity of want.

State the following statements are True or False:
1. Utility depends upon intensity of want.


text:
# [pstricks] pst-Bspline

Michael Sharpe msharpe at ucsd.edu
Thu Nov 27 02:55:14 CET 2008

```Herbert,

I sent the package to CTAN---should appear in a few days.

Michael

On Nov 26, 2008, at 12:12 PM, Herbert Voss wrote:

> Michael Sharpe schrieb:
>> I posted an updated version of pst-Bspline as
>>
>> http://math.ucsd.edu/~msharpe/pst-Bspline.zip
>>
>> In addition to  drawing cubic B-spline curves (open/relaxed or
>> closed)
>> based on a sequence of B-spline control points, it has code to do B-
>> spline interpolation through a sequence of points. The package
>> includes documentation/examples. The interpolation code makes major
>> use of the variant \psLCNodeVar of \psLCNode, and is included as a
>> routine in pst-Bspline.tex.
>
> Michael,
> you should upload the package to CTAN to make it available
> to the public:
>


text:
# With 2 lists of numbers is it possible to get the actual ratio of the sum of both lists with only the ratio of each individual group?

To get the ratio of two lists of numbers, I take the sum of each column and divide the first by the second.

Suppose I only have 'access' to the ratio of each individual number combo. Is it possible to get the 'total' ratio from just the individual ratios?

Taking the average of the individual ratios gets me pretty close but it's no good.

EDIT: Thanks for the re-tagging :)

-
Since the ratio of totals can change while the individual ratios stay the same, the answer is no. –  Henry Apr 12 '12 at 22:32
oh :( Alright. Thanks for your help. If you submit a answer, I'll accept it for you. –  Robo Apr 12 '12 at 22:35

Since the ratio of totals can change while the individual ratios stay the same, the answer is no.

Here is an example where two single ratios (of 3 and 5) can correspond to different ratios of the totals

     num denom ratio
a     3   1      3
b     7   1      7
------------------
sum  10   2      5

num denom ratio
a     9   3      3
b     7   1      7
------------------
sum  16   4      4

-
I went the other way around with my answer (one total ratio can be found by two different ratio vectors), but this is more convenient for an answer. +1 –  Patrick Da Silva Apr 12 '12 at 22:42

So clearly, you are given $a_1, a_2, a_3, b_1, b_2, b_3$. You're computing the individual ratios $r_i = a_i / b_i$, you sum them to form $R_1 = r_1 + r_2 + r_3$. Then you compute another ratio which is $(a_1 + a_2 + a_3)/(b_1 + b_2 + b_3)$, and you expect that given $r_1, r_2, r_3$, you can compute $(a_1 + a_2 + a_3) / (b_1 + b_2 + b_3)$. If I am wrong on this description correct me.

Let's see what we can do. I'll try to express the total ratio in terms of the $r_i$'s and assume they are real numbers. You can compute $$\frac{a_1 + a_2 + a_3}{b_1 + b_2 + b_3} = \frac{r_1 b_1 + r_2 b_2 + r_3 b_3}{r_1 a_1 + r_2 a_2 + r_3 a_3} = \frac{r \cdot b}{r \cdot a}$$ by letting $r = (r_1, r_2, r_3)$, $a = (a_1, a_2, a_3)$, and $b = (b_1, b_2, b_3)$, with the $\cdot$ meaning scalar product (hence my algebra-precalculus tag). You have access to the vector $r$ and wish to know the ratio $(r \cdot b)/(r \cdot a)$. As you can see from linear algebra, you can find two different ratios that will give you the same total ratio. For instance, if $\lambda \in \mathbb R \backslash \{0 \}$, $$\frac{r \cdot b}{r \cdot a} = \frac{(\lambda r) \cdot ((1/\lambda)b)}{(\lambda r) \cdot ((1/\lambda) a)}$$ Therefore, given the ratios $r_1, r_2, r_3$ or the ratios $\lambda r_1, \lambda r_2, \lambda r_3$, you find the same total ratio. This makes no hope of computing the total ratio with only the "component-wise ratios", since you can take component wise ratios as big as you want and still obtain the same total ratio. But then again it doesn't mean that there exist no such formula ; I just have little faith that such a formula exists. It doesn't seem very useful either, to be honest ; why do you want to compute things this way? With more information perhaps I could help on the problem behind the question.

Hope that helps,

-
Unfortunately, the ratio of the components is the only thing that we have access to due to the way the data was processed and stored in a database. I was hoping there was a mathematical solution to this problem so I wouldn't have to re-write the application I'm working on. –  Robo Apr 12 '12 at 22:51


text:
Celsius to Delisle

Calculates the Delisle temperature from the given Celsius or centigrade scale value

Type what you want to convert in the box below or
(autosubmits, max. 1MB)
<- ups! invalid URL! please, delete it or correct it!

Celsius to Delisle

The Celsius or centigrade temperature scale is based on the water tristate, placing the 0 at the point when the liquid water becomes ice, and the 100 at the point when the liquid water becomes gas, both at 1 atm pressure. Its unit is represented with ℃ after the value.

The Delisle temperature scale was defined by Joseph-Nicolas Delisle in 1732. This measure adjust 0 at the point of the boiling water and 150 as the freezing point of the water. The unit is represented by °D (and sometimes as °De) after the value.

Delisle values are calculated based on the formula $$(100 - Celsius) × {3 \over 2}$$


text:
## The Monge-Ampère equation.(English)Zbl 0989.35052

Progress in Nonlinear Differential Equations and their Applications. 44. Boston, MA: Birkhäuser. xi, 127 p. (2001).
In recent years the Monge-Ampère equation $\det D^2u = f \tag{1}$ has received a great deal of attention. Originally this equation arose in differential geometry, in connection with problems involving hypersurfaces of prescribed Gauss curvature, such as the Minkowski problem. More general Monge-Ampère equations also arise in geometry, for example, in the Weyl embedding problem. More recently a connection with mass transportation problems was discovered, and this has led to renewed interest in Monge-Ampère equations in the calculus of variations and optimization theory, as well as in applications such as meteorology, fluid dynamics, and mathematical physics.
In this book the author presents an essentially self-contained exposition of the theory of generalized solutions of (1), starting with some basic facts about convex functions, and ending with the deep interior regularity theory developed by L. A. Caffarelli [Ann. Math., II. Ser. 131, No. 1, 129-134 (1990; Zbl 0704.35045), ibid., 135-150 (1990; Zbl 0704.35044)]. Aspects of the theory dealing with the existence of globally smooth solutions are not discussed. Along the way some fundamental results for solutions of linear elliptic equations are also proved.
Chapter 1 contains a very clear account of the theory of Aleksandrov generalized solutions of (1). This provides a way of interpreting $$\det D^2u$$ as a Borel measure on $$\Omega$$ if $$u$$ is a convex function defined on a convex domain $$\Omega\subset{\mathbb R}^n$$. One can then study the solvability of the Dirichlet problem for (1) in the class of convex functions. A central result of Chapter 1 is Theorem 1.6.2, which states that for any bounded strictly convex domain $$\Omega\subset{\mathbb R}^n$$, any Borel measure $$\mu$$ on $$\Omega$$ with $$\mu(\Omega)<\infty$$, and any $$g\in C(\partial\Omega)$$, the equation $$\det D^2u=\mu$$ has a unique convex generalized solution $$u$$ with $$u=g$$ on $$\partial\Omega$$.
Chapter 1 also deals with the Aleksandrov-Bakelman-Pucci maximum principle. One variant of this states that if $$u\in C^2(\Omega)\cap C(\overline\Omega)$$, then $\max_{\Omega} u \leq \max_{\partial\Omega} u + \omega_n^{-1/n}\text{diam}(\Omega) \left( \int_{{\mathcal C}^*(u)} |\det D^2 u(x)|dx \right)^{1/n},$ where $$\omega_n$$ is the Lebesgue measure of the unit ball in $${\mathbb R}^n$$ and ${\mathcal C}^*(u) = \{x\in\Omega : u(y)\leq u(x) + Du(x)\cdot(y-x) \text{ for all } y\in\Omega \}$ is the upper contact set of $$u$$. This estimate is essentially an estimate for convex solutions of (1). Its variants and extensions to $$u\in C(\overline\Omega)$$ are central to the regularity theory developed later in the book, and are also important in the theory of linear elliptic equations.
The chapter concludes with a discussion of ellipsoids of least volume containing a given bounded convex set $$\Omega\subset{\mathbb R}^n$$ with nonempty interior. The existence of such an ellipsoid is proved, and furthermore, it is shown that if $$E$$ is an ellipsoid of minimum volume containing $$\Omega$$ and centred at the centre of mass of $$\Omega$$, then $$\alpha_nE \subset \Omega \subset E$$ where $$\alpha_n=n^{-3/2}$$ and $$\alpha E$$ denotes the $$\alpha$$-dilation of $$E$$ with respect to its centre. If $$T$$ is an affine transformation such that $$T(E)$$ is the unit ball in $${\mathbb R}^n$$, then the set $$T(\Omega)$$ is said to be normalized by $$T$$. Such normalizations arise naturally when studying the Monge-Ampère equation, because $$\det D^2u$$ is invariant with respect to affine transformations of determinant one. This invariance is strongly exploited throughout the book.
Chapter 2 is concerned with the derivation of the fundamental Harnack inequality of N. V. Krylov and M. V. Safonov [Izv. Akad. Nauk SSSR, Ser. Mat. 44, 161-175 (1980; Zbl 0439.35023)] for solutions of linear elliptic equations of the form $\sum_{i,j=1}^n a^{ij}D_{ij}u = f \tag{2}$ where $$a^{ij}$$ is symmetric, measurable and satisfies $\lambda|\xi|^2 \leq \sum_{i,j=1}^n a^{ij}\xi_i\xi_j \leq \Lambda|\xi|^2 \quad\text{for all}\quad \xi\in{\mathbb R}^n \tag{3}$ for some positive constants $$\lambda,\Lambda$$.
In Chapter 3 sections of convex functions are studied. A section of a convex function $$\phi$$ defined on $${\mathbb R}^n$$ is a set of the form $S_{\phi}(x_0,p,t) = \{x\in{\mathbb R}^n:\phi(x)<\ell(x)+t \}$ where $$t>0$$ and $$\ell(x)=\phi(x_0)+p\cdot(x-x_0)$$ is a supporting hyperplane to the graph of $$\phi$$ at $$(x_0,\phi(x_0))$$. It is assumed that $$\phi$$ is such that $$S_{\phi}(x_0,p,t)$$ is a bounded set for any $$x_0\in {\mathbb R}^n$$ and any $$t>0$$. Several important results are proved. First, a geometric characterization of Monge-Ampère measures $$\mu=M\phi$$ that are doubling with respect to centre of mass is proved. This means that there are constants $$C>0$$ and $$\alpha\in(0,1)$$ such that for all sections $$S_{\phi}(x,p,t)$$ we have $\mu(S_{\phi}(x,p,t)) \leq C\mu(\alpha S_{\phi}(x,p,t)),$ where $$\alpha S_{\phi}(x,p,t)$$ denotes the $$\alpha$$-dilation of $$S_{\phi}(x,p,t)$$ with respect to its centre of mass. It is shown that this is equivalent to the existence of $$\tau,\lambda\in(0,1)$$ such that for all $$x_0\in{\mathbb R}^n$$ and $$t>0$$ we have $S_{\phi}(x_0,p,\tau t) \subset \lambda S_{\phi}(x_0,p,t).$ Furthermore, it is shown that if $$M\phi$$ is doubling with respect to centre of mass, then sections of $$\phi$$ behave like Euclidean balls in certain respects. These technical facts are of fundamental importance in the development of the subsequent regularity theory in particular, in the proof of a key covering lemma of Besicovitch type in Chapter 6.
Although not discussed in the book, the results on sections are also used in the study of the linearized Monge-Ampère equation, for example, in proving an affine invariant analogue of the Krylov-Safonov Harnack inequality for solutions of such equations. In this theory, developed by L. A. Caffarelli and the author [Trans. Am. Math. Soc. 384, No. 3, 1075-1092 (1996; Zbl 0858.35034)], the coefficient matrix $$[a^{ij}]$$ in (2) is the cofactor matrix of the Hessian of a convex function $$\phi$$, and the uniform ellipticity condition (3) is replaced by $0 < \lambda \leq \det D^2\phi \leq \Lambda < \infty. \tag{4}$ Further applications of the theory of sections arise in harmonic analysis.
In Chapter 4 the author gives a proof of a classical result of Jörgens ($$n=2$$), Calabi ($$n\leq 5$$), Pogorelov and Cheng and Yau ($$n\geq 2$$), that an entire convex solution of $$\det D^2u=1$$ in $${\mathbb R}^n$$ must be a paraboloid. Here a proof is given following an argument of Caffarelli, making use of a priori estimates and normalization techniques.
The last two chapters are concerned with the regularity of generalized solutions of (1). Chapter 5 contains a proof of the fact (due to Caffarelli) that if $$u\geq 0$$ is a convex function on a bounded convex open set $$\Omega\subset{\mathbb R}^n$$ such that $0 < \lambda \leq \det D^2u \leq \Lambda < \infty \tag{5}$ in the generalized sense, and $$\Gamma=\{x\in\Omega: u(x)=0 \}$$ is nonempty, then $$\Gamma$$ has no extreme points in the interior of $$\Omega$$. It is then shown that generalized solutions $$u$$ of (5) with $$u=0$$ on $$\partial\Omega$$ are strictly convex and belong to $$C^{1,\alpha}(\Omega)$$ for some $$\alpha>0$$ depending only on $$\lambda,\Lambda$$ and $$n$$. An important point should be made here. Generalized solutions of (1) need not be even $$C^1$$ in the interior of their domain of definition if $$n\geq 3$$, even if $$f$$ is smooth and positive. One needs some control of the boundary behaviour of $$u$$ to prove interior regularity. An important technique for proving regularity of solutions of Monge-Ampère equations in many situations is to show that the solution is strictly convex; one can then easily localize to the case of affine Dirichlet data.
Chapter 6 contains an exposition of the interior regularity theory of Caffarelli for generalized solutions $$u$$ of $\det D^2u = f(x) \quad\text{in}\quad\Omega, \qquad u = 0 \quad\text{on}\quad\partial\Omega.$ The basic result is that if $$f$$ is continuous on $$\Omega$$ and bounded between two positive constants, then $$u$$ belongs to $$W^{2,p}_{\text{loc}}(\Omega)$$ for all $$p<\infty$$. The proof of this is very technical and uses much of the machinery developed in the earlier chapters.
In conclusion, the author has given a careful exposition of some of the main aspects of the theory of the Monge-Ampère equation. The book will be useful both to those seeking an introduction to the subject, and to researchers in the field of nonlinear partial differential equations.

### MSC:

 35J60 Nonlinear elliptic equations 35-02 Research exposition (monographs, survey articles) pertaining to partial differential equations 35J65 Nonlinear boundary value problems for linear elliptic equations 53A15 Affine differential geometry 52A20 Convex sets in $$n$$ dimensions (including convex hypersurfaces) 35D05 Existence of generalized solutions of PDE (MSC2000)

### Citations:

Zbl 0704.35045; Zbl 0704.35044; Zbl 0439.35023; Zbl 0858.35034


text:
# Multiplicative property of equality with decimals Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Multiplicative property of equality with decimals. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - Solve the following equation. Round the answer to the nearest tenth.

14.8 = 5.2x

### Explanation

Step 1:

Using multiplicative property of equality, we divide both sides by 5.2

$\frac{5.2x}{5.2} = \frac{14.8}{5.2}$

Step 2:

So, x = 2.84 ≈ 2.8 is the solution to nearest tenth

Q 2 - Solve the following equation. Round the answer to the nearest tenth.

$\frac{g}{2.4} = -12$

### Explanation

Step 1:

Using multiplicative property of equality, we multiply both sides by 2.4

$\frac{g}{2.4} \times 2.4 = -12 \times 2.4$

Step 2:

So, g = -28.8 is the solution to nearest tenth

Q 3 - Solve the following equation. Round the answer to the nearest tenth.

$\frac{a}{1.2} = -7.5$

### Explanation

Step 1:

Using multiplicative property of equality, we multiply both sides by 1.2

$\frac{a}{1.2} \times 1.2 = -7.5 \times 1.2$

Step 2:

So, a = -9.0 is the solution to nearest tenth

Q 4 - Solve the following equation. Round the answer to the nearest tenth.

- 18s = -19.2

### Explanation

Step 1:

Using multiplicative property of equality, we divide both sides by 18

$\frac{-18s}{18} = \frac{-19.2}{18}$

Step 2:

So, s = 1.07 ≈ 1.1 is the solution to nearest tenth

Q 5 - Solve the following equation. Round the answer to the nearest tenth.

9p = 5.6

### Explanation

Step 1:

Using multiplicative property of equality, we divide both sides by 9

$\frac{9p}{9} = \frac{5.6}{9}$

Step 2:

So, p = 0.62 ≈ 0.6 is the solution to nearest tenth

Q 6 - Solve the following equation. Round the answer to the nearest tenth.

$-0.6 = \frac{n}{14}$

### Explanation

Step 1:

Using multiplicative property of equality, we multiply both sides by 14

$-0.6 \times 14 = \frac{n}{14} \times 14$

Step 2:

So, n = -8.4 is the solution to nearest tenth

Q 7 - Solve the following equation. Round the answer to the nearest tenth.

4.8x = 12.7

### Explanation

Step 1:

Using multiplicative property of equality, we divide both sides by 4.8

$\frac{4.8x}{4.8} = \frac{12.7}{4.8}$

Step 2:

So, x = 2.65 ≈ 2.7 is the solution to nearest tenth

Q 8 - Solve the following equation. Round the answer to the nearest tenth.

$\frac{j}{1.8} = -15$

### Explanation

Step 1:

Using multiplicative property of equality, we multiply both sides by 1.8

$\frac{j}{1.8} \times 1.8 = -15 \times 1.8$

Step 2:

So, j = -27.0 is the solution to nearest tenth

Q 9 - Solve the following equation. Round the answer to the nearest tenth.

3.3 = 2.2y

### Explanation

Step 1:

Using multiplicative property of equality, we divide both sides by 2.2

$\frac{3.3}{2.2} = \frac{2.2y}{2.2}$

Step 2:

So, y = 1.5 is the solution to nearest tenth

Q 10 - Solve the following equation. Round the answer to the nearest tenth.

$-0.9 = \frac{m}{11}$

### Explanation

Step 1:

Using multiplicative property of equality, we multiply both sides by 2.4

$-0.9 \times 11 = \frac{m}{11} \times 11$

Step 2:

So, m = -9.9 is the solution to nearest tenth

multiplicative_property_of_equality_with_decimals.htm


text:
#### Quantum-inspired evolutionary algorithm = 양자 개념을 도입한 진화 알고리즘

Cited 0 time in Cited 0 time in
• Hit : 744
This thesis proposes a novel evolutionary algorithm inspired by quantum computing, called a quantum-inspired evolutionary algorithm (QEA), which is based on the concept and principles of quantum computing, such as a quantum bit and superposition of states. Like other evolutionary algorithms, QEA is also characterized by the representation of the individual, the evaluation function, and the population dynamics. However, instead of binary, numeric, or symbolic representation, QEA uses a Q-bit, defined as the smallest unit of information, for the probabilistic representation and a Q-bit individual as a string of Q-bits. A Q-gate is introduced as a variation operator that drives the individuals toward better solutions. The termination condition of QEA is designed by defining a new measure on the convergence of Q-bit individuals. To analyze the characteristics of QEA, the theoretical analysis of the QEA algorithm as well as the effects of changing parameters of QEA are examined. In particular, some issues of QEA such as the analysis of changing the initial values of Q-bits, a novel variation operator $H_\epsilon$ gate, and a two-phase QEA (TPQEA) scheme are addressed to improve the performance of QEA. To demonstrate the effectiveness and applicability of QEA, experiments are carried out on the knapsack problem, which is a well-known combinatorial optimization problem. The results show that QEA performs well, even with a small number of population, without premature convergence as compared to the conventional genetic algorithms. Moreover, through the experiments on numerical optimization problems, the superior performance of QEA is also verified. These results show that QEA can be applied to a class of numerical as well as combinatorial optimization problems.
Kim, Jong-Hwanresearcher김종환researcher
Description
한국과학기술원 : 전기및전자공학전공,
Publisher
한국과학기술원
Issue Date
2003
Identifier
231142/325007  / 000995393
Language
eng
Description

학위논문(박사) - 한국과학기술원 : 전기및전자공학전공, 2003.8, [ viii, 131 p. ]

Keywords

optimization; probabilistic representation; evolutionary algorithm; quantum-inspired computing; evolutionary computation; 조합 최적화; 수치 최적화; 확률 표현법; 진화 연산; 양자 진화 알고리즘

URI
http://hdl.handle.net/10203/35191


text:
# Homework Help: Conservation of Momentum

1. Jan 18, 2012

### vaironl

Hello form, Vaironl here.

I have some few very basic questions. Why might you ask? I came from a trip about a week ago and in my physics class we are discussing momentum, and elasticity.
I asked my teacher for help but he really is a bit busy at the moment, and the semester/quarter will soon be done.

I have this problem explained better on a image see it below.
http://img703.imageshack.us/img703/9095/problemhm.jpg [Broken]

I tried to solve this, in the following order:
Find the total momentum before : 1kg * 2m/s + 2kg * -2m/s = 2kgm/s + -4kgm/s = -2kgm/s
Find total momentum after (This is were I get stuck): 1kg * -1m/s + 2kg * VEL = -2kgm/s + 2kg

I really don't know what to do, sorry to bother you guys with such basic questions

Last edited by a moderator: May 5, 2017
2. Jan 18, 2012

### bacon

The question is not a bother, it's the purpose of the forum.
Your analysis of the before collision momentum looks very good. The after part is almost correct. The terms to the left of the equals sign are good and on the right the first term is as well. There is a "+ 2kg", is that a typo?

3. Jan 18, 2012

### vaironl

No sorry for that I thought that I should have left it blank indicating I din't know the speed, but I see I created greater confusion.

4. Jan 18, 2012

### 15tungalbert

I think the best way to approach the second problem is to first find out the velocity.

Remember that conservation of momentum

m1v1(initial) + m2(v2)initial = (m1) v1final + m2v2final

You know every single variable except for v2final. Solve for that.

5. Jan 18, 2012

### bacon

That's ok.
Remember that since momentum is conserved, the momentum before the collision is the same as that after the collision. You have correctly calculated the momentum before the collision, -2kgm/s, so that's what the total after collision momentum must be as well. Your unknown speed is the VEL in your second term. Can you solve for that?

6. Jan 18, 2012

### vaironl

I believe now I got it I was confused because I was trying to use two separte variables (1kg*-1m/s)+(2kg * VEL) but I notice that I can just add the kg and find the vel right?

(1kg+2kg)(VEL)
-2kgm/s = 2kg * Vel

Vel = $\frac{-2kgm/s}{2kg}$
= -1m/s

7. Jan 18, 2012

### bacon

The velocities of the two masses are not necessarily the same(most of the time they are not). Look at the equation 15tungalbert posted, the before collision momentum(left side) is equal to the after collision momentum(right side).
You are given the velocity of the 1kg mass post collision. Your only unknown is the velocity of the 2kg mass, post collision.


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# Why is the velocity on the top of a wheel twice the velocity of its axle?

When a wheel is rolling, not skidding, and its axle moves at velocity $\vec{v}$, then a point on the top of its circumference will move at velocity $2\vec{v}$, shown below.

I really don't understand this. I'm quite familiar with the geometry of a circle, but I don't understand how it's being applied on this case. Also:

1. Why doesn't the relation between the velocities depend on the radius of the wheel?
2. When is this relation valid? Does it take friction into account or it doesn't matter? If it doesn't matter, why?
3. A consequence of all this is that the point in the bottom has no velocity. Why? That makes no sense to me.

This is a very confusing topic to me. Here is a very nice page full of pictures and animations about the physics of a wheel but I still don't get it.

-
The really simple way to get #3 is to start from a frame of reference tied to the axle. In this frame, the bottom of the wheel has velocity -v, while the top of the wheel has velocity v. To convert back into the ground's frame, add v to each of these, so they become 0 and 2v. –  Ben Crowell Sep 10 '13 at 4:36

I'll tackle your questions in reverse:

3. The contact point is stationary because the wheel is not slipping. This happens when the force of static friction is able to counter the force of the wheel on the ground. This is what you want for controllable transport. If the wheel starts slipping (because of low friction) that's a skid and you are no longer able to steer or brake. If you like, imagine getting your car stuck in mud. You spin the wheel and fling mud all over the place without getting anywhere - not enough friction. If that doesn't help try taking a wheel, marking a spot on it, and slowly rolling it while carefully watching the point of contact.

2. You need sufficient static friction to enforce the no-slip condition. The relation between the velocity at the top, centre and bottom of the wheel is geometrical and is not affected by friction per-se. If the car wheel spins with angular frequency $\omega$, has a radius $R$ and velocity at the axle of $v$ then the velocity of the wheel at the top and bottom is $$v_{top} = v + \omega R, v_{bottom} = v - \omega R$$

1. If the no slip condition holds then $v_{bottom}=0$, so $v = \omega R$. Using this in the top equation gives $v_{top} = v + v = 2v$, independent of $R$. This is because both $v$ and $v_{top}$ increase in the same proportion as $R$ increases.

(Aside: Is there a way to make markdown do reverse lists?)

-
There isn't any way to do reverse lists with Markdown. I'd use bullets, or you can put a backslash in front of the dot (1\.) to prevent it from automatically formatting. –  David Z Jan 3 '13 at 11:32
Thanks @DavidZaslavsky. Fixed it. –  Michael Brown Jan 3 '13 at 11:37
@MichaelBrown "The relation between the velocity at the top, centre and bottom of the wheel is geometrical". Could you please elaborate a little bit more on the geometry? I know that an inscribed angle in a circumference determines an arc measuring twice its length, for example --and I fully comprehend the triangular similarity involved in that. Does that have anything to do with the wheel? I don't understand how. –  BeetleTheNeato Jan 3 '13 at 20:35
@MichaelBrown I'm also struggling to understand that "The contact point is stationary because the wheel is not slipping.". Isn't the contact point ever changing? If I mark a point on the bottom of the wheel with a chalk, it'll move upwards as the wheel rolls, right? So how is the contact point stationary? Or does that mean that at every moment the one contact point at that time is stationary? –  BeetleTheNeato Jan 3 '13 at 20:41
Notice that at the bottom of the wheel, $\theta=\pi$, $v_h = v_v = 0$. The contact point is stationary. I arranged that by adding just the right amount to the horizontal velocity. As a consequence, at the top of the wheel $v_h = 2v$. For any given piece of the wheel these equations are instantaneous: the velocity changes as it moves around the wheel. This means the material in the wheel is accelerating, which means there must be a force towards the centre (centripetal force), which means there must be spokes or something connecting it to the axle or the wheel must be sufficiently strong. :) –  Michael Brown Jan 4 '13 at 1:57


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# decimate

Version:

Uses a type 1 Chebyshev filter to resample a signal.

## Syntax

c = decimate(a, b)
c = decimate(a, b, n)
c = decimate(a, b, option)
c = decimate(a, b, n, option)

## a

Original signal. a is a real vector.

## b

New sampling rate. b is a positive integer.

## n

Order of the type 1 Chebyshev filter to use. n is a positive integer.

## option

Filter type. option is a string that accepts the following values:

Name Description
'FIR'

Directs MathScript to use an FIR filter to perform the resampling.

'IIR'

Directs MathScript to use an IIR filter to perform the resampling.

## c

Resampled version of a at 1/b times the original rate. The length of c is 1/b times the length of a. c is a real vector.

A = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7];
B = 2;
N = 3;
C = decimate(A, B, N)

Where This Node Can Run:

Desktop OS: Windows

FPGA: This product does not support FPGA devices


text:
# Programmatically creating sprite animations in Unity

Some background: I have a set of animated gifs which I've converted to spritesheets, which are then imported into Unity as multiple sprites. The end result is that each spritesheet in the Unity editor contains a list of sprites named xxxxxxx_yyy_zzz -- for example:

1313000_000_000 to 1313000_000_014, followed by
1313000_001_000 to 1313000_001_008, followed by
1313001_000_000 to 1313001_000_031,
and so on.

Now, I would like to make simple sprite animations out of this spritesheet. In the example above, 1313000.anim should contain all sprites beginning with 1313000 played in numerical order, 1313001.anim should contain all sprites beginning with 1313001, etc. etc.

This isn't hard to do by hand -- highlight desired range of sprites, drag and drop -- but we're talking about a lot of animations. Creating an editor script which lets me right-click a spritesheet and generate a series of .anim files seemed the best solution, but I'm turning up conflicting info on the subject. Getting a sprite array from the spritesheet is simple enough, but programmatically creating an animation is something very few people seem to have discussed.

Essentially, is it possible and practical to create and populate sprite animations from an editor script?

I have a similar situation where I have numbers that animate on and off for a number puzzle game. I have a grid sprite sheet with each of these cells evenly spaced. I give each of these cells a number from 0 to X. From a cell number you can get a row and column like this:

row = cellNumber / cellsWide;
col = cellNumber % cellsWide;

Then I have an XML file that contains the cell numbers for each animation and other information about the sprite sheet. Like the width of a cell in pixels and the total width of the sheet. I load this info from the XML and store it in arrays for each animation.

The instead of using the built in sprite type in Unity, I build the sprite as a mesh that I create in code. Then all I have to do is alter the UV for this mesh to animate what cell is showing on the "sprite".

Actually in my case, I build all the sprites that are on the screen at any given time in a single mesh. This is a much more efficient way of rendering because all the sprites are rendered in a single draw call. Sort of like particles. That also means that my transform for the mesh is 0,0,0. I place each quad where it needs to go on screen.

The UV is calculated from the row and column, the width of the sprite cell, and the total width of the sprite sheet. The way I do it is to use one class to read the XML and create a copy of the unscaled, un-transformed quad. Then another class to get those quads and scale and move them to where they go as I make the full mesh. Then update that mesh each frame if things are moving.

So YES, it is completely possible and practice to create and populate sprite-TYPE animations using a script. But you have to decide which is more work. For me, I have the code down for creating meshes, reading in from XML, etc. And I have other needs that shy me away from the built in sprite system in Unity.

Alternatively you can just create sprite arrays and manually stick in each of the animation cells in there and access those programatically. But sound like that is what you are wanting to avoid.

With my above system you'll still have some manual inputting of animation cells, but it's sticking them in an XML document. Not sure which one will take less time.

There are a few options for replacements to the sprite system in Unity. There is one called Sprite Manager which is free and open source: https://github.com/Dkendal/SpriteManager.

There is also Sprite Manager 2 which is \$150: http://www.anbsoft.com/middleware/sm2/

Or you can easily write you own like I'm doing.


text:
Preprint Article Version 1 Preserved in Portico This version is not peer-reviewed

# On the Eternal Role of Planck Scale in Machian Cosmology

Version 1 : Received: 17 October 2018 / Approved: 18 October 2018 / Online: 18 October 2018 (08:42:57 CEST)
Version 2 : Received: 25 October 2018 / Approved: 25 October 2018 / Online: 25 October 2018 (09:51:27 CEST)
Version 3 : Received: 30 October 2018 / Approved: 2 November 2018 / Online: 2 November 2018 (02:30:57 CET)

How to cite: Seshavatharam, U.V.S.; Lakshminarayana, S. On the Eternal Role of Planck Scale in Machian Cosmology. Preprints 2018, 2018100407 (doi: 10.20944/preprints201810.0407.v1). Seshavatharam, U.V.S.; Lakshminarayana, S. On the Eternal Role of Planck Scale in Machian Cosmology. Preprints 2018, 2018100407 (doi: 10.20944/preprints201810.0407.v1).

## Abstract

Considering Planck mass as a baby universe and evolving universe as a growing Planck ball, a hypothetical spherical model of cosmology can be developed. In addition, by considering the famous relation GM = c2R, a Machian model of quantum cosmology can also be developed. Proceeding further, in all directions, cosmic expansion velocity can be referred to ‘from and about’ the baby universe. This hypothetical scenario can be made versatile by adding a heuristic idea: apart from normal expansion, ‘cosmic temperature seems to be redshifted by a factor $\left({\left({Z}_{T}\right)}_{t}+1\right)\cong \sqrt{1+\mathrm{ln}\left({H}_{pl}/{H}_{t}\right)}$; where (Hpl, Ht) represent Planck scale and time dependent Hubble parameters respectively.

## Subject Areas

Planck scale; Mach’s relation; quantum cosmology; critical density; ordinary matter; dark matter; thermal redshift; expansion velocity; Hubble’s law

Views 0


text:
# Bank Exams :: Quantitative Aptitude :: Mensuration IBPS Recruitment Latest Govt Jobs

## Home Bank Exams / Quantitative Aptitude Mensuration Questions and Answers

1 .
The circumference of two circles is 88 m and 220 m respectively. What is the difference between the area of the larger circle and the smaller circle ?
3422 sq m
3242 sq m
3244 sq m
3424 sq m
None of these
2 .
The length of a rectangle is twice the diameter of a circle. The circumference of the circle is equal to the area of a square of side 22 cm. What is the breadth of the rectangle if its perimeter is 668 cm ?
24 cm
26 cm
52 cm
Cannot be determined
None of these
3 .
The area of a square is 196 sq. cm. whose side is half the radius of a circle. The circumference of the circle is equal to breadth of a rectangle, if perimeter of the rectangle is 712 cm. What is the length of the rectangle ?
196 cm
186 cm
180 cm
190 cm
None of these
4 .
A tank 30 m long, 20 m wide and 12 m deep is dug in a field 500 m long and 30 m wide. By how much will the level of the field rise, if the earth dug out of the tank is evenly spread over the field?
0.33 m
0.5 m
0.25 m
0.4 m
None of these
5 .
The circumference of two circles is 132 m, and 176 m respectively. What is the difference between the area of the larger circle and the smaller circle ?
1048 sq m
1076 sq m
1078 sq m
1090 sq m
None of these
6 .
If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. The perimeter of the original rectangle (in centimetres) is :
44
46
48
50
None of these
7 .
What will be the area (in square metre) of 1.5 m wide garden developed around all the four sides of a rectangular field having area equal to 300 sq. m and breadth equal to threefourth of the length ?
96
105
114
Cannot be determined
None of these
8 .
A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end, and the other end as its base. The volume of the cylinder, hemisphere and the cone are, respectively in the ratio :
2 : 3 : 2
3 : 2 : 1
3 : 1 : 2
1 : 2 : 3
None of these
9 .
Three cubes of a metal are of edges 3 cm, 4 cm and 5 cm. These are melted together and from the melted material, another cube is formed. The edge of this cube is :
8 cm
10 cm
9 cm
6 cm
None of these
10 .
A solid cylinder and a solid cone have equal base and equal height. If the radius and the height be in the ratio of 4 : 3, the total surface area of the cylinder to that of the cone are in the ratio of :
10 : 9
11 : 9
12 : 9
14 : 9
None of these
11 .
The length, breadth and height of a cuboid are in the ratio 1 : 2 : 3. The length, breadth and height of the cuboid are increased by 100%, 200% and 200%, respectively. Then, the increase in the volume of the cuboid will be :
5 times
6 times
12 times
17 times
None of these
12 .
A sphere of radius 3 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 6 cm. If the sphere is submerged completely, then the surface of the water is raised by :
$1 \over 4$ cm
$1 \over 2$ cm
1 cm
2 cm
None of these
13 .
Four equal circles are described about the four corners of a square so that each touches two of the others. If a side of the square is 14 cm, then the area enclosed between the circumferences of the circles is :
24 $cm^2$
42$cm^2$
154$cm^2$
196$cm^2$
None of these
14 .
The largest triangle is inscribed in a semi-circle of radius 14 cm. Find the area inside the semi-circle which is not occupied by the triangle.
100$cm^2$
112$cm^2$
115$cm^2$
113$cm^2$
None of these
15 .
In a quadrilateral, the length of one of its diagonal is 23 cm and the perpendiculars drawn on this diagonal from other two vertices measure 17 cm and 7 cm respectively. Find the area of the quadrilateral.
250$cm^2$
276$cm^2$
300$cm^2$
325$cm^2$
None of these
16 .
A cube of maximum volume (each corner touching the surface from inside) is cut from a sphere. Find the ratio of the volumes of the cube and the sphere.
2 : $\sqrt {3}$p
3 : 2P
5 : 4P
3 : $\sqrt {2}$p
None of these
17 .
A large cube is formed from the material obtained by melting three smaller cubes of dimensions 3, 4 and 5 cm. What is the ratio of the total surface areas of the smaller cubes and the large cube ?
2 : 1
3 : 2
27 : 20
25 : 18
None of these
18 .
A circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 6 : 5. Find the smaller side of the rectangle.
50 cm
60 cm
70 cm
80 cm
None of these
19 .
A right circular cone is exactly fitted inside a cube in such a way that the edges of the base of the cone are touching the edges of one of the faces of the cube and the vertex is on the opposite face of the cube. If the volume of the cube is 343 cc, what, approximately, is the volume of the cone?
80 cc
90 cc
110 cc
105 cc
None of these
20 .
Find the number of coins, 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm :
430
440
450
460
None of these
21 .
A rectangular lawn 80 metres by 60 metres has two roads each 10 metres wide running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling them at ` 30 per square metre :
Rs.39000
Rs.3900
Rs.3600
Rs.36000
None of these
22 .
A hemispherical bowl is filled to the brim with a beverage. The contents of the bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both bowl and cylinder, then the volume of the beverage in the cylindrical vessel will be
66$2 \over 3$%
78.5%
100%
More than 100% (that is, some liquid will still be left in the bowl)
None of these
23 .
If the angle of triangle are in the ratio of 4 : 3 : 2, then the traingle
is obtuse angled triangle
has one angle greater than 80 °
is a right triangle
is acute angled triangle
None of these
24 .
The length of a room is double its breadth. The cost of colouring the ceiling at Rs 25 per sq. m is Rs 5,000 and the cost of painting the four walls at Rs 240 per sq. m is Rs 64,800. Find the height of the room.
4.5 m
4 m
3.5 m
5 m
None of these
25 .
A rectangular piece of iron sheet measuring 50 cm by 100 cm is rolled into cylinder of height 50 cm. If the cost of painting the cylinder is Rs. 50 per square metre, then what will be the cost of painting the outer surface of the cylinder?
Rs.25.00
Rs.37.50
Rs.75.00
Rs.87.50
None of these
26 .
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, then find the edge of the third smaller cube.
10 cm
14 cm
12 cm
16 cm
None of these
27 .
Four horses are tied on the four corners of a square field of 14 m length so that each horse can just touch the other two horses. They were able to graze in the area accessible to them for 11 days. For how many days is the ungrazed area sufficient for them ?
3 days
4 days
5 days
2 days
None of these
28 .
The radius of a cylindrical cistern is 10 metres and its height is 15 metres. Initially the cistern is empty, we start filling the cistern with water through a pipe whose diameter is 50 cm, water is coming out of the pipe with a velocity of 5 m/s. How many minutes will it take in filling the cistern with water?
20
40
60
80
None of these
29 .
A garden is 24 m long and 14 m wide. There is a path 1 m wide outside the garden along its sides. If the path is to be constructed with square marble tiles 20 cm $\times$ 20 cm, the number of tiles required to cover the path is
1800
200
2000
2150
None of these
30 .
An order was placed for supply of carpet of breadth 3 metres. The length of carpet was 1.44 times of breadth. Subsequently the breadth and length were increased by 25 and 40 percent respectively. At the rate of Rs. 45 per square metre, what would be the increase in the cost of the carpet?
1020.6
398.8
437.4
583.2
None of these


text:
# Is the Gateway Arch a Parabola?

By Murray Bourne, 06 May 2010

The Gateway Arch in St Loius, Mo., USA, is part of the Jefferson National Expansion Memorial.

The shape of the arch certainly looks like a parabola, but is it?

The Gateway Arch, St Loius, USA [Image source]

We'll model the arch using a parabola and see how well it fits.

I drew a set of x-y axes and a regular grid on top of the photograph.

Next, I determined the values of y for x ranging from -3 to 3. We can see the resulting points (in magenta color) in the next image.

So my aim is to find the parabola that passes through the Origin (0, 0) and the points (-3, -4) and (3, -4). (I choose these for convenience).

I know the equation for my parabola will have a negative in front, since it is inverted (upside-down).

I can find the equation for this parabola in several different ways. One is to recall the general form for a parabola:

y = ax2 + bx + c

We have 3 points and 3 unknowns, so we just substitute them in:

Substituting (0,0) gives:

0 = 0 + 0 + c

So c = 0.

Next, substitute (−3, −4) and (3, −4) to give 2 equations in 2 unknowns:

−4 = 9a − 3b (1)

−4 = 9a + 3b (2)

−8 = 18a

So a = −4/9

Substituting this back into either (1) or (2) gives:

b = 0

So the parabola we are looking for is:

Here is the graph of that parabola.

It looks quite good. It passes through the Origin and the points (-3, -4) and (3, -4). But how does it match up with the Gateway Arch?

Here is the parabola next to the points we found above (when x = -3, -2, -1, 0, 1, 2, 3):

Our parabola has missed going through the other points (the ones where x = -2, -1, 1 and 2)!

Comparing our parabola to the photo, we can see (especially near the waterline at the bottom) the shape of the arch is not really a parabola at all.

## Another Curve Needed

The Finnish-American designer of the Gateway Arch, Eero Saarinen, knew that a parabola was not the best shape for such an arch.

Arches have been used throughout history for bridge and roof supports, since they are good at directing forces downwards, rather than outwards, thus reducing the chance of collapse.

An arch shape that is often used is the catenary.

A catenary is the shape made by a chain that is freely hanging between 2 supports.

Chain hanging on a log. [Image source]

Paper chains [Image source]

## Catenaries

A caternary is an interesting curve. It is the average of the y-values of an exponentially decreasing curve and an exponentially increasing curve.

A simple example is:

The first part of the numerator, e−x, decreases exponentially as x increases, while the second part, ex, increases exponentially as x increases. (See more on exponential curves.).

Here are the two curves on the same set of axes.

We add the y-values of the 2 exponential curves together and divide by 2 (in other words, we find the average of the y-values) to give us our simple catenary (in black) .

Let's now compare the shape of a catenary (in black) to the shape of a parabola (in magenta).

We see they are not quite the same, and perhaps we are getting somewhere.

## [Aside] Cosh

This is not important for this discussion but there is another way of writing the equation we've just been discussing.

This is an example of a hyperbolic function and you may meet the notation cosh later. It is called "cosh" because in some ways, it behaves similarly to the "cosine" function from trigonometry. The "h" in "cosh" comes from "hyperbolic".

We can write:

See more on hyperbolic functions.

Back to the topic.

## Flattened Catenaries

Now, a catenary is the shape we see when there is a chain of constant thickness hanging between 2 fixed points. (The word "catenary" comes from the same word as "chain".)

But if the chain is thinner in the middle, it has a slightly different shape (it is flatter). Since the Gateway Arch is thinner at the top than it is near the base, the architect chose a flattened catenary, whose equation has the form:

Now, in our example there will be a negative in front of our equation (since the catenary is inverted) and we also will need to add A to our equation so the curve passes through (0,0) (otherwise it will pass through (0, -A)).

So we are looking for an equation of the following form, that passes through (−3, −4) and (−2, −1.42):

Using a computer algebra system, we solve the above system of equations and obtain the following equation.

Here is the plot of that curve superimposed on the Gateway Arch (in light green). The parabola we found earlier (in magenta) is there for comparison. We can see our catenary follows the shape of the bridge quite closely, especially at the bottom.

## Conclusion

This article has shown the Gateway Arch is not a parabola. Rather, it is in the shape of a flattened catenary, which is the shape we see if we hang a chain that is thin in the middle between two fixed points.

We have also seen how to go about modeling curves to find the equation representing such curves.

### 46 Comments on “Is the Gateway Arch a Parabola?”

1. Jessy George says:

This is really intersting am going to share with my students ,as they can adapt this method to check

Thank you Murray

2. Farzin says:

Very interesting. I always thought it was a parabolla.

Thanks

3. Rina King says:

Thanks a ton for this Rika! It is a great example to show learners why in science we need to show a directly proportional relationship,ie resolve it into a straight ine - any other relationship cannot be assumed.

4. Anah says:

We just finished a project in calculus II. The subject was to find the best model for the cumulative new cases of AIDS and the epidemic curve in the US. I wonder if there is an ideal model taking into account different factors such as geographic, bio...Ifsomeone has an idea, I will be interested.

5. Murray says:

Hi Anah. This earlier article may help you: H1N1 and the Logistic Equation.

6. payal says:

Thanks for the information.I have learned lot many new things from this article.I would be happy if you send me more such information.
Thanks again

7. rishabh says:

really very interesting!!!!!!!!!!!!!!!!!

8. abramovich says:

interesting indeed

9. harold2 says:

I really appreciated it!
Congratulations for this excellent post!

Thank you. I have learned a lot. I thought it was a parabola...nice..

11. Nikki says:

Thank you. I was recently reading a book that described my beloved Arch as a "Half-parabola." That can't be right, I thought, I don't believe the Arch is a parabola at all, and it certainly isn't a half of one. You have confirmed my suspicions. Thank you.

12. ?mega£ither says:

D: great, now I have to tell my awesome 7th grade algebra honors teacher she's wrong (advanced class)

great explanation . . . i am expecting much of such these facts instead of boring old concepts. add more and more applications with as much less theory as possible... thank you.

14. Nick says:

Shouldn't you be comparing various curve types against an architectural drawing of the arch, as your photograph will have perspective distortion.

15. Murray says:

@Nick: You're right - there will be some parallax error for the photograph.

Here is the original architectural drawing of the Gateway Arch:

[Source: Library of Congress]

Here's a catenary placed over that drawing:

I drew the catenary using GeoGebra. It turned out to be:

For comparison, here is a parabola through the extremities of the arch.

So while there may have been some parallax issues in my earlier analysis, I believe it was insignificant, especially since the photo I chose had the arch directly facing the camera, and after all, a "squashed" catenary is still (almost) a catenary, and a "squashed" parabola is (almost) still a parabola, as long as the distortion is along the axis of the curve.

16. Francesca says:

What program did you use to place the plane of coordinates in front of the picture?

17. Murray says:

@Francesca: I'm using GeoGebra (and/or Scientific Notebook) to create the coordinate plane image, then I place a transparent screen shot from there over the image using PaintShop Pro, which is similar to Photoshop).

18. Francesca says:

Thanks a lot!

19. Francesca says:

What is the computer algebra system that you used to solve the simultaneous equations for the hyperbolic case?

20. Murray says:

@Francesca: I used Scientific Notebook to solve that system. You could probably use Wolfram|Alpha to get the solution as well.

BTW, it was a "hyperbolic cosine" (or "catenary") case, which is different from a "hyperbola".

21. Francesca says:

What further calculations can you do once you have the function for the arch?

22. Murray says:

@Francesca: Probably the main reason for finding such a function would be to find height of the arch at various points along the ground underneath it.

One practical example could be if you needed to erect scaffolding under one section (for maintenance, or whatever), then how high would it need to be at various points.

23. Husna says:

24. Murray says:

@Husna: If your question is about parabolas or catenaries, it's probably best to ask here, then everyone can benefit.

If not, perhaps one of the categories in the IntMath Forum would be appropriate for your question. Once again, other readers would benefit from the discussion.

25. Me says:

@Murray:I tried to proof the shape of the arch by using the same way as yours, using my own picture. Huhu But I failed to draw the parabolic shape. The size of the parabola is smaller than the arch, I mean very small..may I know, why is this happening

26. Murray says:

Hello again, Husna. What are you using to draw the parabola? What image editing software are you using?

27. Me says:

@murray Oh so I have to use image editing software?sorry didnt know about it😓😓😓im just using geogebra..so sorry

28. Star says:

Hey murray.how could we derive to the flattened catenary formula manually without using software?tq

29. Murray says:

@Star: I don't think there's a "nice" algebraic way to solve it. I believe you would need to do it numerically.

30. Star says:

@Murray: Would you mind to show it how

31. Murray says:

Well, the 2 equations in 2 unknowns from the known values (-3, -4) and (-2, -1.42) are:

Multiplying both through by 2 gives:

Subtract [1] from [2]:

You still have 2 unknowns, but could assume (for now) and solve

Using just a few steps of Newton's Method gives you close to the I had before.

Then you would need to do some juggling (trial and error?) with A.

Much easier and more logical to get a computer to do it.

32. Star says:

tq for showing..just wondering why you choose values (-3,-4) and (-2,-1.42)..could it be other point as well?

33. April says:

Im interested in finding height of the arch at various points along the ground underneath the arch as u I mentioned. But im not sure how can I do it

34. Murray says:

@April: You just need to substitute in values of x) which are distances along the ground) to get the height h at that point.

For example, at we'll be at

This means we are units below the horizontal passing through the top of the arch. Assuming the arch is units, high, the height there is

35. Murray says:

@Star: Yes, I could have chosen any other points around the catenary. They just seemed convenient at the time.

36. Jan says:

Great article.

Constructions without side forces are intrisic strong.

37. The STEM Around Us | Casio Education Blog says:

[…] and trying to mathematically represent - is it a parabola? (In fact, it is NOT a parabola, but rather a flattened catenary). (Cool 3D mathematical model […]

38. Parabola’s in real life – Algebra 2017 says:
39. Flyover Comedy Festival: Angela Smith - St. Louis Independent Comedy says:

[…] as it is wide, you know! Did you know that? That the Arch is as TALL as it is WIDE?? It’s a parabola.”  The Arch should be paying me for this […]

40. Eric says:

Please Guys i will like to ask help on parabola dish Measurement i want a 40 feet dish Guys can you help me with the measurement ? i want to know the Depth of 40 feet dish

Diameter = 40 feet
Depth =
Focal length =

41. Murray says:

@Eric: This would be a good question for the IntMath Forum. Here's the appropriate section: Plane Analytic Geometry forum.

42. Lili says:

Hi! I wanted to know that if the Gateway Arch was a parabola, what would the equation be? For me, the roots came to be (4,0) and (-4,0). The y-intercept was (0,5). My vertex form didn't really work, but would standard form, or factored form work better?
Thank you!!

43. Murray says:

@Lili: Please see the early part of the article where I show how to find the equation of the parabola.

44. Danish says:

Hi murray,

on comment 31, you say
"Well, the 2 equations in 2 unknowns from the known values (-3, -4) and (-2, -1.42) are:"

My question is , how did you get the point (-2, -1.42)?

45. Murray says:

@Danish - The points chosen for the analysis came from overlaying a grid over the image of the arch, and reading it from the grid. The grid I used had a lot more precision than the one I used for the image in the article!

46. The Best Activities for Analyzing Quadratic Equations – ChalkDoc | Math Activities and Assessments says:

[…] P.S. Yes, the header image is of the Gateway Arch in St. Louis.  No, it’s not really a parabola.  It’s admittedly more complicated than that… […]

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