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# Inverters

 Question 1
The output voltage of a single-phase full bridge voltage source inverter is controlled by unipolar PWM with one pulse per half cycle. For the fundamental rms component of output voltageto be 75% of DC voltage, the required pulse width in degrees (round off up to one decimal place) is _________
 A 82.6 B 98.6 C 112.8 D 168.2
GATE EE 2019   Power Electronics
Question 1 Explanation:
\begin{aligned} V_{01,rms}&=\frac{2\sqrt{2}}{\pi}\\ 0.9 \sin d \; V_s&=0.75V_s\\ d&=56.44^{\circ}\\ 2d&=112.88^{\circ}\\ \text{Pulse width }&=112.88^{\circ} \end{aligned}
 Question 2
The figure below shows a half-bridge voltage source inverter supplying an RL-load with $R=40\Omega$ and $L=(\frac{0.3}{\pi })H$. The desired fundamental frequency of the load voltage is 50 Hz. The switch control signals of the converter are generated using sinusoidal pulse width modulation with modulation index. M = 0.6. At 50 Hz, the RL-load draws an active power of 1.44 kW. The value of DC source voltage $V_{DC}$ in volts is
 A 300$\sqrt{2}$ B 500 C 500$\sqrt{2}$ D 1000$\sqrt{2}$
GATE EE 2017-SET-2   Power Electronics
Question 2 Explanation:
\begin{aligned} V_{01 peak}&=M\cdot \frac{V_s}{2} \;\;\;M\rightarrow \text{Modulation index}\\ V_{01peak}&=\frac{M\cdot \frac{V_s}{2}}{\sqrt{2}}=M\cdot \frac{V_s}{2\sqrt{2}}\\ V_{01peak}&=\frac{0.6V_s}{2\sqrt{2}} \end{aligned}

\begin{aligned} V_{01}&=\frac{0.3}{\sqrt{2}}V_s\\ |Z_1|&=\sqrt{R^2+(\omega L)^2}\\ &=\sqrt{40^2+(2 \pi f L)^2}\\ &=\sqrt{40^2+\left (2 \pi \times 50 \times \frac{0.3}{\pi} \right )^2}\\ &=\sqrt{40^2+30^2}=50\\ \phi _1&=\tan ^{-1}\frac{\omega L}{R}=\tan ^{-1}\frac{30}{40}=36.896^{\circ}\\ \text{Active power} &=V_{01}I_{01} \cos \phi \\ &=V_{01}\frac{V_{01}}{|Z_1|} \cos \phi \\ 1.44 \times 10^3&=\left (\frac{0.3}{\sqrt{2}}V_s \right )^2\cdot \frac{1}{50}(\cos 36.896^{\circ} )\\ \therefore \;\;V_s^2&=\frac{1.44 \times 10^3 \times 100}{0.3^2 (0.8)}\\ &=20 \times 10^5\\ V_s&=\sqrt{2} \cdot 10^3=1000\sqrt{2}\\ V_{DC}&=\frac{V_s}{2}=500\sqrt{2}V \end{aligned}
 Question 3
A three-phase voltage source inverter with ideal devices operating in 180$^{\circ}$ conduction mode is feeding a balanced star-connected resistive load. The DC voltage input is $V_{dc}$. The peak of the fundamental component of the phase voltage is
 A $\frac{V_{dc}}{\pi}$ B $\frac{2V_{dc}}{\pi}$ C $\frac{3V_{dc}}{\pi}$ D $\frac{4V_{dc}}{\pi}$
GATE EE 2017-SET-2   Power Electronics
Question 3 Explanation:
$3-\phi \;\text{ VSI }180^{\circ} \text{ mode:}$

\begin{aligned} V_{Rn}&=\frac{6 V_{dc}/3}{n \pi} \sin n\omega t\\ &= \frac{2 \times V_{dc}}{n \pi} \sin n\omega t \\ V_{Rn} &= \frac{2 V_{dc}}{ \pi} \sin \omega t \end{aligned}
 Question 4
In the converter circuit shown below, the switches are controlled such that the load voltage $v_o(t)$ is a 400 Hz square wave. The RMS value of the fundamental component of $v_o(t)$ in volts is _______.
 A 220 B 440 C 198 D 110
GATE EE 2017-SET-1   Power Electronics
Question 4 Explanation:

$V_{01}=\frac{4V_s}{\sqrt{2} \pi}=\frac{4 \times 220}{\sqrt{2} \pi}=198.06V$
 Question 5
A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure below. For a star connected resistive load of 20 $\Omega$ per phase, the load power for 120$^{\circ}$ device conduction, in kW is __________.
 A 18 B 9 C 245 D 90
GATE EE 2017-SET-1   Power Electronics
Question 5 Explanation:

\begin{aligned} V_{rms}&=\sqrt{\frac{1}{\pi}(300)^2\frac{2 \pi}{3}}\\ &=300\sqrt{\frac{2}{3}}=244.94V\\ 3\frac{V^2}{R}&=\frac{3 \times (244.94)^2}{20}=9kW \end{aligned}
 Question 6
A three-phase Voltage Source Inverter (VSI) as shown in the figure is feeding a delta connected resistive load of 30 $\Omega$/phase. If it is fed from a 600 V battery, with 180$^{\circ}$ conduction of solid-state devices, the power consumed by the load, in kW, is __________.
 A 10 B 20 C 24 D 30
GATE EE 2016-SET-2   Power Electronics
Question 6 Explanation:
\begin{aligned} V_L&=V_{ph}=\sqrt{\frac{2}{3}}V_s\\ V_{ph}&=\sqrt{\frac{2}{3}} \times 600\\ P&=3\frac{V_{ph}^2}{R}=\frac{3 \times \frac{2}{3} \times 600^2}{30}\\ &=24kW \end{aligned}
 Question 7
A single-phase full-bridge voltage source inverter (VSI) is fed from a 300 V battery. A pulse of 120$^{\circ}$ duration is used to trigger the appropriate devices in each half-cycle. The rms value of the fundamental component of the output voltage, in volts, is
 A 234 B 245 C 300 D 331
GATE EE 2016-SET-1   Power Electronics
Question 7 Explanation:
\begin{aligned} V_{01(rms)}&=\frac{2\sqrt{2}}{\pi}V_s \cdot \sin d\\ \text{Pulse width} & \text{ where } 2d=120^{\circ}\\ d=60^{\circ}\\ V_{01(rms)}&=\frac{2\sqrt{2}}{\pi}V_s \sin 60^{\circ}\\ &=\frac{2\sqrt{2}}{\pi}V_s \frac{\sqrt{3}}{2}=234V \end{aligned}
 Question 8
The switches T1 and T2 in Figure (a) are switched in a complementary fashion with sinusoidal pulse width modulation technique. The modulating voltage $v_{m}(t) = 0.8 \sin (200 \pi t) V$ and the triangular carrier voltage ($v_{c}$) are as shown in Figure (b). The carrier frequency is 5 kHz. The peak value of the 100 Hz component of the load current ($i_{L}$), in ampere, is ________ .
 A 10 B 100 C 150 D 20
GATE EE 2016-SET-1   Power Electronics
Question 8 Explanation:
\begin{aligned} m_a&=0.8\\ (V_{01})_{peak}&=m_a\frac{V_d}{2}\;\;[m_a\leq 1]\\ &=0.8 \times 250=200V\\ (I_{01})_{peak}&=\frac{(V_{01})_{peak}}{Z_1}\\ &=\frac{200}{\sqrt{R^2}+(\omega L)^2}\\ &=\frac{200}{\sqrt{12^2+16^2}}=10A \end{aligned}
 Question 9
The single-phase full-bridge voltage source inverter (VSI), shown in figure, has an output frequency of 50 Hz. It uses unipolar pulse width modulation with switching frequency of 50 kHz and modulation index of 0.7. For $V_{in}$ =100 V DC, L=9.55 mH, C=63.66 $\mu$F, and R=5 $\Omega$, the amplitude of the fundamental component in the output voltage $V_o$ (in Volt) under steady-state is ________.
 A 12.25 B 40.25 C 62.75 D 82.75
GATE EE 2015-SET-1   Power Electronics
Question 9 Explanation:
The rms value of the fundamental component of output voltage,
\begin{aligned} V_{R1}&=\frac{V_s}{\sqrt{2}}\times m_i \\ &= \frac{100}{\sqrt{2}} \times 0.7=49.49V \end{aligned}
Amplitude of the value is peak value (i.e.)
$V_{R_{peak}}=\sqrt{2} \times 49.49=70V$

$V_{R_{peak}}$ is the amplitude of fundamental voltage at the VSI. $V_{0_{(peak)}}$ is the amplitude of fundamental voltage at the resistor.
\begin{aligned} Z&=\frac{R(-jX_C)}{R-X_Cj} \\ &= \frac{5(-j50)}{5-j50}\\ &= 4.97\angle -5.71^{\circ}\Omega \\ \text{where, } X_C&= \frac{1}{\omega C}\\ &=\frac{1}{2 \pi \times 50 \times 63.66 \times 10^{-6}} \\ &= 50\Omega \\ X_L &=\omega L \\ &=2 \pi \times 50 \times 9.55 \times 10^{-3} \\ &= 3 \Omega\\ V_{0(peak)}&=V_{R_{peak}} \times \frac{4.97\angle -5.71^{\circ}}{(4.97\angle -5.71^{\circ})+j3}\\ &=62.75\angle -32.5^{\circ}\;V \end{aligned}
Therefore, the amplitude of the fundamental component in the output voltage,
$V_0=62.75V$
 Question 10
A single-phase voltage source inverter shown in figure is feeding power to a load. The triggering pulses of the devices are also shown in the figure.

If the load current is sinusoidal and is zero at $0,\pi ,2\pi..,$ the node voltage $V_{AO}$ has the waveform
 A A B B C C D D
GATE EE 2014-SET-3   Power Electronics
Question 10 Explanation:
$\theta \leq \omega t \leq \pi-\theta \;\;(S_{11}S_4\rightarrow ON)$

\begin{aligned} V_A &=V_{DC}, \; V_0=\frac{V_{DC}}{2} \\ V_{AO}&=\frac{V_{DC}}{2}\;\; (\pi-\theta \lt \omega t \lt z) \end{aligned}

\begin{aligned} V_A &=0\\ V_{AO}&=V\cdot 0-\frac{V_{DC}}{2}\\ &=-\frac{V_{DC}}{2} \end{aligned}
$\pi +\theta \leq \omega t\leq 2 \pi$

$V_A=0,\;\; V_0=\frac{V_d}{2}$
There are 10 questions to complete.


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# Paint It And Disassemble It

Geometry Level 1

The above is a wooden cube block of size $$7\times 7\times 7$$, whose outer faces are painted in blue.

Now, the whole block is cut into smaller cubes $$1\times1\times1$$.

How many $$1\times 1\times1$$ cubes have precisely 1 of its faces painted blue?

×

format: plain


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# Triac datasheet values

I have a Triac that I want to use to control low voltage (~16 V) AC. The data sheet is found here. It specifies that the peak gate voltage is 5 V with a peak current rating of 2 A (which seems quite high to me). The datasheet also lists out the Gate trigger voltage as 0.7 V (typical).

I'll be using this opto-isolator to drive the triac gate, which doesn't seem to have much play in my question (but I could be wrong).

UPDATE: After some additional research it turns out that I was wrong here. The opto-isolator's max current is the determining factor for the gate resistor value. The iosolator used here has a max detector current of 1A. 16VAC has a RMS voltage value of 22.62VAC which leads to a minimum resistor value of 22.6 / 1 = ~22 ohms.

My question is which of these values should I consider when choosing a resistor to place before the triac's gate/opto isolator? I understand how a Triac and it's associated driver works but I just need a little assistance with reading the datasheet and which of the values I should be considering.

Also, are there any good resources for learning how to read datasheets or is this something that just comes with experience?

• I am sorry, but I can't see how you plan to use a triac and, basically, PWM driving, without having a proper idea about the circuit you want to use, the devices themselves or simply reading a datasheet. Maybe it's best to either call a friend or wait until you are certain about these things. I am sorry if this sounds harsh, but there is a reason for it. – Vlad Nov 14 '12 at 7:20
• @Vlad I'm not sure I get your perspective or what you comment is trying to accomplish. Maybe I didn't ask the question well. I'm just trying to determine which of the values in the datasheet I should use or a nudge in the right direction for learning. It would be great if I could call someone, but then no one else would benefit from my experience. In any case, Felice was kind enough to answer the question for me. – Matt Ruwe Nov 14 '12 at 12:02
• My intention was not to put you down or to put myself above, I hope this isn't what came out. To me, after reading your post, it seemed that not only you can't read a datasheet (don't know which one to consider, max gate voltage or threshold), that you have pieces but no schematic or can't think about one ("opto-isolator...which doesn't seem to have much play..."), and that you're not familiar with electronics in general ("...good resources for how to read datasheets or...?"). All these made me say that, but I apologize if my answer made you feel bad -- such was my conclusion then. – Vlad Nov 14 '12 at 13:11
• I try to keep my questions short and specific because this typically leads to the best answers (like Felice's below). The focus of my question was around the datasheet (as indicated in the title). I didn't want answers to focus on the parts or the schematic, just the datasheet. I certainly could learn some things about reading datasheets, but when I said that opto-isolator didn't "have much play" I was referring to the fact that it didn't have any effect on the choosing of a resistor value. – Matt Ruwe Nov 14 '12 at 13:42

I think you have to design in term of current, that must be >= to the Gate Trigger current in order to have the device ON. You can use the Gate trigger voltage as a voltage drop when you measure the current flowing in the gate in order to switch the TRIAC.


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# Newbie questions on RF circuit

#### n9jcv

Joined Apr 14, 2016
5
Hello everyone, I am a ham radio operator. I have tinkered with electronics but never took any formal training. Now trying to learn from books. I have a real life troubleshooting case with a HF radio. So many questions.

Below are links to 3 items.
#1 - Schematic (partial) of circuit I am suspect of
#2 - scope photo of oscillator driver output with keydown(transmitting)
#3 - scope photo of oscillator driver output - not transmitting

First, I do have a bk precision 15mhz scope, the circuit is now running about 7mhz. The problem is that I get no output, or lets say less than 200mw output, cause the driver is outputting. The following is not to explain the circuit to you all, as I know you probably know these things, but it is to ask for your confirmation on how I believe it works. One note, with my dvm, I was able to detect that ZD1 was defective, I pulled it and sure enough it was shorted, both ways, so I have replaced that already. My understanding is ZD1 is there to block reflected rf (when the swr is high on the antenna). So if rf comes back down the antenna, it will not kill the finals. Well ZD1 was a dead short. I can't explain what caused it to short. But since it was shorted, I am almost certain there is other damage. And the finals still do not transmit. On the circuit diagram, I have circled and highlighted in pink my measured voltages, all DC, no rf probe used as I don't have one. Any way starting with the driver Q5, it has voltages, and they are within reason of the marked voltages from Heath. I took a pic of the oscillator output. It looks fairly good to me, it is about .5vpp (volt peak to peak if that is how you say it). Then when I transmit(key down), it goes to 3vpp. So I am going to say, I think the oscillator is ok. I did adjust the 40m driver cap for max output at R31 while keeping it clean. At the base of Q6 and Q7, there is no voltage and no oscillator signal. the collectors of q6/q7 are 0 volts. The voltage on the other side of RFC2 is 12.2v. My thought now is that both the final transistors are bad. I am thinking they are open and not shorted. If they were shorted, I think I would have bigger issues basically pulling all the current my power supply could feed.

So am I anywhere on track with this? I am not necessarily asking you to fix it for me. More I would like confirmation that what I have done for troubleshooting is logical and correct. And is my hypothesis that the finals are bad correct. I know I could pull one and test it, but I do not know if I can test it in circuit.

Any help would be greatly appreciated, and by the way I just found this forum and love it. I could read here for months or years and learn a lot I am sure.

Pic of schematic with voltages marked - Pink are what I actually observed, numbers in rectangular boxes next to them are transmit values from heath.

https://goo.gl/photos/mjjGBEXP1oM264aU6

Pic of driver when not transmitting .5vpp at q5 collector

https://goo.gl/photos/c71gPtrQUzGpMcCh6

pic of oscillator output at q5 collector when transmitting about 3vpp

https://goo.gl/photos/vAQc9mnMQXN6bKbC8

Moderators note: uploaded pictures and inserted them in as full image.

Last edited by a moderator:

#### DickCappels

Joined Aug 21, 2008
7,571
If I am reading this correctly, you have 12.2 volts across RFC2. It is quite likely that when the Zener shorted it opened RFC2.

While you are in there with an ohmmeter, you might want to check to see whether there is a short between the bases and collectors of the output transistors too.

#### n9jcv

Joined Apr 14, 2016
5
I have a mpsu05 that is bad. They are no longer made. There are some on ebay for big $$or NTE which is also big money. I am looking for a cheaper generic substitue. Would a 2SC1678 serve my needs. The intended circuit is for RF amplification in the final transmitter stage. A 3 Watt QRP radio 3.5mhz to 21mhz. The relevant part of the circuit can be seen below. Transistors are Q6 and Q7. https://goo.gl/photos/mjjGBEXP1oM264aU6 Thread Starter #### n9jcv Joined Apr 14, 2016 5 Thanks for the input Dick. It appears the RFC is open. I will need to replace that. In measuring the transistors from the base to the collector, both ways, it is near infinite resistance. I will need to pull them to verify, but I am guessing they are bad as well. I am thinking perhaps someone reversed the power input and that is what caused the damage. #### GopherT Joined Nov 23, 2012 8,012 I have a pair of MRF477 you could use. Should be close. Vintage Motorola new old stock. #### ian field Joined Oct 27, 2012 6,539 I have a mpsu05 that is bad. They are no longer made. There are some on ebay for big$$ or NTE which is also big money. I am looking for a cheaper generic substitue. Would a 2SC1678 serve my needs. The intended circuit is for RF amplification in the final transmitter stage. A 3 Watt QRP radio 3.5mhz to 21mhz. The relevant part of the circuit can be seen below. Transistors are Q6 and Q7.

https://goo.gl/photos/mjjGBEXP1oM264aU6
That package style fell by the wayside long ago, but you should be able to find an equally capable part in TO126.

You could probably wriggle a TO220 part in - but I think the pinout might be back to front.


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# The union of family of open disjoint non empty subsets is dense.

The statement I'm gonna write down here is a proposition I conjectured in order to prove a detail in Hartogs theorem; thus I won't write all the context, otherwise it would become the longest post ever in MSE! On the other side this statement is quite general, thus it shouldn't be a problem.

Consider the unitary open disk $\Delta$ in $\Bbb C$. $\Delta$ is a topological space, with topology induced by $\Bbb C$. We'll work in this topological space $\Delta$, thus, "open", "the closure of" etc are all related to this topological space.

Let $\Lambda$ be a family of index more than countable. Let $\{E_{\lambda}\}_{\lambda\in\Lambda}$ be a family of disjoint open non empty subsets of $\Delta$. Can we say that $$\operatorname{cl}_{\Delta}\left(\bigcup_{\lambda\in\Lambda}E_{\lambda}\right)=\Delta\;\;\;\;?$$ I'd say yes, and it seems natural, but honestly I don't know neither where to start to prove it.

• There is no such family of disjoint open sets. (Hence the answer to your question is actually yes, but only because if $P$ is false then anything whatever follows from $P$...) Jul 13 '15 at 18:56
– Joe
Jul 13 '15 at 18:58
• ??? More explicit about what exactly? The fact that any collection of disjoint nonempty open sets is countable, or that silliness about one falsehood implying anything? The second isn't that important, and the first is clear beacuse $\Delta$ is separable... Jul 13 '15 at 19:00
• What is so special about it being the unit disk? Could not all these open sets belong to the disk of radius 1/2? Jul 13 '15 at 19:04
• @DavidP: It's clear that exists such a family but my question is "does my argument work for every such a family?".
– Joe
Jul 13 '15 at 19:32

To see this, let $\Lambda$ be an uncountable index set and suppose that $(E_{\lambda})_{\lambda\in\Lambda}$ is a family of disjoint, non-empty open subsets of $\Delta$. Also, let $$\mathbb Q_{\mathbb C}\equiv\{a+bi\,|\,\text{a and b are rational numbers}\}$$ be the set of numbers on the complex plane with “rational grid points” (that is, both the real and the imaginary parts are rational). It is not difficult to show that $\mathbb Q_{\mathbb C}$ is dense in $\mathbb C$ and also that $\mathbb Q_{\mathbb C}$ is countable.
For each $\lambda\in\Lambda$, pick $x_{\lambda}\in E_{\lambda}$ (you need the axiom of choice here). Since $E_{\lambda}$ is open and $\mathbb Q_{\mathbb C}$ is dense in $\mathbb C$, one can pick $q_{\lambda}\in\mathbb Q_{\mathbb C}$ such that $q_{\lambda}\in E_{\lambda}$. Because of disjointness, all the $(q_{\lambda})_{\lambda\in\Lambda}$ are distinct, so that the function $\lambda\mapsto q_{\lambda}$, mapping from $\Lambda$ to $\mathbb Q_{\mathbb C}$ is injective. This is implies that the set $\mathbb Q_{\mathbb C}$ is uncountable, which is impossible.


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# How do you solve x^2=121?

##### 1 Answer
Mar 10, 2018

$x 1 = - 11 ,$ $x 2 = 11$

#### Explanation:

Move the constant to the left and change its sign

${x}^{2} - 121 = 0$

Using ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$, Factor the expression

$\left(x - 11\right) \left(x + 11\right) = 0$

When the product of factors equals 0, at least one factor is 0

$x - 11 = 0$

$x + 11 = 0$

Solve both equations for $x$

$x = 11$

$x = - 11$

The final solutions are

∴ $x 1 = - 11 ,$ $x 2 = 11$


text:
mersenneforum.org A thought on Division in Mathematics
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2020-06-23, 16:07   #45
BillyB

Jun 2020

29 Posts

Quote:
 Originally Posted by Uncwilly I don't think "great things" fits with "contradicts". If you weren't sure what he meant, you should have have asked. Maybe he was trying to tell you to go find out why your idea fails. Maybe your memories have altered the facts since you left school years ago. If it contradicts everything in the math department, then either your idea is flawed, or all of known math is wrong.
Some feedback I received was in person, on the phone or via email. I still have the emails.

2020-06-23, 16:09   #46
BillyB

Jun 2020

29 Posts

Quote:
 Originally Posted by storm5510 Off-topic, somewhat. Seeing all the fractions in the pizza video reminded me of my father. He worked over 25 years in a road construction machine factory. He could look at a bolt head or a nut and say what size it was. He would get 8th's and 16th's right every time. He was also a part-time carpenter. He could glance at a tape-measure and do the same thing. He struggled with millimeters when it became common much later in his life.
In my opinion this is right on topic. Imagine living your entire life learning and perfecting something one way and then having a new system or way of doing something introduced. It would be difficult for most people to just accept it as is and understand it right away. Thank you for your post.

 2020-06-23, 16:55 #47 CRGreathouse     Aug 2006 10110111001102 Posts It's hard for me to believe we're five pages in and you still haven't defined your operation. You have a function -- not division, but your new take on it -- which takes two arguments and returns a third. Let's call it BillyB(x, y) = z. For which pairs of x and y is BillyB(x, y) defined? When BillyB(x, y) is defined, how can BillyB(x, y) = z be computed? Are there some pairs x and y for which BillyB(x, y) = x/y or BillyB(x, y+1) = x/y?
2020-06-23, 17:37   #48
paulunderwood

Sep 2002
Database er0rr

22·72·17 Posts

Quote:
 Originally Posted by retina 1/0 is how many times can we cut things of size 0 from a thing of size 1.
is that bigger than $$\aleph_{1000}$$?

2020-06-23, 18:25   #49
BillyB

Jun 2020

29 Posts

Quote:
 Originally Posted by CRGreathouse It's hard for me to believe we're five pages in and you still haven't defined your operation. You have a function -- not division, but your new take on it -- which takes two arguments and returns a third. Let's call it BillyB(x, y) = z. For which pairs of x and y is BillyB(x, y) defined? When BillyB(x, y) is defined, how can BillyB(x, y) = z be computed? Are there some pairs x and y for which BillyB(x, y) = x/y or BillyB(x, y+1) = x/y?
You are on the right track! I will reply more later tonight as I don’t have the time this minute to go further into detail. Thank you for your post!

2020-06-23, 18:26   #50
BillyB

Jun 2020

29 Posts

Quote:
 Originally Posted by CRGreathouse It's hard for me to believe we're five pages in and you still haven't defined your operation. You have a function -- not division, but your new take on it -- which takes two arguments and returns a third. Let's call it BillyB(x, y) = z. For which pairs of x and y is BillyB(x, y) defined? When BillyB(x, y) is defined, how can BillyB(x, y) = z be computed? Are there some pairs x and y for which BillyB(x, y) = x/y or BillyB(x, y+1) = x/y?
It is a function but I was told by a math professor that I have the utmost respect for to call it a new thought on division so I stuck with that until I realized the formula that it created. You can use it to graph coordinates as well.

2020-06-23, 18:26   #51
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

172×29 Posts

Quote:
 Originally Posted by Uncwilly 45
I rechecked and I think 65 is a closer value.

Some dude that likely lives in rural 'merica at the end of a gravel road thinks he found something new with regards to basic math. It has been known for ~1500 years that division by zero is absurd.

2020-06-23, 19:30   #52
BillyB

Jun 2020

29 Posts

Quote:
 Originally Posted by Uncwilly I rechecked and I think 65 is a closer value. Some dude that likely lives in rural 'merica at the end of a gravel road thinks he found something new with regards to basic math. It has been known for ~1500 years that division by zero is absurd.
Why are you so offended by this? And no, you have me all wrong. How would anything new be created if it weren’t for new ideas?

2020-06-23, 19:47   #53
xilman
Bamboozled!

May 2003
Down not across

10,163 Posts

Quote:
 Originally Posted by BillyB Why are you so offended by this? And no, you have me all wrong. How would anything new be created if it weren’t for new ideas?
1) I am not in the least offended.
2) I agree whole-heartedly.

That said, please define the function which has been called "BillyB".

Here is a first take, based on what I have read so far. I am very far from infallible so my definition may be completely wrong.

Let x and y be real numbers.

BillyB(x, y) = x/y for all x, y where x and y are both non-zero.
BillyB(x,0) = 1 for all x not equal to zero.

At the moment I have no idea of the value of the only remaining case, viz BillyB(0,0).

Please enlighten me. This is especially true for the third case, "0/0".

Last fiddled with by xilman on 2020-06-23 at 19:54 Reason: Remove otiose comma

2020-06-23, 20:48   #54
bsquared

"Ben"
Feb 2007

3,271 Posts

Quote:
 Originally Posted by xilman Here is a first take, based on what I have read so far. I am very far from infallible so my definition may be completely wrong. Let x and y be real numbers. BillyB(x, y) = x/y for all x, y where x and y are both non-zero. BillyB(x,0) = 1 for all x not equal to zero.
I agree with the first but my take on his "explanation" so far is that BillyB(x,0) = x. i.e., a thing of size x cut zero times remains size x. So BillyB(0,0) = 0 by that definition.

but that assumes some definitions... namely that x is the quantity or size of a cuttable object and y is the number of times to cut it into equal pieces. Then BillyB(x,y) = x / (y+1). Physically you can only cut something zero or more times so the domain of the function is >= 0. So, by that definition, there is no dividing by zero possible, by the everyday/ordinary definition of "divide". BillyB is not a division operation but some kind of splitting operation.

[edit-edit]
I suppose that must mean that x, y are in $\script{N}^0$. The guess at a definition doesn't work otherwise....

Last fiddled with by bsquared on 2020-06-23 at 21:06

2020-06-23, 23:08   #55
BillyB

Jun 2020

2910 Posts

Quote:
 Originally Posted by bsquared I agree with the first but my take on his "explanation" so far is that BillyB(x,0) = x. i.e., a thing of size x cut zero times remains size x. So BillyB(0,0) = 0 by that definition.  but that assumes some definitions... namely that x is the quantity or size of a cuttable object and y is the number of times to cut it into equal pieces. Then BillyB(x,y) = x / (y+1). Physically you can only cut something zero or more times so the domain of the function is >= 0. So, by that definition, there is no dividing by zero possible, by the everyday/ordinary definition of "divide". BillyB is not a division operation but some kind of splitting operation. [edit-edit] I suppose that must mean that x, y are in $\script{N}^0$. The guess at a definition doesn't work otherwise....
Excellent interpretation!

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# Determinant of a matrix

This online calculator computes determinant of a matrix, using it's definition

The site engine now allows input of long texts, so this is first calculator to use this feature, and it is used to enter matrix, and calculator computes the determinant of entered matrix.

It uses determinant definition, which is recursive calculation, and, in theory, quite resouce consuming, but for our case, with matrix, entered manually, I believe it is enough.

Some notes about determinant are below the calculator, for those who forgot.

### Determinant of a matrix

Digits after the decimal point: 2
Determinant of a matrix

Save the calculation to reuse next time, to extension embed in your website or share share with friends.

Some recursive formulae:

$det A=\begin{vmatrix} a_{11}\end{vmatrix} = a_{11}$

• determinant of a matrix 1x1

$det A=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}$

• determinant of a matrix 2x2

$det A=\sum_{j=1}^n (-1)^{1+j} a_{1j}\bar M_j^1$

• determinant of a matrix nxn, where n > 2
$\bar M_j^1$ - minor of $a_{1j}$.
Minor of $a_{1j}$ - is the determinant of a (n–1) × (n–1) matrix that results from deleting the 1-th row and the j-th column of A. That's why this is recursive definition.

For example, here is the determinant of a matrix 3x3
$det A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}-a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}+a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$


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## Pieri and Giambelli Formulas

It’s been a few weeks, but now I’m back and today we’ll talk about the multiplication in the cohomology ring of Grassmannians.  Though we won’t talk about the Littlewood-Richardson rule in its full glory, we will howver discuss the special cases of the Pieri rule and the Giambelli formula.

Last time, we saw that the cohomology of a Grassmannian had a basis parameterized by Young diagrams.  The real powerhouse is the Pieri rule, for one simple reason: it completely determines the multiplication.  Once it’s been determined, we can steal formulas from anything else that happens to use the Pieri rule, and there are good candidates, which we’ll discuss later in the post.  For now, here’s the Pieri rule:

Pieri Rule: Let $\lambda$ be any Young diagram and let $k$ be the Young diagram with only one row of boxes of length $k$.  Then $\sigma_\lambda \sigma_k=\sum \sigma_\nu$ where $\nu$ runs over the the Young diagrams obtained from $\lambda$ by adding $k$ boxes, no two in the same column.

We note quickly that diagrams that don’t fit into the box given by the Grassmannian we’re working on just give Schubert class zero, and that with this the rule is perfectly fine without reference to any specific Grassmannian.  We’ll make a bit of use of this later.  For now, we’ll prove this:

Proof: To prove this, the first thing we need is the dual Young diagram.  Given a Young diagram $\lambda$ inside of an $m\times n$ rectangle, we define the dual, $\lambda^*$, to be the diagram obtained by deleting $\lambda$ from the rectangle, and then rotating 180 degrees.  This is an extremely useful notion in the Schubert calculus, for the following reason:

Claim: Let $\lambda,\mu$ be two Young diagrams contained in an $m\times n$ rectangle such that $|\lambda|+|\mu|=mn$.  Then $\sigma_\lambda\cup \sigma_\mu=\delta_{\mu,\lambda^*}$ in the top cohomology group.

This claim can be seen by choosing a basis and writing out matrices as we did a few posts ago, and noting that the dual can be represented by the matrix with ones in teh same positions, but with the zeros and stars switched, so the two spaces must intersect at a unique point.

Now, to show this, we introduce the opposite flag, $\tilde{F}$, whose $k$th term is given by the span of the last $k$ basis vectors for our vector space, and write the Schubert variety found using that flag as $\tilde{\Omega}_\lambda$.    (Note, working the matrices out for these carefully is what proves the above claim)

Now, set $A_i=F_{n+i-\lambda_i}$, $B_i=\tilde{F}_{n+i-\mu_i}$ and $C_i=A_i\cap B_{r+1-i}$ for given $\lambda,\mu$.  These are useful when working with the intersection of $\Omega_\lambda$ and $\tilde{\Omega}_\mu$.

Now, to show Pieri’s formula, we just need to show that both sides have the same intersection numbers with all $\mu$ which have $|\mu|=rn-|\lambda|-k$ in our Grassmannian.  So that means that we’d need $\sigma_\mu\sigma_\lambda\sigma_k=1$ whenever we have that $\mu\subset \lambda^*$, and that the boxes in $\lambda^*$ but not in $\mu$ are all in different columns.  This is just the condition $n-\lambda_r\geq \mu_1\geq n-\lambda_{r-1}\geq\mu_2\geq\ldots\geq n-\lambda_1\geq \mu_r\geq 0$.

So now we take $\Omega_\lambda,\tilde{\Omega}_\mu$ and define $\Omega_k(L)$ by taking $L$ to be a general linear subspace of dimension $n+1-k$.  Now, we’ve just used the word “general” and we’re going to talk much more about it next time.  The idea is that on the Grassmannian of these things, there might be a few $n+1-k$ planes that need to be avoided, but they form subvarieties of positive codimension, so we have an open, dense set to work with.  Pretty much we can choose a plane at random, and it will, with probability 1, be ok.

Now, let $C$ be the span of the $C_i$, and set $A_0=B_0=0$.  Then, we have that $C=\cap_{i=0}^r (A_i+B_{r-i})$, $\sum_{i=1}^r \dim(C_i)=r+k$ and that $C=C_1+\ldots+C_r$ is a direct sum iff we have the situation with the Young diagrams of $\lambda$ and $\mu$ as above.  Now, if that condition fails, the triple intersection $\Omega_\lambda\cap \tilde{\Omega}_\mu\cap \Omega_k(L)=\emptyset$.  We just need to show that it’s a single point if it does hold.

Now, any point in $\Omega_\lambda\cap \tilde{\Omega}_\mu$ will have $\dim(V\cap A_i)\geq i$ and $\dim(V\cap B_{r+1-i})\geq r+1-i$, by definition, and so we will get $\dim(V\cap C_i)\geq i+(r+1-i)-r=1$.  So now, if the $C_i$ are linearly independent, then $V\supset \oplus (V\cap C_i)$, which has big enough dimension that they are equal.  So $V$ is a direct sum of $V\cap C_i$, one for each dimension of $V$, and none of them being trivial.  So all must have dimension 1.

This tells us that $C=\oplus C_i$, and a generic $L$ will meet it in a line $\mathbb{C}\cdot v$, with $v=u_1\oplus\ldots\oplus u_r$ with $u_i\in C_i$.  Now, $V\subset C$, and so must contain $v$, and so $u_i\in V$, so $V$ is the subspace spanned by the $u_i$, and is the unique point in which the three Schubert varieties intersect, giving us the Pieri rule.  $\Box$.

The upshot from our point of view is that this lets us start doing some computations.  Look at $Gr(2,4)$, and say we want to compute $\sigma_1\sigma_1$.  The Pieri rule says that this will be $\sigma_2+\sigma_{1,1}$, very quickly.  Next week, we’ll use this fact to start solving actual problems.  In the meantime, there are other formulas that are worth mentioning, as well as connections to things that I understand a bit less well.

The big theoretical thing the Pieri Rule does for us is that it tells us that there’s a surjective ring homomorphism $\Lambda\to H^*(Gr^n(\mathbb{C}^m),\mathbb{Z})$, where $\Lambda$ is the ring of symmetric functions.  The map is given by taking the Schur function $s_\lambda$ to the cohomology class $\sigma_\lambda$.  Now, I know very little about Schur functions, so I’m not even going to bother defining them.  The upshot is that there are lots of formulas known in $\Lambda$, and we can just use them in our cohomology calculations.  One of the better consequences is the following:

Giambelli Formula: $\sigma_\lambda=\det(\sigma_{\lambda_i+j-i})_{1\leq i,j\leq r}$

This formula says that we can represent any Schubert class in terms of the “special” Schubert classes that the Pieri rule tells us how to multiply.  So with these two formulas we can, in principle, perform any multiplication we could possibly desire in the cohomology ring of the Grassmannian.  And example of the Giambelli formula in $Gr(2,4)$ is that we can write $\sigma_{1,1}=\det\left(\begin{array}{cc}\sigma_1&\sigma_2\\ \sigma_0&\sigma_1\end{array}\right)=\sigma_1^2-\sigma_2$.  This formula isn’t terribly exciting because we already obtained it by using Pieri on $\sigma_1^2$, but Giambelli is useful in more difficult situations as well.

All this culminates in the fact that there are numbers called the Littlewood-Richardson coefficients, $c_{\lambda\mu}^\nu$, which are defined completely combinatorially (there may be a guest post on this) which are the structure constants.  That is, they’re positive integers such that $\sigma_\lambda\sigma_\mu=\sum_\nu c_{\lambda\mu}^\nu \sigma_\nu$.


text:
# VESC Tool manuals (non video based)

Submitted by frank on Tue, 2017-10-03 19:42

# In this section you will find detailed, screen shot based setup guidance.

### VESC Tool

What kind of Operating System on my computer do i need to use the VESC Tool?

Greetings

Schlawiner

### Operating System

Windows 7 or later, LINUX.

Frank

### Detection Failure

I am currently setting up my ESC using motor setup wizard, however, when asked to run the detection for the parameters, my motor does not cooperate. I have gone through both FOC and BLDC, however neither are working. Any advice for how to make it work. I am using a torque board motor 6353/190KV and my battery that I am trying to set it up with is 4 cells 5.2 A, 14.8V. I plan on running my longboard with two of these LiPo batteries

### Hi mjmarchetto

Hi mjmarchetto

maybe you got it working in the meantime, but I guess you are facting the same problem as I that your setup has a high inertio, so the overcurrent protection got in between. Try to set the eRPM (500afaik) in the detection lower for FOC. For BC (called BLDC here) I did not get mine working at all; as I got not feedback how to tweak the setup I was unable to get further on that front, but as FOC works as it should (so far) it's OK for me.

Cheers

Sam

Pedelec usage in combination with a GoldenMotor MagicPie3 BLDC

### Electrical

Hi , It is worth re-checking the integrity of all electrical Power and sensors. Tighten any connections or secure where possible, effectively you have to look at all your wires and see which one is duff, enjoy :)

808

### hope you develop VESC Application to work on the iOS or OSX

I only own Apple Hardware... I may have to buy a Windows or Linux emulator program or another computer to access VESC tools.. thx

### Hi, You can use whats called

Hi, You can use whats called a "Virtual Machine" to run linux, mac, or windows OS's on your computer. I have a PC, and I run Linux through a virtual machine. My friend has a mac, and he runs Windows 10 on his mac on a VM. The software is called virtual box. Its made by a pretty big company named Oracle. It is free! Get ready to watch some youtube videos on the subject if you're unfamiliar with it.  https://www.virtualbox.org/

### vesc-tool 1.08 connection

Hi, i seem to be having trouble simply connecting to my dual FOCBox setup.... it says something about delay semaphore expired could it have something to do with baud rate?

Zoom zoom

### Installation crashes

I am trying to install bldc tool on windows 10 but instalation crashes on startup is there any trick to fix this issue?

### I have the same issue, but it doesn't seem common

Been googling around for VESC tool errors and I'm not coming up with much. There might be some specific compatability issues here. What kind of error code were you getting btw (if you still have it, I realize this was 6 months ago.)  I'm gettting a 0xc5

### regarding control of motor speed

can i run the vesc motor at full speed for all time without using transmitter and reciever.

### Vesc tool 2.02 not starting on windows 10

Unfortunately I can't even start the tool on windows,

I copied exe file to desktop and when I double click it, it shows an error message."vesc 2.02 exe stopped working" Windows will close the program.

Any ideas?

Mac Motor

### Windows wont open tool

I have installed the tool properly and when I click to open it begins to open and immediately closes and disappears, I removed all security settings leaving my computer wide-open but still it doesn't work and I see other people have the same problem so it would be real nice if somebody would respond with a fix.

it would be nice if somebody replied with a fix in July 2020

ernest sanne

### Hi!   Нave you solved this

Hi!   Нave you solved this problem?

### Motor Types

which types of motors are supportred by the VESC controller

can we use it with all AC and DC motors?

can we use it with PMSM?

or only BLDC?

Thanks,

Mohamed

### не могу купить vesk tool .

добавляю vesc tool в корзину, нажимаю кнопку (оформить заказ) из корзины всё пропадает, и сообщает что (vesc tool из корзины удалён ) соответсвенно покупки не происходит! подскажите как быть?

Сергей

### Has anyone solved this

Has anyone solved this problem? Can you quote this below? jigsaw puzzle

### Que problema puede haber si mi esc no se conecta con vesc tool ?

Que problema hay si mi esc no conecta con vesc tool ????

### 1v1.lol unblocked

I installed the tool correctly, and when I click to open it, it opens for a few seconds before closing and disappearing. I removed all security settings, leaving my computer wide open, but it still doesn't work, and I see that other people are having the same issue, so it would be great if someone could respond with a fix. 1v1.lol unblocked

### Hi good afternoon.

Hi good afternoon.
I have a request for you, I have already registered on your vesc tool page.
However, I do not get an e-mail with the program that I can program my mountainboard vesc.
In this regard, I would ask you if you would send me the program by email.
Kind regards
Gardener Gabriel.

Name .Gardener Gabriel
e-mail. ggleba@yahoo.de

### Yy device cannot connect vesc

This article has piqued my interest since it is both beautiful and instructive. If you have time, we can play retro bowl and discuss it more.

Your VESC motor controller's integration with an electrical system: WARNING: To familiarize yourself with the equipment, read the entire instruction booklet. 1v1 battle  build now gg

### on my macb i=ook pro wionbt coinnect

to the usb cables and ion my windows computer it is windows 10 an d it will start to open r\then  the screen goes grey

### You will like playing drift

You will like playing drift hunters online if you enjoy playing racing games.

### What programming language do

What programming language do you need to know to program a microcontroller. I only learned Python when I worked on the chase routing number project. Once upon a time, I worked with Arduino microcontrollers.

### It's pretty safe to say that

It's pretty safe to say that Elden Ring is a tough game. Frequent ambushes and enemies that can kill you with a single hit can make the first few hours very difficult.

### First way is Automatically

First way is Automatically installation:After WinXP system, Windows has built-in function of automatically search driver on the internet, so when you connect VESC to your computer with USB cable, if current Windows has no VESC driver, Windows can search related driver and install automatically, at this moment, Windows has a hint of showing “new device is discovered meanwhile automatically search and install driver”, you just need to wait for the installation finished, it takes about 1-2 minutes or longer. Don't forget that if you are too stressed in the installation, or join Paper Minecraft to have a really refreshing spirit.

### To prevent damage to your

To prevent damage to your battery, which could occur if the current exceeded its capacity, you can set a maximum current of X Amps. pacman 30th anniversary

### Do I have to have windows to get into vesc bin trying for days

After days of trying to get into vesc I just saw summing about using windows to do this you wood think that wood be the first thing on this web site I just want to start my EBike with my calibike Bluetooth dongle can't believe what you have done with something that was so simple to get into bin trying to get into my yuyangking app for days now I can believe this I make sum of the fastest EBikes and I can't get this older bike that I made to even start with what you have done


text:
Reference documentation for deal.II version 9.4.0
Polynomials::HermiteLikeInterpolation Class Reference

#include <deal.II/base/polynomial.h>

Inheritance diagram for Polynomials::HermiteLikeInterpolation:
[legend]

## Public Member Functions

HermiteLikeInterpolation (const unsigned int degree, const unsigned int index)

double value (const double x) const

void value (const double x, std::vector< double > &values) const

void value (const Number2 x, const unsigned int n_derivatives, Number2 *values) const

void values_of_array (const std::array< Number2, n_entries > &points, const unsigned int n_derivatives, std::array< Number2, n_entries > *values) const

unsigned int degree () const

void scale (const double factor)

void shift (const number2 offset)

Polynomial< double > derivative () const

Polynomial< double > primitive () const

Polynomial< double > & operator*= (const double s)

Polynomial< double > & operator*= (const Polynomial< double > &p)

Polynomial< double > & operator+= (const Polynomial< double > &p)

Polynomial< double > & operator-= (const Polynomial< double > &p)

bool operator== (const Polynomial< double > &p) const

void print (std::ostream &out) const

void serialize (Archive &ar, const unsigned int version)

virtual std::size_t memory_consumption () const

## Static Public Member Functions

static std::vector< Polynomial< double > > generate_complete_basis (const unsigned int degree)

## Protected Member Functions

void transform_into_standard_form ()

## Static Protected Member Functions

static void scale (std::vector< double > &coefficients, const double factor)

static void shift (std::vector< double > &coefficients, const number2 shift)

static void multiply (std::vector< double > &coefficients, const double factor)

## Protected Attributes

std::vector< double > coefficients

bool in_lagrange_product_form

std::vector< double > lagrange_support_points

double lagrange_weight

## Subscriptor functionality

Classes derived from Subscriptor provide a facility to subscribe to this object. This is mostly used by the SmartPointer class.

void subscribe (std::atomic< bool > *const validity, const std::string &identifier="") const

void unsubscribe (std::atomic< bool > *const validity, const std::string &identifier="") const

unsigned int n_subscriptions () const

template<typename StreamType >
void list_subscribers (StreamType &stream) const

void list_subscribers () const

static ::ExceptionBaseExcInUse (int arg1, std::string arg2, std::string arg3)

static ::ExceptionBaseExcNoSubscriber (std::string arg1, std::string arg2)

using map_value_type = decltype(counter_map)::value_type

using map_iterator = decltype(counter_map)::iterator

std::atomic< unsigned intcounter

std::map< std::string, unsigned intcounter_map

std::vector< std::atomic< bool > * > validity_pointers

const std::type_info * object_info

static std::mutex mutex

void check_no_subscribers () const noexcept

## Detailed Description

Polynomials for a variant of Hermite polynomials with better condition number in the interpolation than the basis from HermiteInterpolation.

In analogy to the proper Hermite polynomials, this basis evaluates the first polynomial $$p_0$$ to 1 at $$x=0$$ and has both a zero value and zero derivative at $$x=1$$. Likewise, the last polynomial $$p_n$$ evaluates to 1 at $$x=1$$ with a zero value and zero derivative at $$x=0$$. The second polynomial $$p_1$$ and the second to last polynomial $$p_{n-1}$$ represent the derivative degree of freedom at $$x=0$$ and $$x=1$$, respectively. They are zero at both the end points $$x=0, x=1$$ and have zero derivative at the opposite end, $$p_1'(1)=0$$ and $$p_{n-1}'(0)=0$$. As opposed to the original Hermite polynomials, $$p_0$$ does not have zero derivative at $$x=0$$. The additional degree of freedom is used to make $$p_0$$ and $$p_1$$ orthogonal, which for $$n=3$$ results in a root at $$x=\frac{2}{7}$$ for $$p_0$$ and at $$x=\frac{5}{7}$$ for $$p_n$$, respectively. Furthermore, the extension of these polynomials to higher degrees $$n>3$$ is constructed by adding additional nodes inside the unit interval, again ensuring better conditioning. The nodes are computed as the roots of the Jacobi polynomials for $$\alpha=\beta=4$$, which are orthogonal against the square of the generating function $$x^2(1-x)^2$$ with the Hermite property. Then, these polynomials are constructed in the usual way as Lagrange polynomials with double roots at $$x=0$$ and $$x=1$$. For example with $$n=4$$, all of $$p_0, p_1, p_3, p_4$$ get an additional root at $$x=0.5$$ through the factor $$(x-0.5)$$. In summary, this basis is dominated by nodal contributions, but it is not a nodal one because the second and second to last polynomials that are non-nodal, and due to the presence of double nodes in $$x=0$$ and $$x=1$$. The weights of the basis functions are set such that the sum of all polynomials with unit weight represents the constant function 1, similarly to Lagrange polynomials.

The basis only contains Hermite information for degree>=3, but it is also implemented for degrees between 0 and two. For the linear case, the usual hat functions are implemented, whereas the polynomials for degree=2 are $$p_0(x)=(1-x)^2$$, $$p_1(x)=2x(x-1)$$, and $$p_2(x)=x^2$$, in accordance with the construction principle for degree 3.

These two relaxations improve the condition number of the mass matrix (i.e., interpolation) significantly, as can be seen from the following table:

Condition number mass matrix
degree HermiteInterpolation HermiteLikeInterpolation
n=3 1057 17.18
n=4 6580 16.83
n=5 1.875e+04 15.99
n=6 6.033e+04 16.34
n=10 9.756e+05 20.70
n=15 9.431e+06 27.91
n=25 2.220e+08 43.54
n=35 2.109e+09 59.51

This polynomial inherits the advantageous property of Hermite polynomials where only two functions have value and/or derivative nonzero on a face advantageous for discontinuous Galerkin methods but gives better condition numbers of interpolation, which improves the performance of some iterative schemes like conjugate gradients with point-Jacobi. This polynomial is used in FE_DGQHermite.

Definition at line 726 of file polynomial.h.

## ◆ HermiteLikeInterpolation()

 Polynomials::HermiteLikeInterpolation::HermiteLikeInterpolation ( const unsigned int degree, const unsigned int index )

Constructor for the polynomial with index index within the set up polynomials of degree degree.

Definition at line 1175 of file polynomial.cc.

## ◆ generate_complete_basis()

 std::vector< Polynomial< double > > Polynomials::HermiteLikeInterpolation::generate_complete_basis ( const unsigned int degree )
static

Return the polynomials with index 0 up to degree+1 in a space of degree up to degree.

Definition at line 1405 of file polynomial.cc.

## ◆ value() [1/3]

 double Polynomials::Polynomial< double >::value ( const double x ) const
inlineinherited

Return the value of this polynomial at the given point.

This function uses the most numerically stable evaluation algorithm for the provided form of the polynomial. If the polynomial is in the product form of roots, the evaluation is based on products of the form (x - x_i), whereas the Horner scheme is used for polynomials in the coefficient form.

Definition at line 107 of file polynomial.h.

## ◆ value() [2/3]

 void Polynomials::Polynomial< double >::value ( const double x, std::vector< double > & values ) const
inherited

Return the values and the derivatives of the Polynomial at point x. values[i], i=0,...,values.size()-1 includes the ith derivative. The number of derivatives to be computed is thus determined by the size of the array passed.

This function uses the Horner scheme for numerical stability of the evaluation for polynomials in the coefficient form or the product of terms involving the roots if that representation is used.

Definition at line 120 of file polynomial.cc.

## ◆ value() [3/3]

 void Polynomials::Polynomial< double >::value ( const Number2 x, const unsigned int n_derivatives, Number2 * values ) const
inlineinherited

Return the values and the derivatives of the Polynomial at point x. values[i], i=0,...,n_derivatives includes the ith derivative. The number of derivatives to be computed is determined by n_derivatives and values has to provide sufficient space for n_derivatives + 1 values.

This function uses the most numerically stable evaluation algorithm for the provided form of the polynomial. If the polynomial is in the product form of roots, the evaluation is based on products of the form (x - x_i), whereas the Horner scheme is used for polynomials in the coefficient form.

The template type Number2 must implement arithmetic operations such as additions or multiplication with the type number of the polynomial, and must be convertible from number by operator=.

Definition at line 142 of file polynomial.h.

## ◆ values_of_array()

 void Polynomials::Polynomial< double >::values_of_array ( const std::array< Number2, n_entries > & points, const unsigned int n_derivatives, std::array< Number2, n_entries > * values ) const
inlineinherited

Similar to the function above, but evaluate the polynomials on several positions at once, as described by the array argument points. This function is can be faster than the other function when the same polynomial should be evaluated on several positions at once, e.g., the x,y,z coordinates of a point for tensor-product polynomials.

The template type Number2 must implement arithmetic operations such as additions or multiplication with the type number of the polynomial, and must be convertible from number by operator=.

Definition at line 160 of file polynomial.h.

## ◆ degree()

 unsigned int Polynomials::Polynomial< double >::degree
inlineinherited

Degree of the polynomial. This is the degree reflected by the number of coefficients provided by the constructor. Leading non-zero coefficients are not treated separately.

Definition at line 170 of file polynomial.h.

## ◆ scale() [1/2]

 void Polynomials::Polynomial< double >::scale ( const double factor )
inherited

Scale the abscissa of the polynomial. Given the polynomial p(t) and the scaling t = ax, then the result of this operation is the polynomial q, such that q(x) = p(t).

The operation is performed in place.

Definition at line 180 of file polynomial.cc.

## ◆ scale() [2/2]

 void Polynomials::Polynomial< double >::scale ( std::vector< double > & coefficients, const double factor )
staticprotectedinherited

This function performs the actual scaling.

Definition at line 270 of file polynomial.cc.

## ◆ shift() [1/2]

 void Polynomials::Polynomial< double >::shift ( const number2 offset )
inherited

Shift the abscissa oft the polynomial. Given the polynomial p(t) and the shift t = x + a, then the result of this operation is the polynomial q, such that q(x) = p(t).

The template parameter allows to compute the new coefficients with higher accuracy, since all computations are performed with type number2. This may be necessary, since this operation involves a big number of additions. On a Sun Sparc Ultra with Solaris 2.8, the difference between double and long double was not significant, though.

The operation is performed in place, i.e. the coefficients of the present object are changed.

Definition at line 199 of file polynomial.cc.

## ◆ shift() [2/2]

 void Polynomials::Polynomial< double >::shift ( std::vector< double > & coefficients, const number2 shift )
staticprotectedinherited

This function performs the actual shift

Definition at line 277 of file polynomial.cc.

## ◆ derivative()

 Polynomial< double > Polynomials::Polynomial< double >::derivative
inherited

Compute the derivative of a polynomial.

Definition at line 205 of file polynomial.cc.

## ◆ primitive()

 Polynomial< double > Polynomials::Polynomial< double >::primitive
inherited

Compute the primitive of a polynomial. the coefficient of the zero order term of the polynomial is zero.

Definition at line 212 of file polynomial.cc.

## ◆ operator*=() [1/2]

 Polynomial< double > & Polynomials::Polynomial< double >::operator*= ( const double s )
inherited

Multiply with a scalar.

Definition at line 218 of file polynomial.cc.

## ◆ operator*=() [2/2]

 Polynomial< double > & Polynomials::Polynomial< double >::operator*= ( const Polynomial< double > & p )
inherited

Multiply with another polynomial.

Definition at line 224 of file polynomial.cc.

## ◆ operator+=()

 Polynomial< double > & Polynomials::Polynomial< double >::operator+= ( const Polynomial< double > & p )
inherited

Definition at line 230 of file polynomial.cc.

## ◆ operator-=()

 Polynomial< double > & Polynomials::Polynomial< double >::operator-= ( const Polynomial< double > & p )
inherited

Subtract a second polynomial.

Definition at line 236 of file polynomial.cc.

## ◆ operator==()

 bool Polynomials::Polynomial< double >::operator== ( const Polynomial< double > & p ) const
inherited

Test for equality of two polynomials.

Definition at line 242 of file polynomial.cc.

## ◆ print()

 void Polynomials::Polynomial< double >::print ( std::ostream & out ) const
inherited

Print coefficients.

Definition at line 248 of file polynomial.cc.

## ◆ serialize()

 void Polynomials::Polynomial< double >::serialize ( Archive & ar, const unsigned int version )
inlineinherited

Write or read the data of this object to or from a stream for the purpose of serialization using the BOOST serialization library.

Definition at line 257 of file polynomial.h.

## ◆ memory_consumption()

 std::size_t Polynomials::Polynomial< double >::memory_consumption
virtualinherited

Return an estimate (in bytes) for the memory consumption of this object.

Definition at line 263 of file polynomial.cc.

## ◆ multiply()

 void Polynomials::Polynomial< double >::multiply ( std::vector< double > & coefficients, const double factor )
staticprotectedinherited

Multiply polynomial by a factor.

Definition at line 283 of file polynomial.cc.

## ◆ transform_into_standard_form()

 void Polynomials::Polynomial< double >::transform_into_standard_form
protectedinherited

Transform polynomial form of product of linear factors into standard form, $$\sum_i a_i x^i$$. Deletes all data structures related to the product form.

Definition at line 291 of file polynomial.cc.

## ◆ coefficients

 std::vector Polynomials::Polynomial< double >::coefficients
protectedinherited

Coefficients of the polynomial $$\sum_i a_i x^i$$. This vector is filled by the constructor of this class and may be passed down by derived classes.

This vector cannot be constant since we want to allow copying of polynomials.

Definition at line 301 of file polynomial.h.

## ◆ in_lagrange_product_form

 bool Polynomials::Polynomial< double >::in_lagrange_product_form
protectedinherited

Stores whether the polynomial is in Lagrange product form, i.e., constructed as a product $$(x-x_0) (x-x_1) \ldots (x-x_n)/c$$, or not.

Definition at line 307 of file polynomial.h.

## ◆ lagrange_support_points

 std::vector Polynomials::Polynomial< double >::lagrange_support_points
protectedinherited

If the polynomial is in Lagrange product form, i.e., constructed as a product $$(x-x_0) (x-x_1) \ldots (x-x_n)/c$$, store the shifts $$x_i$$.

Definition at line 313 of file polynomial.h.

## ◆ lagrange_weight

 double Polynomials::Polynomial< double >::lagrange_weight
protectedinherited

If the polynomial is in Lagrange product form, i.e., constructed as a product $$(x-x_0) (x-x_1) \ldots (x-x_n)/c$$, store the weight c.

Definition at line 319 of file polynomial.h.

The documentation for this class was generated from the following files:


text:
0
TECHNICAL PAPERS

# On the Normal Component of Centralized Frictionless Collision Sequences

[+] Author and Article Information
Pieter J. Mosterman

The MathWorks, Inc., 3 Apple Hill Dr., Natick, MA 01760

Notation is derived from (3). Contact points will be superscripted, whereas reference to bodies around contact points and their variables are subscripted.

During continuous behavior, $v1,Ei=v1,Ai$ and $v2,Ei=v2,Ai$, as the left and right limit values of a point on a continuous $(C0)$ curve have to be equal.

Note that $gi$ is not indexed by either $E$ or $A$ as it is a $C0$ variable.

Alternatively, to achieve the same effect, the clause can be replaced by $Ik−1P=∅$.

The MATLAB code of the collision models is available upon request.

J. Appl. Mech 74(5), 908-915 (Jan 04, 2001) (8 pages) doi:10.1115/1.2712237 History: Received December 05, 2000; Revised January 04, 2001

## Abstract

A typical assumption for rigid body collisions with multiple impact points is that all collisions occur simultaneously and are synchronized in their compression/expansion behavior, a useful assumption given the microscopic time scale at which collisions occur. In the case in which collisions are dependent upon one another, however, there is interaction between and within compression and expansion phases. Instead of treating the collisions as separate consecutive impacts or by activating all constraints at the same time, a rule is presented that orders the collisions as a sequence of interacting events at a point in time to handle the normal component of the collisions.

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## Figures

Figure 2

Collision between two bodies and a floor

Figure 3

Resulting velocities for Newton’s collision rule: (a) As a function of ϵ12; (b) as a function of ϵ12 with m3 large; (c) as a function of ϵ23

Figure 4

Resulting velocities for Poission’s collision rule: (a) As a function of ϵ12; (b) as a function of ϵ12 with m3 large; (c) as a function of ϵ23

Figure 1

A sequence of dependent collisions

Figure 5

Different elasticity leads to differences in relative velocity

Figure 6

Resulting velocities for the new collision rule: (a) As a function of ϵ12; (b) as a function of ϵ12 with m3 large; (c) as a function of ϵ23

Figure 7

A sequence of four colliding bodies

Figure 8

Resulting velocities for the new collision rule: (a) As a function of ϵ12; (b) as a function of ϵ23; (c) as a function of ϵ34

Figure 9

Final velocities for m2=0.1 and 0≤ϵ23≤1

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• This is an assessment test.
• To draw maximum benefit, study the concepts for the topic concerned.
• Kindly take the tests in this series with a pre-defined schedule.

## Arithmetic: Time and Work Test-4

Congratulations - you have completed Arithmetic: Time and Work Test-4. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
Ayesha can complete a piece of work in 16 days. Amita can complete the same piece of work in 8 days. If both of them work together in how many days can they complete the same piece of work?
 A 6 days B $\displaystyle 4\frac{2}{5}$ days C $\displaystyle 5\frac{1}{3}$ days D 12 days
Question 1 Explanation:
Let the work be 16 units.
Ayesha does 1 unit/day.
Amita does 2 units / day.
Ayesh+ Amita does (1+2) = 3 units/ day.
Total number of days required is 16/3 = 5 1/3 days.
 Question 2
Two cyclists start on a circular track from a given point but in opposite directions with speeds of 7 m/s and 8 m/s respectively. If the circumference of the circle is 300 m, after what time will they meet at the starting point?
 A 20 s B 100 s C 300 s D 200 s
Question 2 Explanation:
The relative speed is 8-7 =1 m/s.
Therefore, 300 m can be traveled in 300 secs.
Thus in 300 secs, the cyclist will meet for the first time.
 Question 3
42 women can do a piece of work in 18 days. How many women would be required do the same work in 21 days?
 A 36 B 24 C 30 D 44
Question 3 Explanation:
In 18 days the number of women required to finish the work is 42.
In 1 day the number of women required to finish the work is 42 x 18.
Therefore, in 21 days the number of women required to finish the work is 42 x 18/21= 36 days.
 Question 4
A is twice as good a workman as B and together they finish a piece of work in 14 days. The number of days taken by A alone to finish the work is:
 A 11 B 21 C 28 D 42
Question 4 Explanation:
Let A do 2p units work in 1 day and thus B does p units in 1 day.
Together they finish the work in 14 days @(1+2)=3p units of work per day.
Total work = 14 x 3p = 42p units of work.
A does 42x units of work in 42p/2p= 21 days.
 Question 5
A team of workers was employed by a contractor who undertook to finish 360 pieces of an article in a certain number of days. Making four more pieces per day that was planned, they could complete the job a day ahead of schedule. How many days did they take to complete the job?
 A 8 days B 10 days C 9 days D 12 days
Question 5 Explanation:
Let us do this by options.
We must understand that whatever be the answer (say X) it must be a factor of 360 .
Therefore X and X-1 must be factors of 360.
Let us see the factors of 360 = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20…..
We can see that 9 and 10 are factors of 360.
And we knew that X and X-1 are the days.
Therefore X=10.
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
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