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# Powerchair

We wanted Alejandro to have a power wheelchair. We wanted him to be able to move through the world by his own volition and it’s clear that the ability to independently explore his environment is important for his cognitive development on multiple levels. We tried going through the typical process of obtaining a power wheelchair but met numerous obstacles:

1. A new power wheelchair was much too expensive for our family to purchase out of pocket.
2. Blue Cross Blue Shield of Illinois wouldn’t pay for one because they already paid for a medical stroller when Alejandro was 9 months old. We were told that we needed to wait 5 years. This was much too long to wait, especially during these early years which are so important developmentally.
3. Recent cuts to the Illinois Medicaid program have made it very difficult to obtain funding for such devices.
4. Alejandro is very weak and it was difficult to find an off the shelf access method that he could use. While there are a number of sensitive switches, joysticks and sensors available none of them seemed feasible as they all required a precise and delicate positioning that would be lost the moment his wheelchair hit a bump. Furthermore, finding this positional sweet spot took too much time for practical use.

Our best option was to for me to try to build something myself.

While looking for motors and wheels on Ebay I stumbled across a used adult Invacare TDX3 with MK5 electronics for relatively cheap (\$800, shipping included, batteries not). I decided that starting with this was a signifiant head start on the project and was well worth the money. I purchased the wheelchair with the intention of using the motorized base and controller and then add new seating and a customized access method. I also purchased a wheelchair programmer, a device that allows you to configure and modify the wheelchair’s internal drive and control parameters.

## Seating

The chair from Alejandro’s EasyS stroller fits him well and provides decent support as well as being reclinable and adjustable. It can detach from the stroller base and attach to a ‘Hi-Lo’ base for use around the house and it also allows you to attach a tray.

I fabricated a mating piece to be fixed to the power wheelchair base, allowing the EasyS chair to clip into it the same way it clips into the stroller base or the Hi-Lo base. Below is a photo of the piece that I built from plywood which is sturdy and adequate although at some point I’ll have someone weld up a metal version now that I have a working prototype.

Due to the adjustable and reclinable nature of the EasyS there are two sliding extruded aluminum rails that extend below and behind the seat. They protrude the most when the chair is in an upright position, which is necessary for driving. I was able to acquire a pair of shorter versions of this piece which allowed me to get the entire seat assembly as low and small as possible. I wanted the seat to be low for stability and also because I want my kid to move through the world at a height which is typical for people his age.

## MK5 Interface

None of the available methods that I investigated seemed to be feasible for Alejandro to be able to consistently and reliably control a wheelchair. It was pretty clear that I’d have to come up with something myself and I would somehow have to interface this with the wheelchair electronics in order to drive it. The Invacare MK5 electronics communicate with the user controls via a proprietary serial protocol. Other’s have tried to hack this without success and I really don’t have time to try my hand at it so I decided to find a more simple route.

I originally purchased a 1554M5 Sip & Puff interface on Ebay which connects the the MK5 electronics. The 1554M5 has a mini-DIN connector for a Proportional Attendant Control and also a DB9 connector for switch inputs. The DB9 connector pinout is documented and has inputs for forward, reverse, left and right. I tested this by connecting some pushbuttons but was disappointed with its functionality. Pressing the forward button moves the chair forward at the programmed speed and pressing the left button rotates the chair in place as well. Unfortunately pressing the forward and left buttons simultaneously moves the chair in the expected direction, but at a speed which seems to be the sum of the programmed forward and turning speeds. This is a speed which is greater than both the forward and turning speeds resulting in a somewhat jerky operation when switching headings. I played with the drive parameters for quite awhile but was never able to get any reasonably smooth transition from forward to a turn. Besides this I really wanted Alejandro to have a proportional control if at all possible.

The pinout for the Proportional Attendant Control is not published but I was able to figure it out by poking around and making some educated guesses based on the documented functionality and by following circuit board traces. There are three analog inputs for X/Y and an overall speed control, as well as a switch input to override the user control and enable the attendant control. Unfortunately the the MK5 electronics do not allow you to provide drive profile settings for the attendant control. There are no adjustments for acceleration, braking, torque etc., these are all set at some pre-programmed values. I wanted to be able to program different drive profiles for indoor, outdoor, learning etc., so my search continued.

What I finally settled on was a modified Invacare 1812 Joystick. The 1812 is a proportional (analog) joystick which connects to the MK5 electronics and can be assigned to the drive profile(s) of your choice. The 1812 contains an analog joystick and a round circuit board which measures the joystick X/Y voltages and transmits them via proprietary serial protocol to the MK5 electronics. The analog joystick outputs voltages that vary between 0V and 5V, where 2.5V is the center of each axis. I found that the 1812 that I purchased had very poor sensitivity and linearity in the left direction, even after performing a joystick throw calibration. I initially thought that I had received a broken unit so I got another one, but the second unit that I received exhibited the exact same behavior. I’m pretty sure that there isn’t a problem with the MK5 electronics or the wheelchair motors as the MPJ joystick that came with the wheelchair drives with great linearity in all directions. At this point I’m chalking it up to a firmware bug in either the 1812 or the MK5 electronics. I was able to work around this by exchanging the X and Y axes in the drive profile. This moves the poor sensitivity/linearity from the left to the reverse direction which is much less of an issue as I’m primarily concerned with steering symmetry and poor sensitivity in the reverse direction can be compensated for by adjusting the reverse speed in the drive profile.

I modified the 1812 module by replacing the analog joystick with a small circuit board that I designed which simulates its output. The circuit board uses a microcontroller and D/A converter to receive serial data packets and converts them to analog X/Y voltages. These voltages are fed to the 1812 electronics where they are converted back to digital and sent to the MK5 controller. This works great and allows me to send proportional drive commands to the wheelchair without having to deal Invacare’s proprietary protocol.

I call my 1812 adapter board SERDAC (serial D/A converter). The firmware expects to see a serial X/Y data packet every 100ms or faster. If 100ms passes without the receipt of a valid packet the SERDAC immediately outputs neutral X/Y voltages (2.5V) which stops the wheelchair. This could happen if the control data source malfunctions or if the serial data cable becomes disconnected. The SERDAC will continue to output neutral X/Y voltages until it has received valid control data for a full two seconds. The last thing I want is a runaway wheelchair.

In place of the joystick I mounted a 5-pin XLR connector on the 1812 module which connects GND, serial RX and also passes along 15V power from the wheelchair for the controller and regulated 5V from the 1812 PCB.

To test it all out I wrote a simple Max patch to convert mouse movements to X/Y serial data packets and used a Bus Pirate to interface my laptop to the modified 1812 joystick module. Success! The serdac schematic and source code can be downloaded here.

## Access

At the time of this writing Alejandro is able to move his left index finger, his right thumb and some toes on both feet, all with very little force. About six months ago he had the opportunity to demo a power wheelchair. At that point in time his right big toe was relatively strong and he was able to deflect a small, snesitive, very carefully placed joystick in both the up and down directions. His movement was essentially planar but we had hopes that he might somehow learn to also move the joystick left and right too. As it turned out, a couple of weeks later, over the course of a couple of days he suddenly lost the ability to move that toe at all. Such is the progression of a degenerative neuromuscular condition.

We’ve always hoped that Alejandro would somehow be able to drive a wheelchair with a proportional access method. It’s clear that this allows much greater control and maneuverability. It’s difficult to find proportional access methods for people who are only able to perform planar movement and it seems generally accepted that in these cases switched access is the only option. Nevertheless I was determined to do my best to find a proportional access method that he could use. Alejandro’s toes, while quite weak, seem to respond faster to his motor intentions than his two working fingers, so I decided to focus my efforts there.

I was reminded of how tracked vehicles like tanks and bulldozers are driven. In the simplest form there are two levers, each which control the speed and direction of the associated track. The vehicle is driven using these two controls and can go forward, reverse, turn or even rotate in place. I planned to emulate some variation of this method of operation using variable amounts of force applied with Alejandro’s toes.

We have a digital kitchen scale which we use primarily to weigh out Alejandro’s feeding formula. About the time that I was thinking about all of this the scale broke. I decided to take it apart to see if I could fix it, which I couldn’t, but it gave me an idea. An electronic scale contains something called a ‘load cell’. This is a machined aluminum bar which is supported at one end and has a pair of strain gages epoxied to it. Placing items on the scale applies force to the unsupported end of the load cell and causes the bar to bend by a microscopic amount. When the bar bends the strain gages stretch and compress which modulates their resistances. Using a wheatstone bridge and high gain amplifier a voltage can be generated which is proportional to the amount of force applied. With the appropriate load cell and amplifier even fractions of a gram of force can be easily measured.

I thought that a load cell and amplifier might be a good way to measure the small amounts of force that Alejandro is able to willfully apply with his toes. I took the load cell out of our broken kitchen scale and began experimenting with it. The results were good. The next challenge was to find some way to easily and repeatably position his toes to press on the sensor. What I came up with are a pair of sensor pedals that his feet strap into, pictured below.

I built the pedals by gluing a pair of sandals to platforms made of plexiglass. The top level of the platform supports the bulk of the sandal and the bottom level supports one end of a load cell. The unsupported end of the load cell has a nylon standoff attached to it which protrudes through a hole in the top level of the platform and makes physical contact with the toe of the sandal. The nylon standoff was filed down to be the precise length necessary. When force is applied to the toe of the sandal it causes the load cell to flex by a measurable yet visually imperceptible amount.

Now that I had a pair of pressure sensitive pedals that Alejandro could operate I had to somehow connect them via serial interface to the power wheelchair so that he could drive it. While I was at it I also wanted to have the option of using these same pedals to control a computer via HID protocols. I designed a circuit board which includes the load cell amplifiers, a microcontroller with A/D converters, bluetooth and USB interfaces, and power supply with support for rechargeable battery powered operation.

The controller PCB also includes a 3 axis accelerometer. The pedals are very sensitive and are capable of moving the wheelchair with only a few grams of force. I was concerned that  driving the wheelchair on bumpy surfaces could potentially cause unwanted forces to be applied to the pedals creating oscillations and instability in the system. The accelerometer was included to to offer the possibility of applying an intelligent damping algorithm to prevent this. At the time of this writing this has not been implemented or necessary.

I also designed a second daughter PCB with LED indicators and a pair of pushbutton switches. The LEDs are used to provide a visual display the forces applied to each pedal and the computed heading. The pushbutton switches are used to put the device in ‘standby’ mode, reset the system, perform calibration, and initiate a USB bootloader for firmware changes.

The wheelchair is driven using a virtual joystick who’s position is computed in polar coordinates from the combination of the forces applied to the pedals. The distance (speed) component is simply taken as the greater of the two forces. The angle (heading) component is computed from the balance of the forces applied to the pedals using the following equations:

$$\Large balance=\frac{L}{L+R} – \frac{R}{L+R}$$

$$\Large angle=\frac\pi2 + (balance \times range)$$

The result is an angle which is +/- range radians from the positive Y axis. I chose a range value of about 1.9 radians (109º).

The polar coordinates of the virtual joystick position are converted to cartesian (X/Y) coordinates and filtered using a second order lowpass filter with an fc of about 0.7 Hz, before being transmitted via serial interface to the SERDAC board mounted in the 1812 joystick housing. The lowpass filter corner frequency was chosen experimentally as the best compromise between responsiveness and smooth operation.

The 218º range of steering allows a center wheel drive wheelchair to not only rotate in place, but at the very extremes of the range it will actually make a turn while also moving in reverse. Although it is not possible to drive straight in reverse the user can still back out of a position using a ‘tacking’ type of motion. Obviously a 360º range of steering would be ideal but 218º seemed to be adequate, considering the challenges. My feeling is that a wider range of steering while providing more maneuverability also creates a less intuitive experience for the driver. At some point, once Alejandro has gained some level of mastery I may try further increasing the steering range if it seems advantageous.

Prior to each use, after Alejandro’s feet have been strapped into the pedals, a simple calibration must be performed to establish a baseline force while his toes are relaxed. When Alejandro is not intentionally applying any force to the pedals a button is pressed which stores the current values for each pedal. These values represent an offset and are subtracted from the continuously measured pedal values.

Thresholding with hysteresis is applied to the offset pedal values. This prevents small amounts of mechanical or electrical noise from causing the wheelchair to ‘creep’ while stopped. Additionally, while the current pedal values are below the threshold, the previously established offsets are very slowly adjusted to approach the current (below threshold) pedal values. This compensates for any long term mechanical or electrical drift in the system. In practice this technique has worked quite well.

I also wrote a Max patch which monitors the system via bluetooth connection. This allows me to observe the sensor values in realtime while Alejandro operates the wheelchair and make adjustments to the control algorithm.

## Success

So far the system has been working great and Alejandro has quickly been learning to drive his power wheelchair. His right foot is somewhat stronger than his left and at first he had a tendency to drive in circles, continually turning right. I was tempted to increase the sensitivity of the left pedal to compensate, but under advice from his physical therapist I resisted this temptation and eventually he has learned to compensate on his own and apply less pressure with his right foot in order to go straight.

It has been truly thrilling for us as parents to watch as Alejandro persistently works at learning to control his own movement in space. His motivation is amazing. For a kid like Alejandro there is nothing quite like finally being able to explore your environment all by yourself.

• c.o'd

nice ‘work’ 😉

• TMC

Really amazing work you’ve done. Your son is beautifully positioned and the foot switch is quite unobtrusive.

• rdie

Excellent

• Kenneth

Alejandro has a caring, knowledgable, and talented Dad. I have to assume that there are many similar parents out there who just as caring, but who lack the knowledge and talent to do something similar for their children. Is there a way to team up persons who have the ability to replicate something like this with with parents/children that have the need? I can’t think of anything more worthwhile than being able to help expand a child’s horizons like this. Surely groups for this purpose already exist? Pointers? Thanks very much.

This is something that my wife and I think about often. Each time our child achieves something amazing and new with the help of some contraption that I come up with our hearts are filled with a bittersweet joy, knowing that there are 1000 more children (or adults for that matter) out there that could benefit from the same, but the technology is out of reach for them. While there are growing number of low cost, open-source technologies available for people with disabilities, it seems to be difficult to find the people with the skills and motivation to help apply these technologies to the lives of individuals. The only organization that currently I’m aware of is the Speak Your Mind Foundation, which focuses on providing assistive technologies to people with disabilities so that they can communicate. Hopefully there are other similar groups, but I haven’t found them yet. In most cases, the requisite technologies are cheap and simple enough that they could be implemented by an engineering student or electronics hobbyist. I would very much like to see a website that serves as a repository for low cost open-source assistive technologies, as well as a way to connect people with the skills necessary to apply these technologies with the people who need them. Unfortunately I myself don’t have the time to spearhead such a project right now, but would be excited to contribute however possible if such a thing were to exist.

I also just learned about the Tetra Society.

• Sean

There is something almost like what you are talking about. http://thecontrollerproject.com/ aims to help people with disabilities by pairing makers with them to help them figure out a solution.

• I am glad that you created this hack so your son could travel through the world. Also, I wanted to tell you that your hack look very professional.

An idea that I think would be useful is combining the feet control that you created together with an eye tracking device and a neuro-headset. That could add more options to your son in order to interact with the world around him.

I wish the best for you and your family.

There are two other projects that I’m currently working on. The first is a low cost wireless EEG brain computer interface and the other is implementing the pupil headset as an augmentative communication device. Things always move more slowly than I would like around here, life can be complicated, but I’ll be sure to post what I come up with.

• I also read this and thought of Google Glass and of the brain controlled chair we had at my hackerspace. Hmmmm.

• Carlos

Your work is truly inspirational. Well done, sir!

• SkinnyV

You should be proud of the work you’ve done. This is really touching and heartwarming.

• AbdoWakileh

This is amazing. I work in medicine, and we need more people like you!!!!

• ronald bakker

i used to work for a dutch company responsible for maintaining that brand of wheelchairs amongst others for some of the states in our countrie and even for us pro’s it was realy hard to get any support from the manufacturer or even a key to get intoo the electronics/programming to adjust it for speed ramp up and down and such
hat of to you for being able to break intoo the system and get it to work with the system you build for it

• c.o'd

Some of your photos are beautiful unto themselves, page layout concise..Thank you for taking the time to explain and inspire.

• Mitch

Bless your heart for this hack! Truly wonderful and really made my day to see this, so thank you and God bless!

• Impressed

You sir, are amazing. Alejandro has one of the coolest dads ever. The amount of knowledge needed to do what you have done is astounding. God Bless.

• “Thresholding with hysteresis”, what a novel way to approach the issue. Hats off to you as a Dad and inventor. Are you an engineer? I am the the son and father of engineers. Dad got his degree in Zurich 1936, Son Boulder Co. 2011

• This is freaking awesome. Thank you for doing such a great writeup and including all the details and photos! Fantastic! Would you like to repost this on Hackabilityblog.com? A bunch of us also have a mailing list for tinkering, modding, inventing, and hacking mobility and accessibility equipment over here: https://www.noisebridge.net/mailman/listinfo/hackability

• Elizabeth B

Fantastic. Brilliant.

• My baby nephew diagnosed a year ago with a very rare type of muscular dystrophy has been hospitalized for a year now at Hopkins. This gives me hope. Thank-you for your invention and for your devotion to your son!

• dave sharley

Truly remarkable mate. loved this story and your passion. if there were more like you, the world would be a wonderful place. inspirational.

• I am truly inspired by your creativity. God bless you and your family.

• Ivan

Bravo!

• Very well done! Wish you and your family all the best!

• dogon

faith in humanity restored. you are a father, a real father. this an is inspiration for everyone. thank you so much for this!

• flo

you should visit wheelchairdriver.com! a lot of powerchairs hacks in there!

• Excellent

• Wow de verdad que eres un padre fascinante, eso es lo que se debe hacer para su amado hijo, de verdad muy buen trabajo, saludos y mucho amor y suerte para tu hijo y toda tu familia, Cuídate.

• this is a big wow…you are one talented guy and a wonderful parent, love to you
dutch, husband of blmeyer

• Eetu

You, sir, are a fine human being. Inspiring

• Marta

Well done! wish you and your family all the best !

• You remind me of my husband, Dan, Shea. He is most impressed, as we all are at Dragonmobility. Lots of our Dragonmobility followers on facebook have joined in the praise! Our project started over 30 years ago when Dan responded in a similar way as a father. We are also an SMA family, so I know how pressured you life must be just in caring for him. My love and prayers for you and your wife and to Alejandro, together with great admiration and congratulations. You have given your son a great gift.

• mburu

wow.

• S.

Congrats on your hack!. Im trying to do same hack on VR2 controller. Please, post schematics and code. Thanks!

• I was suggested this web site by means of my cousin.
I’m now not sure whether or not this put up is written by
means of him as no one else recognize such unique
approximately my trouble. You are incredible!
Thank you!

• Jeff Kang

“the other is implementing the pupil headset as an augmentative communication device”

Just in case there’s any interest:

Eye Tracking subreddit: http://www.reddit.com/r/EyeTracking

• Jeff Kang

Lastly, I’m not sure if this will help, but here’s an interesting Indiegogo project that has backing from the United Cerebral Palsy organization:

Google Glass Eye-Control App for Wheelchairs

I will create controls for powered wheelchairs using the Google Glass eye-tracking technology to provide mobility for disabled persons, with support from UCP.

Technology

The patent Google issued for eye-tracking technology as part of their Google Glass product (US 8,235,529) indicates there will be rather precise tracking of the right eye’s movement relative to objects presented in the user’s line of sight. With this level of precision, users will be able to seamlessly use subtle eye motions to select controls that change the speed and direction of the chair by staring at a particular button for a determined period of time just long enough for the app to distinguish between a casual glance and a deliberate action. Being able to rely on subtle eye motions for button-like control will also allow the app to track for more extreme eye movements to provide system overrides, such as an emergency stop.

• Tomris

This is the most beatiful thing I have ever seen for a while thank you to fulfill my heart with hope! God bless you, hope your son will be in better condition for next years. Keep up the good work!

• Great job … the embedded engineers at my office think it’s awesome, so do i.

• you must be the best Dad ever.

• Mateus Caruccio

Good job!

tip: put the video on start of the post.

• Ryan Christensen

Really great project and super awesome human, you are a great person and father to do this.

I hope to see this on kickstarter or a way to keep iterating on it with funding. Donations and crowdsourced funding for projects like this should be plenty. Hope to see you do more since you are very skilled at it.

• You do awesome work I need to learn from you.
Your son is blessed for having you as a parent!!!!!

• Ben

This is fantastic.

On the one hand you could say that it’s a pitty that you could not afford a proper wheelchair. On the other hand I think it’s great to have a challenge like this and being able to make it.

Fate often challenges people to prove their power. And you have a lot of power. Your son must be very proud of you!

I hope you will have more great ideas like this for your son. There are not many people to face this kind of challenge!

Keep up the great work!

• Armel beaudry

wooow amazing. This is awesome work.

• Wow, well done! I am a wheelchair user myself (and I was a wheelchair using child ages ago). To move around independently is so important for children. It’s a great thing you did for your son. Keep up the good work! 🙂

• The fact that you could not afford a wheelchair, hacked used one, and then gave away the process for others for free makes me tear. Hats off and all the best for you and your kid.

• This URL is an article just published in Medical Design Technology. The first paragraph explains the experiments being conducted. This may be a future solution for your son.
Excerpt from MDT online: An experimental device is letting paralyzed people drive wheelchairs simply by flicking their tongue in the right direction.


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# 400 Years of Logarithms

The logarithm was first presented in John Napier’s 1614 book Mirifici Logarithmorum Canonis Descriptio (Description of the Wonderful Canon of Logarithms). Last week I was celebrating 400 years of logarithms at the Napier 400 workshop held at the ICMS in Edinburgh and organized by NAIS. The previous such celebrations had been in 1914 and, as one speaker remarked, it is nice to participate in an event held only once every 100 years.

This one-day workshop included talks by Mike Giles on computing logarithms and other special functions on GPUs, and Jacek Gondzio on the history of the logarithmic barrier function in linear and nonlinear optimization.

My interest is in the matrix logarithm. The earliest explicit occurrence that I am aware of is in an 1892 paper by Metzler On the Roots of Matrices, so we are only just into the second century of matrix logarithms.

In my talk The Matrix Logarithm: from Theory to Computation I explained how the inverse scaling and squaring (ISS) algorithm that we use today to compute the matrix logarithm is a direct analogue of the method Henry Briggs used to produce his 1624 tables Arithmetica Logarithmica, which give logarithms to the base 10 of the numbers 1–20,000 and 90,000–100,000 to 14 decimal places. Briggs’s impressive hand computations were done by using the formulas $\log a = 2^k \log a^{1/2^k}$ and $\log(1+x) \approx x$ to write $\log_{10} a \approx 2^k \cdot \log_{10}e \cdot (a^{1/2^k} - 1)$. The ISS algorithm for the matrix case uses the same idea, with the square roots being matrix square roots, but approximates $\log(1+x)$ at a matrix argument using Padé approximants, evaluated using a partial fraction expansion. The Fréchet derivative of the logarithm can be obtained by Fréchet differentiating the formulas used in the ISS algorithm. For details see Improved Inverse Scaling and Squaring Algorithms for the Matrix Logarithm (2012) and Computing the Fréchet Derivative of the Matrix Logarithm and Estimating the Condition Number (2013).

As well as the logarithm itself, various log-like functions are of interest nowadays. One is the unwinding function, discussed in my previous post. Another is the Lambert W function, defined as the solution $W(z)$ of $W(z) e^{W(z)} = Z$. Its many applications include the solution of delay differential equations. Rob Corless and his colleagues produced a wonderful poster about the Lambert W function, which I have on my office wall. Cleve Moler has a recent blog post on the function.

A few years ago I wrote a paper with Rob, Hui Ding and David Jeffrey about the matrix Lambert W function: The solution of S exp(S) = A is not always the Lambert W function of A. We show that as a primary matrix function the Lambert $W$ function does not yield all solutions to $S \exp(S) = A$, just as the primary logarithm does not yield all solutions to $e^X = A$. I am involved in some further work on the matrix Lambert W function and hope to have more to report in due course.

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# Problem 2-1(c) Average Speed - Part 3 - B

Determine the average speed in metres per second. Express the final answer in scientific notation using the correct number of significant digits.
(C) The moon takes $27.3$ days to complete one orbit around the earth. The average earth-moon distance is $\mathrm{3.84 \times 10^5\; km.}$

Correct!

Therefore the distance the moon travels in one orbit is:

(A)   $2.413 \times10^9\; m$

(B)   $2.413 \times 10^6\; m$

(C)   $2.413 \times 10^{12} \;m$


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# Publications

2019
A. De, A. Stewart-Height, and D. E. Koditschek, “Task-Based Control and Design of a BLDC Actuator for Robotics,” RAL, pp. 8, 2019. Publisher's VersionAbstract
This paper proposes a new multi-input brushless DC motor current control policy aimed at robotics applications. The controller achieves empirical improvements in steady-state torque and power-production abilities relative to conventional controllers, while retaining similarly good torquetracking and stability characteristics. Simulations show that non-conventional motor design optimizations whose feasibility is established by scaling model extrapolations from existing motor catalogues can vastly amplify the effectiveness of this new control-strategy.
2018
A. Shamsah, A. De, and D. E. Koditschek, “Analytically-Guided Design of a Tailed Bipedal Hopping Robot”. IEEE, pp. 2237–2244, 2018. Publisher's Version
A. De and D. E. Koditschek, “Averaged anchoring of decoupled templates in a tail-energized monoped,” in Robotics Research, Springer, 2018, pp. 269–285.
A. De, S. A. Burden, and D. E. Koditschek, “A hybrid dynamical extension of averaging and its application to the analysis of legged gait stability,” The International Journal of Robotics Research, vol. 37, no. 2-3, pp. 266-286, 2018. Publisher's VersionAbstract
We extend a smooth dynamical systems averaging technique to a class of hybrid systems with a limit cycle that is particularly relevant to the synthesis of stable legged gaits. After introducing a definition of hybrid averageability sufficient to recover the classical result, we illustrate its applicability by analysis of first a one-legged and then a two-legged hopping model. These abstract systems prepare the ground for the analysis of a significantly more complicated two legged model: a new template for quadrupedal running to be analyzed and implemented on a physical robot in a companion paper. We conclude with some rather more speculative remarks concerning the prospects for further extension and generalization of these ideas.
A. De and D. E. Koditschek, “Vertical hopper compositions for preflexive and feedback-stabilized quadrupedal bounding, pacing, pronking, and trotting,” The International Journal of Robotics Research, vol. 37, no. 7, pp. 743-778, 2018. Publisher's VersionAbstract
This paper applies an extension of classical averaging methods to hybrid dynamical systems, thereby achieving formally specified, physically effective and robust instances of all virtual bipedal gaits on a quadrupedal robot. Gait specification takes the form of a three parameter family of coupling rules mathematically shown to stabilize limit cycles in a low degree of freedom template: an abstracted pair of vertical hoppers whose relative phase locking encodes the desired physical leg patterns. These coupling rules produce the desired gaits when appropriately applied to the physical robot. The formal analysis reveals a distinct set of morphological regimes determined by the distribution of the body’s inertia within which particular phase relationships are naturally locked with no need for feedback stabilization (or, if undesired, must be countermanded by the appropriate feedback), and these regimes are shown empirically to analogously govern the physical machine as well. In addition to the mathematical stability analysis and data from physical experiments we summarize a number of extensive numerical studies that explore the relationship between the simple template and its more complicated anchoring body models.
2017
A. De, “Modular Hopping and Running via Parallel Composition,” University of Pennsylvania, 2017. Publisher's VersionAbstract
Though multi-functional robot hardware has been created, the complexity in its functionality has been constrained by a lack of algorithms that appropriately manage flexible and autonomous reconfiguration of interconnections to physical and behavioral components.Raibert pioneered a paradigm for the synthesis of planar hopping using a composition of parts'': controlled vertical hopping, controlled forward speed, and controlled body attitude. Such reduced degree-of-freedom compositions also seem to appear in running animals across several orders of magnitude of scale. Dynamical systems theory can offer a formal representation of such reductions in terms of anchored templates,'' respecting which Raibert's empirical synthesis (and the animals' empirical performance) can be posed as a parallel composition. However, the orthodox notion (attracting invariant submanifold with restriction dynamics conjugate to a template system) has only been formally synthesized in a few isolated instances in engineering (juggling, brachiating, hexapedal running robots, etc.) and formally observed in biology only in similarly limited contexts.In order to bring Raibert's 1980's work into the 21st century and out of the laboratory, we design a new family of one-, two-, and four-legged robots with high power density, transparency, and control bandwidth. On these platforms, we demonstrate a growing collection of $\{$body, behavior$\}$ pairs that successfully embody dynamical running / hopping gaits'' specified using compositions of a few templates, with few parameters and a great deal of empirical robustness. We aim for and report substantial advances toward a formal notion of parallel composition---embodied behaviors that are correct by design even in the presence of nefarious coupling and perturbation---using a new analytical tool (hybrid dynamical averaging).With ideas of verifiable behavioral modularity and a firm understanding of the hardware tools required to implement them, we are closer to identifying the components required to flexibly program the exchange of work between machines and their environment. Knowing how to combine and sequence stable basins to solve arbitrarily complex tasks will result in improved foundations for robotics as it goes from ad-hoc practice to science (with predictive theories) in the next few decades.
V. Vasilopoulos, O. Arslan, A. De, and D. E. Koditschek, “Sensor-based legged robot homing using range-only target localization”. IEEE, pp. 2630–2637, 2017.
2016
G. Kenneally, A. De, and D. E. Koditschek, “Design Principles for a Family of Direct-Drive Legged Robots,” IEEE Robotics and Automation Letters, vol. 1, no. 2, pp. 900-907, 2016.
G. Wenger, A. De, and D. E. Koditschek, “Frontal plane stabilization and hopping with a 2DOF tail”. IEEE, pp. 567–573, 2016. Publisher's Version
T. T. Topping, V. Vasilopoulos, A. De, and D. E. Koditschek, “Towards bipedal behavior on a quadrupedal platform using optimal control”. pp. 98370H, 2016. Publisher's Version
2015
A. De and D. E. Koditschek, “Parallel composition of templates for tail-energized planar hopping”. IEEE, pp. 4562–4569, 2015. Publisher's Version
A. De and D. E. Koditschek, The Penn Jerboa: A Platform for Exploring Parallel Composition of Templates. 2015. Publisher's Version
A. L. Brill, A. De, A. M. Johnson, and D. E. Koditschek, “Tail-assisted rigid and compliant legged leaping”. IEEE, pp. 6304–6311, 2015.
2014
A. De, K. S. Bayer, and D. E. Koditschek, “Active sensing for dynamic, non-holonomic, robust visual servoing”. IEEE, pp. 6192–6198, 2014. Publisher's Version
A. De and D. E. Koditschek, Parallel Composition of Templates for Planar Hopping - Technical Report. University Of Pennsylvania, 2014.
2013
A. De, A. Ribeiro, W. Moran, and D. E. Koditschek, “Convergence of Bayesian histogram filters for location estimation”. IEEE, pp. 7047–7053, 2013.
M. M. Ankarali, H. T. Sen, A. De, A. M. Okamura, and N. Cowan, “Haptic Feedback Enhances Rhythmic Motor Control By Reducing Variability, Not Convergence Rate,” Journal of Neurophysiology, pp. jn.00140.2013, 2013. Publisher's VersionAbstract
Stability and performance during rhythmic motor behaviors such as locomotion are critical for survival across taxa: falling down would bode well for neither cheetah nor gazelle. Little is known about how haptic feedback, particularly during discrete events such as the heel-strike event during walking, enhances rhythmic behavior. To determine the effect of haptic cues on rhythmic motor performance, we investigated a virtual paddle juggling behavior, analogous to bouncing a table tennis ball on a paddle. Here, we show that a force impulse to the hand at the moment of ball-paddle collision categorically improves performance over visual feedback alone, not by regulating the rate of convergence to steady state (e.g. via higher gain feedback or modifying the steady-state hand motion), but rather by reducing cycle-to-cycle variability. This suggests that the timing and state cues afforded by haptic feedback decreases the nervous system's uncertainty of the ball's state to enable more accurate control, but that the feedback gain itself is unaltered. This decrease in variability leads to a substantial increase in the mean first passage time, a measure of the long-term metastability of a stochastic dynamical system. Rhythmic tasks such as locomotion and juggling involve intermittent contact with the environment (i.e. hybrid transitions), and the timing of such transitions is generally easy to sense via haptic feedback. This timing information may improve metastability, equating to less frequent falls or other failures depending on the task.
A. De and D. E. Koditschek, “Toward dynamical sensor management for reactive wall-following”. IEEE, pp. 2400–2406, 2013.
2011
A. De, G. Lynch, A. Johnson, and D. Koditschek, “Motor sizing for legged robots using dynamic task specification,” 2011 IEEE Conference on Technologies for Practical Robot Applications (TePRA). pp. 64 -69, 2011.Abstract
We explore an approach to incorporating task and motor thermal dynamics in the selection of actuators for legged robots, using both analytical and simulation methods. We develop a motor model with a thermal component and apply it to a vertical climbing task; in the process, we optimally choose gear ratio and therefore eliminate it as a design parameter. This approach permits an analytical proof that continuous operation yields superior thermal performance to intermittent operation. We compare the results of motor sizing using our proposed method with more conventional techniques such as using the continuously permissible current specification. Our simulations are run across a database of commercially available motors, and we envision that our results might be of immediate use to robot designers for motor as well as gearbox selection.
2010
A. De, “Neuromechanical Control of Paddle Juggling,” Johns Hopkins University, 2010.


text:
Sat 2021-Dec-25

# Today the Webb Space Telescope Launched

Tagged: Physics

Today the Webb Space Telescope launched successfully. Happy Christmas.

## Webb space telescope launch!

Today the James Webb Space Telescope launched from the European Spaceport in French Guiana, aboard an Ariane 5 rocket. The official video from NASA is livestreamed on the NASA YouTube channel. [1] For those of you too impatient to wade through the preliminaries, the launch itself starts around 1:21:14, which is where the video embedded here will start. (But really, should you be so brave as to listen to the whole thing, you might find yourself in grave danger of learning something.)

The official NASA site for the JWST is interesting, too. [2]

While I’m slightly sad to see that the US now has limited launch capability for such a mission (6164kg launch weight!), I’m happy to see that the Europeans have a fine space facility.

The Webb is an interesting beast: conceived as a successor to the Hubble in the early 90s, construction begain in 2004, launch was scheduled for 2018, and now finally it’s on its way in 2021. It’s been a long time coming. The upside is that there’s been relentless testing of this complex instrument and how it will unfold in space and move to its final destination. Everything has to work properly the first time, since it will operate far away from Earth and can never be serviced or have its resources replenished.

Unlike the visible-light Hubble, the Webb will operate in the infrared. This is appropriate: very old things, which are now very far away because of the expansion of the universe, are now heavily red-shifted and so IR is where the action is:

• the universe is about 13.8 billion years old
• the earliest and most distant galaxy the Hubble can see in visible light was formed at about 0.4 billion years into the universe’s life
• the earliest stars seem to have appeared when the univers was about 0.1 billion years old.

So by going to infrared, we can see about 300 million years further back than we can in visible light, to probe what’s jokingly called the “Dark Ages” whose visible light is no longer visible because it’s redshifted to infrared.

And, of course, serendipity: when you look with a new method, you almost always see something new. As your humble Weekend Editor is fond of saying:

You can’t plan serendipity — give chance a chance!

But that means the JWST detectors need to operate at cryogenic temperature, and thus have an elaborate heat shield to block the sun (and for that matter, the earth) which are sources of sufficient heat to be annoying.

The solution is to operate (sort of) at the earth-sun Lagrange L2 point, which puts both the earth and sun in the same direction, behind the heat shield. The L2 point is unstable, so it will orbit about that point, with some station-keeping ability. The picture here (click to embiggen), from Scientific American, shows how that works.

It keeps the major heat sources (the sun and the earth) behind the heat shield, is relatively gravitationally stable, and close enough to the earth for reasonably high-bandwidth communication (S-band up: 16 kbit/s, S-band down: 40 kbit/s, Ka-band down: up to 28 Mbit/s). But… it’s a million miles away, about 4 times as far as the moon. Never gonna get serviced, like the Hubble was. Whatever’s there has to last the entirety of the 10 year mission.

So it’s important that the launch go perfectly: any use of fuel to correct mistakes here is fuel lost to the rest of the mission (insofar as the booster and the telescope propulsion interact at all).

The result so far:

• successful launch, with insertion into a highly elliptic earth orbit (to do the unfolding and testing before insertion to a transfer orbit to L2)
• the telescope is confirmed to be receiving power from its solar array
• it will take about a month for it to unfold and deploy completely, test systems, and move into position at the L2 point with double-checked communication with Earth.

Wish the JWST some good luck. It’s a Happy Christmas kind of thing, for scientists.

## Notes & References

1: National Aeronautics and Space Administration, “James Webb Space Telescope Launch — Official NASA Broadcast”, NASA YouTube Channel, 2021-Dec-25.


text:
# Circuits acting on subsets of qubits in other circuits

I am trying to figure out what is the best way of dealing with registers while constructing circuits from abstract blocks. Consider the following subcircuit:

$$$$U|x\rangle|y\rangle|z\rangle = \ldots \,,\\$$$$ where neither the sizes nor the order of registers are not known in advance. Assume I want to append $$U(x,y,z)$$ and $$U(y,z,w)$$ to an existing circuit with $$x,y,z,w$$ registers. The following construction would be ideal for me, but it is not supported:

def U(
x : QuantumRegister,
y : QuantumRegister,
z : QuantumRegister,
) -> QuantumCircuit:
qc = QuantumCircuit( list(x) + list(y) + list(z)  )
qc.x(0)
qc.y(1)
qc.cx(z[0],z[1])
return qc

qc = QuantumCircuit(4)
input_circuit = QuantumCircuit( qc.qubits ) # This is more realistic than just qc
x,y,z,w = [ qc.qubits[0] ],\
[ qc.qubits[1] ],\
[ qc.qubits[2], qc.qubits[3] ],\
[ qc.qubits[3], qc.qubits[2] ]
print( U( x, y, z ) )
print( U( x, y, w ) )

However , note that the following code fails to run:

print(
input_circuit + U( x, y, z ) + U( x, y, w )
)

Yet this code works (but with a DeprecationWarning suggesting that I should instead use inequivalent compose(), see below):

print(
qc + U( x, y, z ) + U( x, y, w )
)

In other words, qiskit only works correctly with circuits having a QuantumRegister. But there is no way of assembling QuantumRegister from existing Qubits. Neither can one combine Qubits into a QuantumRegister in an existing circuit.

Instead of + one can try using compose(), but compose() ignores qubit ordering unless provided explicitly. Moreover, the qubits from the second circuit are mapped on those from the first one simply according to the order of qubits in the first one, but not according to their names — unlike the solution with + (which only works if the QuantumRegister is present in the first circuit).

Anyways, let me try to formulate the question precisely. Assume I'm starting from some input_circuit.

What is the best way of coding up functions which return circuits acting on particular subsets of input_circuit.qubits?

You can you QuantumCircuit.append(). This way you can distinguish between how the function U is called, and how the returned circuit will be embedded:
input_circuit.append( U( x, y, z ), x + y + z)

• To my understanding, the way append() interprets the second argument is similar to the way it's done in compose(): the names of qubits in the second circuit are simply ignored, and the qubits are mapped in their order of appearance in the second circuit. Unlike that, "+" is matching qubits with identical names. So I was wondering what would be the best way to implement this behavior. May 25 at 18:25


text:
### Chiral symmetry and Fermion mass non-divergences

07Mar08

One of the main motivators for new physics at the TeV scale is the quadratic divergence in the Higgs mass from radiative corrections:

$m_H^2 \sim \int d^4k \frac{1}{k^2} \sim \Lambda^2$

This means that for a natural UV completion of the Standard Model we would expect new physics entering not too far above the Higgs mass. This is a general feature of scalar particles which have renormalisable (i.e. “good at low energies”) four-point interactions.

Fermion linear divergence…

But what about fermions? Naively, doing the same analysis as above, we have the following diagrams:

The fermion needn’t interact with a vector (e.g. it could interact with a scalar), but Lorentz invariance forces the interaction to be with a boson and hence the propagators lead to the following radiative corrections to the fermion mass:

$m_f \sim \int d^4k \frac{1}{k^3} \sim \Lambda + m_f^{(0)} \log \Lambda$

So we should expect a linear divergence in the fermion masses that would point to new physics, shouldn’t we? (This, of course, would be a bit problematic since the fermions themselves span a range of masses.) Why don’t we ever talk about the fermion linear divergence?

… or not!

Because it happens to not be there. We know that in the limit of zero fermion mass, chiral symmetry protects fermion masses. Namely, $\psi \rightarrow \exp(i\alpha \gamma_5)\psi$ prevents mass terms from being generated at loop level. (The mass term violates the symmetry, and there’s no way to produce symmetry-violating terms out of symmetry-abiding Feynman rules.)

But what good is this if we know that there are explicit mass terms in the Standard Model? I.e. chiral symmetry is broken! Here’s something that might be surprising: even though the symmetry is broken, it’s still useful! (This is an example of a “life lesson from physics.”)
Chiral symmetry is broken in such a way that it is restored in the limit where all the masses go to zero. Compare this to the divergence structure above. In the limit $m_f^(0) \rightarrow 0$, we expect the one-loop mass to be zero. The logarithmic divergence indeed vanishes, but the linear divergence seems to stick around! What does this apparent inconsistency mean? The linear divergence must not be there!!

You can attach more words to this and say something about being continuously connected to the chiral theory, but the point is that consistency in the unbroken limit requires the divergence structure to only be logarithmic. Don’t get me wrong, the theory we’re talking about has broken chiral symmetry — we’re just able to extract useful information because chiral symmetry is broken in a specific way (by explicit mass terms only).

Further discussion: so what?

I should say a few further words about this and why it’s kind of neat. First of all, this is an example of what Professor Lykken calls “symmetry naturalness” (see SSI04). The fermion mass parameters may unnaturally light’ compared to the scale of the SM cutoff, but the [particularly broken] chiral symmetry gives a reason why radiative corrections don’t push the masses to this scale.

Secondly, we should note that the above discussion only holds for Dirac mass terms, $m_D\bar\psi\psi$. Majorana mass terms, $m_M\psi^\dag\psi$, are chirally invariant! This means that they are not protected against radiative corrections. We would hence expect Majorana masses to naturally live at the appropriate cutoff scale. Update (9 Mar 08): This is incorrect! Majorana masses are protected by a chiral symmetry. See my next post.

This has particular relevance in the neutrino seesaw mechanism. In the seesaw one balances a light mass against a very heavy mass. For neutrinos, the light masses come from the chirally protected Dirac mass terms from electroweak symmetry breaking. Observations of neutrino oscillations (and hence estimates of neutrino masses) suggest that the high scale must be very heavy. This is okay since this high scale comes from a Majorana mass term (allowed since the right-handed neutrino is a gauge-singlet), which is unprotected and pushed up to the cutoff scale. Viewed this way, neutrino masses are an indication of an interesting high scale where we expect new physics. Update (9 Mar 08): The reason why the Majorana mass term is large isn’t because it’s not chirally protected, but rather that it is generated by physics at a different scale.

Ok, so we’ve mentioned the scalars and fermions. What about vector bosons? Gauge symmetry helps us here in the same way chiral symmetry did for fermions. One might be clever and ask: what about vector particles that aren’t gauge bosons? Because the gauge symmetry is constructed’ rather than inherent,’ we can make any theory of vector bosons gauge-invariant, see Prof. NAH’s lecture, and hence the construction carries over. (This, by the way, is why you always see vectors as gauge particles.)

More generally, the general lesson is that symmetries that are broken in a particular way can still end up being very useful. This is what is meant by the accidental symmetries’ of the standard model. An example is the $U(1)^5$ flavour symmetry that is broken by Yukawa couplings. (See this older post, or this newer post.) Higher-dimensional operators can break such symmetries, but they are suppressed by powers of the cutoff.

I’ll make one last note, since I gave my split-SUSY talk recently. For natural theories, the Higgs mass divergence is a problem. Finely-tuned (a.k.a. “split”) supersymmetry turns this problem on its head by using it as a feature. Split-SUSY pushes the SUSY-breaking scale far above the TeV scale, which drags up the masses of the unprotected SUSY scalars. These scalars, which mediate troublesome processes like loop-level proton decay, then decouple from the low energy theory. On the other hand, the SUSY fermions, which give us nice things like a dark matter candidate, are chirally protected and can continue to hang out at the phenomenologically-interesting TeV scale.

#### 7 Responses to “Chiral symmetry and Fermion mass non-divergences”

1. 1 piscator

On split susy: aren’t the gauginos in susy all Majorana fermions? So what is it that can protect their mass, and why should they all be light?

2. Ah! Thanks for catching me on a mistake, Piscator! The gauginos are indeed Majorana fermions. They *are* protected by chiral symmetry, I was just sloppy in my statements about Majorana fermions! I’ll correct this shortly. See: http://fliptomato.wordpress.com/2008/03/09/correction-majorana-fermions-and-chiral-symmetry/

3. Funny post adventure. A further fast thought I got while reading it: Given that the failure of preservation of axial vector current is proportional to the Dirac mass, we could reverse the interpretation and *define* Dirac masses as a measurement of the failure of this preservation.

4. Hi Alejandro! Since I’ve had a bad track record with quick responses, I’ll not say anything that commits me to a particular (non)factual statement. Do Majorana mass tersm also fail to preserve the axial current? The Majorana mass term is actuall of the form nu^c nu, where the c refers to charge conjugation. There’s a sigma^2 hidden in the charge conjugation matrix, so I’m not sure. (The answer is probably simple, but that’s what I thought about my previous corrections… and subsequent corrections-to-corrections…)

5. 5 Alejandro Rivero

In any case, your series on chirality is better than the wikipedia (check the entries for Spinos, Majorana Fermion, and Majorana Equation. No, better dont check them; it is painful, and not only for extra dimensions).
About majorana mass, the great thing is that it is a mass in any case, because it appears in the equation of Klein-Gordon derived from Majorana Equation. Ex. 3.4(a) in Peskin-Schroeder, and eq 2.31 ff in Zee. There I see that the major source of notational confusion is that a solution of the Majorana’s Equation is not immediate or forcefully a Majorana spinor; the later applies when one requests the solution to be also a solution of Dirac’s Equation.

6. 6 Alejandro Rivero

I took a pause to go to the cinema () and I did not answered the question in #4. Worse, I introduced more confusiong by referring to Majorana “equation”. Kind of ashamed I am; so I will also try a non-commitment by appealing to authority. Alvaro de Rujula in http://ccdb4fs.kek.jp/cgi-bin/img/reduced_gif?198109208+4+16 says that “the axial current… becomes a conserved current in the majorana m –> 0 limit”. This limit rescues the conservation of lepton number.


text:
# Is there a maximum distance from a planet that a moon can orbit?

Given a planet that orbits a star, and a moon that orbits that planet, is it possible to define a maximum orbital radius of that moon, beyond which the moon would no longer orbit the planet, but the star instead?

I initially (naively) thought this point would be where the star's gravity outweighed that of the planet:

$$d_\text{max} = d_\mathrm p - d_\mathrm px$$

$$x = \frac{1}{\sqrt{\frac{m_\mathrm p}{m_\mathrm s}}+1}$$

Where:

$$d_\text{max} =$$ maximum orbital radius of the moon (around the planet), $$d_\mathrm p =$$ orbital radius of the planet (around the sun), $$m_\mathrm p =$$ mass of the planet, $$m_\mathrm s =$$ mass of the star.

But I quickly realised this assumption was wrong (unless my shoddy maths is wrong, which is very possible), because this gives a value of $$258\,772\ \mathrm{km}$$ using values of the Sun, Moon, and Earth. $$125\,627\ \mathrm{km}$$ closer to the Earth than the Moon's actual orbital radius (values from Wikipedia).

Is there a maximum orbital distance? How can it be calculated?

• OK, I'm not a physicist or a mathematician, but isn't the question flawed? Although we think of moons orbiting planets and planets orbiting stars, don't they actually orbit around their common centres of gravity. Therefore, don't moons already orbit the star (because they are all orbiting each other)? – Gareth Jun 20 '19 at 8:20
• @Gareth It's not that simple. The Moon doesn't orbit the Sun, but the Earth-Moon system does. As long as the Earth and the Moon are gravitationally bound, neither is individually orbiting the Sun. We tend to say that the Earth orbits the Sun, but that can be taken two ways - 1) As massive as the Moon is, it's still just a tiny fraction of the mass of Earth, so it can usually be ignored, just like all the other junk orbiting Earth, 2) We say "Earth", but really mean the "Earth-Moon system" (including all the other tiny bits of stuff gravitationally bound to the Earth). – Luaan Jun 20 '19 at 8:56
• @Gareth If you look at trajectories of Earth and the Moon relative to the Sun, they're very similar - the distance between the Earth and the Moon is tiny compared to their distance to the Sun. Both wobble, with the Earth's wobble being much smaller than the Moon's, but both of those wobble's amplitudes are too small to have much impact relative to the distance to the Sun. So in this sense, both the Earth and the Moon orbit the Sun, with some perturbations. However, all this is OT, since the OP obviously cares about when the moon stops orbiting the planet, not starts orbiting the star :P – Luaan Jun 20 '19 at 9:17
• @Luann Fair enough. And perhaps this needs a question of it's own, but if the moon were to become a trojan or greek would it be considered to be orbiting the star or the planet? Or neither? – Gareth Jun 20 '19 at 9:29
• That all seems like a matter definitions and choices what is a moon and planet etc. If you don't care about technicalities and astronomical vocabulary exactness (like is Pluto a planet) really you just have to look at the constraints necessary to get definitive answers. Like say given a velocity, masses, then it is determined if it is a stable orbit, hyperbole (escape velocity) or parabolic trajectory. Or basically different eccentricity cases, which there are three possibilities for any body in question. – marshal craft Jun 20 '19 at 12:56

The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $$m$$ is in a roughly circular orbit of radius $$a$$ about a star of mass $$M$$, then the radius of this "sphere" is given by $$r_H = a \sqrt[3]{\frac{m}{3M}}.$$ For the Sun-Earth system, this yields $$r_H \approx 0.01 \text{ AU}$$, or about 1.5 million kilometers.

The calculation given in the Wikipedia article shows how to derive this in terms of rotating reference frames. But for a qualitative explanation of why your reasoning didn't work, you have to remember that the moon and the planet are not stationary; both of them are accelerating towards the star. This means that it's not the entire weight of the moon that matters, but rather the tidal force on the moon as measured in the planet's frame. This effect, along with the fact that the centripetal force needed for the star to "steal" the planet is a bit less when the moon is between the star and the planet, leads to the expression given above.

As pointed out by @uhoh in the comments, the L1 and L2 Earth-Sun Lagrange points are precisely this distance from the Earth. These are precisely the points where the gravitational forces of the Earth and the Sun combine in such a way that an object can orbit the Sun with the same period as the Earth, but at a different radius. In a rotating reference frame, this means that the influences of the Earth, the Sun, and the centrifugal force are precisely canceling out; any closer to Earth than that, and the Earth's forces dominate. Thus, the L1 and L2 Lagrange points are on the boundary of the Hill sphere.

• I love answers where you learn something new! I'd never heard of the Hill Sphere before... Also, it seems to be a common theme in orbital dynamics to forget that everything is moving. The slingshot maneuver makes no sense at all until you realise this. – Oscar Bravo Jun 20 '19 at 8:27
• Nice answer! +1 I think that I would say "For the Sun-Earth system..." rather than the "Earth-Moon system"; the Earth's Hill sphere is defined and exists whether or not The Moon exists, it's an artifact of the Sun-Earth system. You might also mention that the Earth-Moon Lagrange points L1 and L2 are also at about 1.5 million kilometers. It's not the same thing but it's somewhat related. – uhoh Jun 20 '19 at 9:17
• Michael, note that @uhoh stated "It's not the same thing but it's somewhat related." The distances between the Sun-Earth L1 and L2 points and the Earth are the real solutions to two slightly different fifth order polynomials, neither of which has a closed form expression. The expression $a\sqrt[3]{\frac m {3M}}$ is a close approximation of these distances. Two additional comments: (1) The so-called Hill sphere isn't really a sphere, and (2) An orbit with a radius that is more than half the Hill sphere radius is most likely unstable. – David Hammen Jun 20 '19 at 13:04
• @DavidHammen: True, but the radius of the Hill "sphere" is also only approximately given by $a\sqrt[3]{\frac m {3M}}$; in fact, the exact expressions for the bounds of the Hill "sphere" are the roots of the same fifth-order polynomial that gives the distances to L1 and L2; see the derivation in the Wiki article. And yes, it's not actually a sphere in the geometric sense (hence the scare quotes on "sphere" in my second sentence), though I suppose you could argue that it's a "sphere" in the sense of a "sphere of influence." – Michael Seifert Jun 20 '19 at 13:47

What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.

• Sigh, posted while I was typing. – Maury Markowitz Jun 19 '19 at 21:01
• Heh, that's happened to me more times than I can count... – Michael Seifert Jun 19 '19 at 21:02
• It's always reassuring to have two concurring opinions ;-). – Peter - Reinstate Monica Jun 20 '19 at 15:41

There are several different ways of characterizing whether something is "really" in orbit around the sun or Earth. One is asking which gravitational force is larger. As you found, by that standard, the moon primarily orbits around the sun, with its orbit being next to Earth's. Another criteria would be looking at which body's gravitational potential energy dominates: if you wanted to fly a rocket from the moon to outside the solar system, which body's gravity would contribute more to the energy you would need? For that question, the moon is even more clearly in orbit around the sun rather than Earth. But you can also ask at what point would Earth's gravity be so weak that perturbations from the sun would take the moon out of a stable orbit, which gives you the Hill sphere, of which the moon is well within. There are other pairs of bodies, such as Pluto and Charon, for which these different definitions also give different answers.


text:
# Michelson-Morley Experiment

I've looked everywhere and I cannot find a complete derivation that includes the step I'm looking for lol...hopefully this will add another link for google.

So the full time for light to travel both directions with the ether then against as follows:

$$T = \frac{L}{(c+v)}+\frac{L}{(c-v)}\tag 1$$ then you get by adding fractions: $$T = \frac{2Lc}{(c^2-v^2)}\tag 2$$

Then everything I read online and my textbook says that equals: $$T=\frac{2L}{c}\frac{1}{1-\frac{v^2}{c^2}}\tag 3$$

then of course: $$T=\frac{2L}{c(1-\frac{v^2}{c^2})}\tag 4$$

I'm not understanding how they get from $(2)$ to $(3)$.

-
Hint: $c=\frac{1}{\frac{1}{c}}$ – John Douma Jun 12 '13 at 20:35

You just divide numerator and denominator by $c^2$:
$$T = \frac{2Lc}{(c^2-v^2)}\tag 2=\frac{\frac{2Lc}{c^2}}{\frac{(c^2-v^2)}{c^2}}=\frac{2L}{c}\frac{1}{1-\frac{v^2}{c^2}}$$
That's a very common strategy, specially in special relativity, you try to get that $v/c$ coefficient as much as you can so you're constantly dividing by $c$ :) – MyUserIsThis Jun 12 '13 at 20:47


text:
Encyclopedia > 1 E7 m

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1 E7 m

To help compare different orders of magnitude, this page lists lengths starting at 107 m (10,000 km). See also lengths of other orders of magnitude.

External link Conversion Calculator for Units of LENGTH (http://www.ex.ac.uk/cimt/dictunit/ccleng.htm)

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text: If area of quadrant of a circle is 38.5 cm 2 then find its diameter (use π = 22/7) cbse; class-10; Share It On Facebook Twitter Email. Here we are going to see the how to find area and perimeter of quadrant. The circumference is defined as the distance around the edge of a circle C. The area is defined as the number of square units contained inside the circle. Question 35: Ungrazed area. Find the area of a quadrant Q of a circle C with radius 8 cm. What is the volume generated? The area of the sector can be obtained by multiplying the circle's area by the ratio of the angle θ (expressed in radians) and 2 π (because the area of the sector is directly proportional to its angle, and 2 π is the angle for the whole circle, in radians): A = π r 2 θ 2 π = r 2 θ 2. Sector area calculator - when it may be useful? Circumference = 2πr. Learn to find the Perimeter of half and quarter circles email: scott.peter.smith@det.nsw.edu.au website: scottpetersmith.wix.com/flippedlearning Answer: First double 3 feet to give a diameter of 6 feet. (ii) Area = πr 2 or πd 2 /4 (iii) Area of semi-circle = πr 2 /2 (iv) Area of a quadrant of a circle = πr 2 /4 The area enclosed by the ellipse (x^2)/16 + (y^2)/9 = 1 on the first and 2 nd quadrant is revolved about the x … Let’s take a look at a few examples on working out the area of quadrants: Work out the area of this quadrant (radius 8cm). Question: What is the area of a quadrant with a radius of 6cm, given in terms of Pi? Divide this Area by 4: A = … Given: r = 7 cm. As quadrant is one fourth part of a circle. Let’s take a look at a few examples on working out the area of quadrants: Quadrant area = Circle area / 4 = πr² / 4. What is the quarter circle's area? Finally divide 254.34 by 4 to give 63.6 to 1 decimal place. Now divide this answer by 4 to give 153.9 cm^2 to 1 decimal place (or 49Pi). Area of quarter circle = (1/4) Π r². Try it now. = π.r/2 +2r = (22×14)/(7×2) +2×14 = 22 + 28 =50 cm. Area of Circle = 3.14 * 4. Quadrant refers to the four quarters in the coordinate plane system. First work out the area of the whole circle by substituting the radius of 8cm into the formula for the area of the circle: = 64π (leave the answer as an exact solution as this need to be divided by 4). Find the area of the shaded region. Now, Area of quadrant =1/4pi"r"^2=1/4xx22/7xx (7/2)^2=77/8 "cm"^2 Hence, the area of the quadrant of the circle is 77/8 "cm"^2. Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Answer: i) Area of quadrant ii) Area of shaded region . Equation: A = (pi)r^2. b. Area of quadrant, A1 = area of circle / 4, Your email address will not be published. A quadrant is a quarter of a circle. Perimeter of the quadrant of a circle= 2.π.r/4 + r +r. Alternatively, you could substitute the radius of the quadrant directly into the formula A = ¼ πr². (i.e.,) circumference/perimeter of a circle = 2πr, where π= 3.14159. that is, pi (π) multiplied by the radius squared (r2). Solution: A = (pi)(7 cm)^2. Try it now. 7.3 Centroid of a quadrant of a circle Area of a quadrant=1/4.pi.r^2 (1/4)(22/7).r^2=154 r^2=154×7×4/22 r=14 cm. Circumference = 12.56 m. (iii) Area of a Quadrant of the Circle. Question: If the wheel of a gate is 3 feet from the wall and it turns over 90 degrees, what is the distance covered by the wheel? , Answer Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. So to work out the area of a quadrant, first work out the area of the whole circle (use the formula A = π ×r²) and then divide the answer by 4. Problem 24 Find the area of the largest rectangle that can be cut from a circular quadrant as in Fig. Now, to calculate the area of a quadrant, divide the area of a circle by 4 (as four quadrants make a circle). A circular arc of 90°; one fourth of the circumference of a circle. We get. Area of quadrant = 1/4× area of circle = 1/4×("π"r2 ) Now, we need to find r. It is given that Circumference = 2πr 22 = 2πr 22/2 = πr 11 = πr 11/ = r The area on the first quadrant of the circle x^2+y^2=36 is revolved about the line y+10=0. So area of quadrant is given by \frac { 1 }{ 4 } × π { r }^{ 2 } where “r” is the radius of the circle. answered Oct 17, 2019 by Deepak01 (58.6k points) selected Oct 17, 2019 by Vikash Kumar . Find the area of a quadrant of a circle whose circumference is 22 cm. 1 Answer +1 vote . (r=3 mm, Pi = 3.14). Next multiply 3.14 by 6 to give the circumference of the whole circle which is 18.84 feet. The perimeter of a figure is the total distance around the edge of the figure. Area of a Quadrant = π * 2 2 / 4 We get, Area of a quadrant, A= (πr 2)/4 Square units. rant (kwŏd′rənt) n. 1. asked Mar 20 in Areas Related To Circles by Mohini01 (67.7k points) areas related to circles; class-10; 0 votes. Area of a Quadrant = πr 2 / 4. asked Aug 2, 2018 in Mathematics by vikashsoni (11.0k points) areas related to circles; class-10 ; 0 votes. Area of a Quadrant = πr 2 / 4 (i) Area of circle. 7.3 CENTROID OF A QUADRANT OF A CIRCLE. Now divide 28.26 by 4 to give 7.065 mm^2. A quadrant is a quarter of a circle. A circle is defined as the locus of all the points that are equidistant from the center. Let us see an example. If r is the radius of a circle, then (i) Circumference = 2πr or πd, where d = 2r is the diameter of the circle. In this article, let us discuss what a quadrant is, how to calculate the area of the quadrant with examples in detail. quadrant synonyms, quadrant pronunciation, quadrant translation, English dictionary definition of quadrant. Area of a quadrant, A=  (πr2)/4 Square units. Area of circle= πr 2 = 3.14(8) 2 = 200.96 cm 2. Area of a Quadrant Example. We know, we know: "why do we need to learn that, we're never ever gonna use it". As you can see it gives exactly the same answer as method 1. Formula for finding Area of Quadrant: Area of circle is π { r }^{ 2 } . To calculate the area of a quadrant of a circle, we must know the area of a circle. So to work out the area of a quadrant, first work out the area of the whole circle (use the formula A = π ×r²) and then divide the answer by 4. Question: Can you find the area of a quadrant whose radius is 9cm? If the radius of the circle is r, then the area of the quadrant is 0.25*pi*r^2 Question: Is the area of a quarter-circle supposed to be (8² x π) /4 ? Question: The radius of a quarter circle is 3 millimeters. to anywhere on the perimeter of the circle. Answer: First square the radius of 6 to give 36. Illustrated definition of Quadrant (circle): A quarter of a circle (made by two radiuses at right angles and the connecting arc) Answer: First, find the radius of the circle by dividing the circumfernece by Pi and halving the answer to give 3.501 to 3 decimal places. Answer: All you need to do is divide 100 by 4 to give 25 cm^2. Find the area of a quadrant of a circle whose circumference is 22 cm. If is , find the area of the i) quadrant an ii) shaded region. Center: a point O of a circle from where all other points are equidistant. In this video, we have solved the question "Find the area of a quadrant of a circle whose circumference is 22 cm. The formula for the area of a circle is π x radius 2, but the diameter of the circle is d = 2 x r 2, so another way to write it is π x (diameter / 2) 2.Visual on the figure below: π is, of course, the famous mathematical constant, equal to about 3.14159, which was originally defined as the ratio of a circle's circumference to its diameter. Now, to calculate the area of a quadrant Q of circle C, divide the area of the circle by 4. 1 answer. Best answer. I think you are showing an example when the radius of the quarter circle is 8. 25 • Length and area of a quadrant of a circle 2 4 1 2 360 90 2 r r r π π π = × = × = l Area of quadrant of a circle = 4 2 πr unit 2 Perimeter of a quadrant = + 2r 4 2πr ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = + = 2 2 π r 2r 2 πr unit Examples : 1 Find the area of a sector of a circle whose radius is … Area of 4 corners as quadrants of circle Area of remaining part = (area of rectangle – area of four quadrants of circles) = (300 – 38.5) m 2 = 261.5 m 2. Question: Can you find the area of quadrant of a circle whose circumference is 22? Question: What is the area of the quadrant with a radius of 14cm ? A circular arc of 90°; one fourth of the circumference of a circle. AREA AND PERIMETER OF QUADRANT. Substitute r = 8 directly into the formula A = ¼ πr². Substitute r = 3.8m directly into the formula A = ¼ πr². Like example 1, begin by substituting the radius of 3.8m into the formula for the area of the circle: = 14.44π (leave the answer as an exact solution as this need to be divided by 4). Area of Circle = 12.56 m 2 (ii) Circumference of the Circle. Question 40: For each of the two opposite corners of a square of side , a quadrant of a circle … The perimeter of a circle is known as its circumference. A = (pi)(49 cm^2) A = 153.86 cm^2. n. 1. When it comes to circle, the quarter of a circle is called a quadrant, which is a sector of 90 degrees. Answer: The area of the whole circle is Pi times 14 times 14 which gives 615.75... cm^2. Quadrant means the circle is being cut into 4 equal pieces, so use the Area of A Circle Equation and then divide your answer by 4. Method 1 (using the area of a whole circle and dividing by 4). Find the area of a quadrant of a circle whose circumference is 22 cm. Now, a quadrant is one-fourth section of a circle which is obtained when a circle is divided evenly into four sections or rather 4 quadrants by a set of two lines which are perpendicular in nature. Now divide the answer by 4 as 90 degrees is 1/4 of the whole circle to give 4.7 feet to 1 decimal place. Perimeter of a quadrant=2r+1/4. a. (2pi.r) = 2×14 +(1/4)(2×22×14/7) =28+22 =50 cm. Quadrant definition: A quadrant is one of four equal parts into which a circle or other shape has been... | Meaning, pronunciation, translations and examples 1 answer. Quadrant's central angle is a right angle (π/2 or 90°), so you'll quickly come to the same equation: Quadrant area = α * r² / 2 = πr² / 4. Each of the four sections is called a quadrant. To find the area of circle C, we need to know the following terms. A quadrant is exactly one fourth of any circle. Perimeter of a quadrant = ( (Π/2) + 2)r. It has 90 degree at the center. Alternatively, you could substitute the radius of the quadrant directly into the formula A = ¼ πr². Solution: Given, Radius, r = 8 cm. Mathematics a. Circumference = 4 * 3.14. Answer: Work out 0.25 mutlplied by Pi multiplied by 4.3^2 to give 14.5 cm^2 rounded to 1 decimal place. The plane area bounded by such an arc and two perpendicular radii. 76. Answer: Yes, the formula can be written as (radius² x π) /4. Area of quadrant: (Radius X Radius X pi(3.14 or 22/7)) dividedby 4. Area of shaded region = Area of semicircle BEC – (Area of quadrant ABDC – Area of Δ ABC) Area quadrant ABDC Radius = 14 cm Area o Ex 12.3, 15 In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. 2229.54 c. 2593.45 d. 2696.50 370. Question: If the area of a circle is 100 cm2, what's the area of one of its quadrants? ), we can see a circle with one of the quadrants ABO coloured in green colour and angle AOB makes a right angle (90o) at the centre O. In the above figure (Fig 1. Mathematics a. Question: What is the area of a quadrant with a radius of 4.3cm? Area of Circle = 3.14 * 2 2. The total area of a circle is πr2. Area of the quadrant calculator uses Area=(pi*(Radius)^2)/4 to calculate the Area, Area of the quadrant is a part of any of the four areas of the circle into which a … The area of the element is approximated as the area of the triangle given by Fig. Find the area of a quadrant of a circle whose circumference is 22 cm. Question 39: In the adjoining figure, is a quadrant of a circle with center and radius . Answer. b. Again, all you need to do now is divide the answer by 4: Area of a quadrant = 14.44π ÷4 = 16π = 11.3 m² to 3 significant figures. 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Figure shows a quadrant of a circle of radius R. The differential element shown by shaded portion subtends an angle dӨ at the centre and is located at Ө with x-axis. Question: What is the Area for 1/4 circle with radius of 6? Circumference = 2 * 3.14 * 2. When four such quadrants are joined, the structure that we get is nothing but a circle. Work out the area of this quadrant (radius 3.8m). Define quadrant. Area of Circle = πr 2. Question: What is the formula for working out the area of a quadrant? Radius: is defined as length of a line segment R from the centre point O, Diameter: is defined as a line segment D twice as the length of the. Example: Find the area of a quadrant Q of a circle C with radius 8 cm. Required fields are marked *. ⓘ Perimeter of a quadrant [P] Meter Millimeter Kilometer Decimeter Centimeter Nanometer Micrometer Mile Yard Foot Inch Exameter Light Year Terameter Petameter Gigameter Megameter Hectometer Decameter Micron Parsec Astronomical Unit League Nautical League UK Nautical League (int) Your email address will not be published. Ex 12.2, 2 Find the area of a quadrant of a circle whose circumference is 22 cm. De très nombreux exemples de phrases traduites contenant "quadrant of circle" – Dictionnaire français-anglais et moteur de recherche de traductions françaises. Generally area of a quadrant = 1/4*π*r², and circumference of a quadrant = (πr)/2 + 2r, Where r is radius, Assume π = 3.14 Given that circumference of quadrant = 22 cm => πr + 4r = 44 cm, => r(π+4) = 44cm => r(7.14) = 44 => r is nearly 6.16 cm, If we can ignore the decimal value r = 6cm, Area of a circle formula. 3924.60 b. 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Of one of its quadrants cm 2 triangle given by Fig points are... Square units is defined as the locus of all the points that are equidistant two perpendicular radii know . Give 14.5 cm^2 rounded to 1 decimal place =50 cm which is feet. Is 3 millimeters this quadrant ( radius 3.8m ) discuss What a whose... Find the area of a circle whose circumference is 22 cm to be ( 8² X π ) /4 defined. Exactly the same answer as method 1 ( using the area of circle C with radius of a circle Mohini01! X radius X radius X radius X pi ( 3.14 or 22/7 ) ) dividedby 4 the center from! = π.r/2 +2r = ( ( Π/2 ) + 2 ) r. it has 90 degree at the.... When four such quadrants are joined, the formula can be cut from a circular quadrant as in Fig to.: Yes, the structure that we get is nothing but a circle learn that, we must know area! Is pi times 14 which gives 615.75... cm^2 ( Π/2 ) + 2 ) /4 email. 3.14 or 22/7 ) ) dividedby 4 get is nothing but a circle circumference... Circle which is a quadrant of any circle / ( 7×2 ) +2×14 = 22 28! 18.84 feet 8 400 400 400 400 ) par bhi such an arc and two perpendicular radii phrases contenant! This article, let us discuss What a quadrant of a circle when four quadrants. X^2+Y^2=36 is revolved about the line y+10=0 be cut from a circular arc of 90° ; one fourth part a... In Fig 2×14 + ( 1/4 ) ( 49 cm^2 ) a = ( 22×14 ) (. 6Cm, given in terms of pi of one of its quadrants points equidistant. 2019 by Deepak01 ( 58.6k points ) selected Oct 17, 2019 by Vikash Kumar π.r/2! Π.R/2 +2r = ( pi ) ( 7 cm ) ^2 14 which gives...! And dividing by 4 to give 36π with center and radius the plane area bounded by an. Part of a quadrant whose radius is 9cm quadrant: ( radius X pi ( or..., Your email address will not be published ( πr 2 ) r. has... X^2+Y^2=36 is revolved about the line y+10=0 as its circumference dictionary definition of quadrant: ( radius X pi 3.14. 8 ) 2 = 3.14 ( 8 400 400 ) par bhi at the.! Comes to circle, we know:  why do we need to know the area of a whole which! The whole circle and dividing by 4 to give a diameter of 6 4, email! Selected Oct 17, 2019 by Deepak01 ( 58.6k points ) selected Oct 17, 2019 Deepak01! A circle is 3 millimeters ) selected Oct 17, 2019 by Vikash Kumar ( )! The quadrant directly into the formula for working out the area of circle... Cm^2 ) a = 153.86 cm^2 ever gon na use it '' 8.. ( pi ) ( 49 cm^2 ) a = 153.86 cm^2 circle / 4, Your address! Be ( 8² X π ) /4 Square units adjoining figure area of quadrant of circle is a quadrant of a circle defined., area of one of its quadrants be cut from a circular arc of 90° ; one of! Figure is the formula a = ( 1/4 ) π r² is known as its circumference 17, by... Now, to calculate the area of a quadrant of a circle is 8 never! 2×14 + ( 1/4 ) π r² 3.8m ) 36, and multiply it by π to give cm^2! Be useful give the circumference of a quadrant Q of circle / 4 is as... 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# when given voltage, reverse voltage according to scale [closed]

I am given a DC line with a variable voltage along a set scale: let's say 0-10V. I would like to send an output DC line with a voltage reversed along that scale.
For example:

• 10 Volts in becomes 0 Volts out,
• 9 Volts in becomes 1 Volt out,
• 5 Volts in becomes 5 Volts out,
• 0 Volts in becomes 10 Volts out.

Note that the polarity is not reversed.

My question is: is there device I can buy or a relatively simple wiring scheme that can get me there? My intention is to take a signal from a PID tension-controller that controls a brake, and instead power a motor when needed. Oof, good question on the current. Let's say it is one amp.

• 1) If you want an answer, ask a question. 2) Need more context; What exactly are you trying to achieve, and what for? – Dampmaskin Sep 21 '17 at 14:37
• Also mention how much current is needed. 0-10V is a standard control voltage for LED dimmers so are we talking about that? – peufeu Sep 21 '17 at 14:42
• I guess OP wants us to design a circuit for that. That's not what we do here. You can propose a circuit and we can comment on that and suggest improvements. But since you have no experience with electronics all I can tell you is: forget this. If this needs to be done get someone with experience in electronics. There are simply hundreds of things you would need to know even if we gave you a circuit design. – Bimpelrekkie Sep 21 '17 at 14:42

If you need the voltage just for indication or measurement then a simple DC differential amplifier with unity gain will do the job, because the output voltage is 10V minus input voltage: $V_{out} = 10V - V_{in}$. :


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HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# A uniform solid sphere has radius 0.2 m and density 8 x 10^3 kg/m^3. Find the moment of inertia about the tangent to its surface. - HSC Science (Electronics) 12th Board Exam - Physics

ConceptPhysical Significance of M.I (Moment of Inertia)

#### Question

A uniform solid sphere has radius 0.2 m and density 8 x 103 kg/m3. Find the moment of
inertia about the tangent to its surface. (π = 3.142)

#### Solution

Solution:
Given: R = 0.2m, ρ= 8000 Kg/m3
To find: Moment of inertia (I)

Formulae:

i. I_0=I_c+MR^2

I_0=2/5 MR^2 + MR^2

I_0=7/5 MR^2

ii. Mass (M) = volume x density

from formula 2

M=Vrho=(4/3piR^3)rho

from formula 1

I=7/5(4/3piR^3rho)R^2

=28/15 piR^5rho

=28/15xx3.14xx(2xx10^-1)^5xx8000

I=15.02 kgm2

Is there an error in this question or solution?

#### APPEARS IN

2016-2017 (July) (with solutions)
Question 2.4 | 3.00 marks
Solution A uniform solid sphere has radius 0.2 m and density 8 x 10^3 kg/m^3. Find the moment of inertia about the tangent to its surface. Concept: Physical Significance of M.I (Moment of Inertia).
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This is another important part of aerodynamics. This is a phenomenon that occurs when a solid body like an airplane travels faster then the speed of sound. The speed sound varies with temperature, humidity and pressure.

It plays a very important role in aerodynamic equations and is represented by a Mach-number. The Mach number represents the speed of object relative to its surrounding atmosphere. A mach number of 1 roughly represents a speed of 1 220 km/h.

# A concorde

Which way do you want to go?


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# Boundary conditions in a heat transfer problem

One end of a small insulated rod is attached to a support which is maintained at a constant temperature of 473 K,and once the rod has reached a uniform temperature in its entire length the insulation is removed. Now the surrounding air is at 298 K with convective heat transfer coefficient h.

I am required to find the time elasped when the temperatures at some length of rod reaches a given temperature.The radius,thermal conductivity and thermal diffusivity are all given

This is a unsteady state heat conduction and convection problem.I got the following pde $$k \frac{\partial^2 T }{\partial z^2} - \frac k\alpha \frac{\partial T}{\partial t}=\frac{2h}{R} (T-T_0)$$

where $$k$$ is thermal conductivity $$\alpha$$ is thermal diffusivity and $$T_0$$ is surrounding ambient temperature.I am facing trouble finding out the two boundary conditions.The initial condition is that at $$t=0$$ the rod is at a uniform temperature of 473K.The first boundary condition is at $$z=0$$ the temperature is 473K.However i am unable to find out the second boundary condition.The tip of the fin is not insulated and is neither maintained at a constant temperature.

• The "second boundary condition" is the prescribed value of $T_0(x,t)$. In your problem it is a constant for all $x$ and $t$, but in general it could be an arbitrary function of both $x$ and $t$. – alephzero Feb 18 '19 at 21:38
• You have a temperature BC at one end of the rod. How does the other end of the rod behave? It has heat flux. Express that mathematically. This becomes your second required BC in $z$. – Jeffrey J Weimer Feb 18 '19 at 23:23
• @user1318499 I couldn't solve the pde with one initial and one boundary condition.can you suggest a way to solve it using finite difference method to find time elasped when temperature at 80cm becomes 323 K and the rod is 160 cm long. – user471651 Feb 19 '19 at 3:36
• Using explicit finite difference we need to extend the interval to 160+1.6 cm and 160+3.2 if we choose 16 mm as our differential length. The initial condition for the extended interval should be 298K and the boundary condition for all time it is at 298 K.Am I right? @user1318499 – user471651 Feb 19 '19 at 3:39
• @alephzero sir the temperature of the fin is not maintained at 298K and it will take time to reach that temperature.Is my approach in the comments above correct?However i am getting a time limit exceeded error when trying to implement it in c++. – user471651 Feb 19 '19 at 4:04

## 1 Answer

The BC at the far end is the flux condition.

$$-k \left. \frac{dT}{dx}\right|_{x=L} = h(T - T_o)$$

See Chapter 17 in the book by Welty etal for examples of extended surfaces and their solutions at steady state. Finally, just for fun, here is an example temperature profile of an extended beam at steady state with the heat transfer BC. Your final answer will end at a comparable profile when $$m^2 = 10$$.