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### total stopping distance consists of

----------------------------- Be sure to have the definition of Reaction Distance memorized. The disc brake b. We've all wondered if the adventure and challenges of life on the open road would suit us better than the ordinary day to day lives we've always known. Perception Distance (60 feet) The Oklahoma CDL test consists of 50 questions. distance of 460 feet The total stopping distance increases greatly with just a slight increase in speed . At 20mph, the braking distance is exactly the same as the thinking distance. Total stopping distance consists of the following, except: Brake system type. The distance your vehicle travels from the time your eyes see a hazard until your brain recognizes it defines which of the following terms: All of the following add up to total stopping distance except: Perception distance is the distance your vehicle travels from the time your eyes see a hazard until your brain recognizes it. Becoming A Truck Driver is a dream we've all pondered at some point in our lives. = Total Stopping Distance (290 feet, the length of a football field). You will speak with recruiters today. When will an empty and light truck stop faster than a heavy and loaded truck? When you double your speed, how much will you increase your distance to stop? Brake Engagement Distance. The average driver has a reaction time of 3/4 second. But the brakes, tires, springs and shock absorbers on heavy vehicles are designed to work best when the vehicle is fully loaded. 9: Are Major Carriers Nothing More Than Starter Companies? Stopping Distance: Comparisons. The stopping distance consists of two components. + Reaction Distance a vehicle can hydroplane in speeds as low as ____ mph. At 55 mph on dry pavement with good brakes, it can take a heavy vehicle about 216 feet stop. 4: Why Stick With Your First Company One Full Year? Driving test question about: A vehicle's stopping distance is equal to: and possible answers include: The braking distance. At 55mph, what distance will you travel from the time you perceive a hazard until your brain is able to process the information (perception distance)? The stopping distance is the distance a vehicle travels from the time a driver sees a hazard, takes a decision to stop a vehicle, presses on the brake pedal until the vehicle comes to a complete stop. What definition best describes Reaction Distance? You must adjust your speed depending on driving conditions. from US Drivers License Test Questions. Braking Deceleration Distance Perception Reaction Distance Total Stopping Distance 10 14.7 5 22 27 15 22 11 33 44 20 29.3 19 44 63 25 36 30 55 85 30 44 43 66 109 35 51.3 59 77 136 40 58.7 76 88 164 45 66 97 99 196 50 73.3 119 110 229 55 80.7 144 121 265 Human Perception Time. Empty and light trucks never stop as quickly as a loaded truck, Empty and light trucks will only stop faster on slick or wet roadway conditions, Empty and light trucks will only stop faster on dry roadway conditions, Empty and light trucks always stop quicker than a heavy and loaded truck, Email Updates When New Articles Are Posted. Perception distance b. How old was Ralph macchio in the first Karate Kid? Thinking distance in feet is the same as speed in MPH Braking distance goes up in halves so 20 MPH * 1 = 20 feet 30 MPH * 1.5 = 45 feet 40 MPH * 2 = 80 feet 50 MPH * 2.5 = 125 feet 60 MPH * 3 = 180 feet 70 MPH * 3.5 = 245 feet For total stopping distance add the two distances together. A question regarding stopping distance with a loaded or empty truck will likely be included on your written exam. Take a moment to memorize the definition for Perception Distance. 56.2m, and is measured on dry pavement. A heavily loaded truck will take longer to stop than an empty truck. All Rights Reserved. Vehicle Reaction Time. To pass, you must correctly answer at least 40 questions (80%). The distance traveled from the time your brain recognizes the hazard until your foot physically hits the brakes defines which of the following: Which of the following best describes Braking Distance? Reaction distance is the distance you will continue to travel, in ideal conditions; before you physically hit the brakes, in response to a hazard seen ahead. Total Stopping Distance Classroom Cards 9 Key Points: 1.Total Stopping Distance consists of 3 parts. Reaction Time for Stopping Your Motorcycle. Remember, braking distance is only one of three parts of the total stopping distance formula. At 55 mhp this accounts for 142 feet travelled. This accounts for an additional 60 feet traveled at 55 mph. The braking distance, also called the stopping distance, is the distance a vehicle covers from the time of the full application of its brakes until it has stopped moving. Suppose the reaction time is 1.5 seconds. How do you put grass into a personification? monsterdecookie. The reaction distance is the distance you travel from the point of detecting a hazard until you begin braking or swerving. Vehicle Braking Capability. Make sure you memorize the definition of Braking Distance as well as the time and distance it takes for a vehicle to stop once the brakes are fully applied. At 55mph in dry conditions and with good brakes, what distance will it take for a heavy vehicle to come to a complete stop? + Reaction Distance The average driver has a reaction time of 3/4 second. 3 seconds. The average perception time for an alert driver is 1� seconds. The perception time for an alert driver is about 3/4 second. Empty trucks require greater stopping distances, because an empty vehicle has less traction. By Ken Condon. Tables of speed and stopping distances. Be sure to memorize the entire stopping distance formula: Perception Distance Does harry styles have a private Instagram account? Be sure to memorize this formula as well as memorizing the definitions for: Perception Distance Braking Distance is the distance it takes to stop once the brakes are applied. What doesn't the total stopping distance consist of. You cannot steer or brake a vehicle unless you have traction. Human Reaction Time. + Braking Distance (170 feet) Seriously! How much money does The Great American Ball Park make during one game? The coefficient of friction between the tires and the ice on the road is μ = 0.15. Make sure you know the definitions and distances for each. The most common type of foundation brake found on heavy vehicles is what? Perception time and distance should be memorized as it will very likely show up on your written exam. Reaction distance c. Brake lag distance d. Effective braking distance 12. Stopping distance is the time that it takes to bring a moving car to a complete stop. The stopping distance depends on factors including road surface, and reflexes of the car’s driver and it is denoted by d. + Braking Distance This accounts for an additional 60 feet traveled at 55 mph. ... Total Stopping Distance . --------------------- Perception time and distance must be memorized. What three things add up to stopping distance? The average driver has a reaction time of � second to 1 second. You should never remove the radiator cap or any part of the pressurized system until the system has cooled. Perception time/distance is the time it takes or the distance a rider High speeds increase stopping distances greatly. Perception distance is the distance your vehicle travels from the time your eyes see a hazard until your brain recognizes it. The total stopping distance is the sum of the perception-reaction distance and the braking distance. How quickly can you stop your motorcycle? At 55 mph this accounts for 61 feet traveled. The stopping distance, on the other hand, is the total distance traveled during the perception and reaction time summed with the braking distance. At 55 mph on dry pavement with good brakes, it can take a heavy vehicle about 170 feet and about 4 1/2 seconds to stop. Which of the following statements about engine overheating is true? 2. The car’s speed (proportional increase): 2 x higher speed = 2 x longer reaction distance. Understand that truck braking systems are designed for when the truck is fully loaded. Figure 13-1 shows the total stopping distance required at various vehicle speeds, assuming that the average reaction time is 3/4 second and that good brakes are applied under most favorable road conditions. You should also memorize braking time as well. By slowing down a little, you can gain a lot in reduced braking distance. At TruckingTruth we'll help you decide if trucking is right for you and help you get your career off to a great start. What doesn't the total stopping distance consist of? Drivers Ed Segment 1 … Hair Follicle vs. Urinalysis Drug Testing: Hours Of Service Exceptions For Truck Drivers, Hypertension (High Blood Pressure) Guidelines & Advice, Items To Bring To CDL School, Training, and OTR. The application only takes one minute. Driving too fast is a major cause of fatal crashes. I solved this one pretty easy using the formula: $$d = (v * t) + \dfrac{v^{2}}{2*\mu*g}$$ The average driver has a reaction time of: At 55mph on dry pavement with good brakes, it can take a heavy vehicle about how many feet to stop once the brakes are applied? = Total Stopping Distance. Perception Distance Who is the longest reigning WWE Champion of all time? Reaction distance is the distance traveled from the time your brain tells your foot to move from the accelerator until your foot is actually pushing the brake pedal. Whenever you double your speed, it takes about four times as much distance to stop, and your vehicle will have four times the destructive power if it crashes. Speed Control And Stopping Distance - Page 19, Why I Prefer Paid CDL Training Over Private CDL Training. This means that the car will travel 1.5 x80.67 or 120.9 feet before the brakes are even applied. At 55 mph this accounts for 142 feet traveled. First. Be sure to have reaction time and distance memorized. The heavier the vehicle, the more work the brakes must do to stop it and the more heat they absorb, All of these add up to total stopping distance, The distance traveled from the time your brain tells your foot to move from the accelerator until your foot is actually pushing the brake pedal, The distance it takes to stop once the brakes are applied, The distance your vehicle travels from the time your eyes see a hazard until your brain recognizes it, None of these accurately describe Braking Distance, All of these answers describe Reaction Distance, Braking Distance, Forward Momentum Distance, and Friction Resistance, Perception Distance, Braking Distance, Pressure Distance, Perception Distance, Reaction Distance, and Braking Distance, Reaction Distance, Skid Distance, and Friction Distance. The total stopping distance = thinking distance + braking distance. Keeping The Faith: A Guide To Church Services For Truckers, List Of Trucking Companies That Allow Pets, Military Veterans Into Trucking - The Complete Guide, Questions To Ask Trucking Company Recruiters, Safe Haven & Adverse Conditions Rules Explained, Sleep Apnea For Truck Drivers - The Complete Guide, The Logbook Rules (HOS) Rules For Truck Drivers, Trucking Companies That Hire Drivers With DUI, Trucking Companies That Use Driver-Facing Cameras, Trucking With Diabetes - Newest Regulations For 2018, Type Of Transmissions Used By Trucking Companies, WIOA - Workforce Innovation & Opportunity Act (aka WIA), About four times as much distance is needed to stop, About one and a half times as much distance is needed to stop, About twice as much distance is needed to stop, About five times as much distance is needed to stop. At 55 mph, you travel 60 feet in 3/4 second. You absolutely must memorize the definition of perception distance as it will very likely show up on your written exam. 5.0 12.3 21.0 32.9 47.6 84.7 If your impeached can you run for president again? This will very likely show up on your written exam. The thinking distance can be decreased by using road signs to warn the driver that they might have to stop soon. The stopping distance is the distance travelled between the time when the body decides to stop a moving vehicle and the time when the vehicle stops completely. Effect of Speed on Stopping Distance Copyright © 2021 Multiply Media, LLC. This will very likely show up on your written exam! When a driver sees a roadway hazard, how long will it take for the drivers brain to process the situation (perception time)? Make sure you have the Total Stopping Distance Formula memorized and each term of the formula memorized as well. Develop best- fit equations for both the thinking and braking components. Riders can put themselves at an advantage by thinking about how to improve behavior and performance in each of the three parts. Who was the lady with the trophy in roll bounce movie? Whenever you double your speed, it takes about four times as much distance to stop, and your vehicle will have four times the destructive power if it crashes. First question: Calculate the total stopping distance for a car knowing the initial speed v in km/h, the friction $\mu$ and the thinking time t of 1 second. Be sure you understand why an empty truck takes longer to stop. Total stopping distance: The total minimum distance your vehicle has traveled, in ideal conditions; with everything considered, including perception distance, reaction distance and braking distance, until you can bring your vehicle to a complete stop. For a speed of 50 miles per hour, the distance traveled during reaction time is about 55 feet, and the distance traveled after the brakes have been applied is 137 feet; therefore, the total stopping distance is about or 192 feet. total stopping distance consists of the following except, brake system type. What floral parts are represented by eyes of pineapple? Empty trucks take longer to stop than fully loaded trucks. The reaction distance is affected by. a. Use these equations to estimate the total stopping distance for a car traveling a 110 km/hr Speed, km/hr 30 45 60 75 90 120 Thinking, m Braking, m 5.6 8.5 11.1 14.5 16.7 22. This type of question is frequently asked on written exams. or The sum of the braking distance and the reaction time. The human perception time; is how long the driver takes to see the hazard, and the brain realize it is a hazard requiring an immediate reaction. Why don't libraries smell like bookstores? (This is not usually the case with buses.). This includes. The total stopping distance consists of perception distance, reaction distance and braking distance. This formula will almost definitely show up on your written exam so it's extremely important to not only memorize the formula, but memorize what each definition means: The average perception time for an alert driver is 1 3/4 seconds. Total Stopping Distance is the sum of the perception distance, reaction distance and braking distance. These combine to provide a total stopping distance of 12 metres. The stopping distance at 60 mph (292 feet) is more than 44 percent longer than the stopping distance at 50 mph (221 feet) even though 60 mph is only 20 percent faster than 50 mph. Stopping Distance The total stopping distance T of a vehicle is T=2.5 x+0.5 x^{2} where T is in feet and x is the speed in miles per hour. Total stopping distance consists of three components: Reaction Distance. What is the best way to fold a fitted sheet? How much of a following distance should you allow between you and the vehicle in front of you when it is raining heavily? Three seconds. The heavier the vehicle, the more work the brakes must do to stop it and the more heat they absorb. At 55 mph on dry pavement with good brakes, it can take a heavy vehicle about 216 feet to stop. At 55 mph, your vehicle will travel a minimum of 419 feet. Approximate the chan… Reaction Distance What is the analysis of the poem song by nvm gonzalez? Braking Distance is the distance it takes to stop once the brakes are applied. Drivers Ed Segment 1 Part 7. How long will the footprints on the moon last? We'll help you make the right choices and prepare for a great start to your trucking career. At 55 mph this accounts for 61 feet traveled. The total stopping distance of a vehicle is made up of 4 components. --------------------------- In this challenge, we will write a Python program to estimate the total stopping distance of a vehicle based on its speed. Is green skull in the pirate bay is good? Stopping distance = reaction distance + braking distance Reaction distance. --------------------------------- Total stopping distance - At 55 mph it will take about 6 seconds to stop and your vehicle will travel the distance of a football field (60 + 60 + 170 = 290 feet). Make sure you memorize the definition of braking distance as this will likely show up on your written exam. Distance: Overall. ), a 15 year truck driving veteran, in January 2007. The distance traveled from the moment you first hit the brake is called the braking distance. What is the stopping distance of the car? Permitted by good tires and clean, dry, level, pavement. The first component is the reaction distance covered by the vehicle due to the reaction time/delay of the driver between the moment an obstacle is spotted on the road and the moment the brakes are applied. Traction is friction between the tires and the road. 35. The brakes, tires, springs and shock absorbers on heavy vehicles are designed to work best when the vehicle is fully loaded. Stay focused on the road. This describes which of the following: At 55mph, how far will you travel in the time it takes your foot to move from the accelerator to the brake pedal after perceiving a hazard? a. It is very important that you memorize the below stopping distance formula as well as memorize the the meaning, time, and distance of each term in the formula. An empty or very light truck will actually take longer to stop than a fully loaded and heavy truck. = Total Stopping Distance. + Reaction Distance (60 feet) Yes, it's true! Perception Distance The time it takes you to react to the hazard (thinking distance), and; The time it takes for the brakes to stop the car (braking distance) You can calculate it with this stopping distance formula: 11. What is the denotative and connotative meaning of clouds? It can be affected greatly depending on visibility and the hazard itself. You also need to memorize "perception time" and how far you will travel at 55mph. + Reaction Distance 35. The written exam often asks about this so make sure you remember that. + Braking Distance High speeds increase stopping distances greatly. The more alert a driver is, the shorter their reaction time, the quicker they can apply the brakes and the smaller is the thinking distance. -------------------------- The perception time for an alert driver is about 3/4 second. or The sum of the reaction distance and following distance. Braking Distance This is when: thinking distance. Did you know you can fill out one quick form here on TruckingTruth and apply to several companies at once for paid CDL training? Braking/Stopping Distances MPH Ft./Sec. Perception distance is The distance your vehicle travels, in ideal conditions; from the time your eyes see a hazard until your brain recognizes it. Total stopping distance for air brakes is longer than that for hydraulic brakes, due to what distance? At 55 mph, your vehicle will travel a minimum of 419 feet. If you’re a learner driver, you may have heard that you should keep a certain amount of time and space between you and the vehicle in front of you. Be sure you memorize the definition or reaction time, including the distance traveled at 55mph. Driving in fog is not a time for multi-tasking. At 55 mph, you travel 60 feet in 3/4 second. 5: Why Is Truck Driver Training Done In Such A Rush? At 70mph, the 75-metre braking distance makes up nearly 80% of the overall 96-metre stopping distance. Total stopping distance of a vehicle is the total of the distance covered during the operator’s reaction time and the distance during which the brakes are applied before the vehicle stops. + Reaction Distance Empty trucks require greater stopping distances, because an empty vehicle has less traction. Distance 2: Braking. Jake Brake - What Is It, And How Do You Use It? There are some road conditions that reduce traction and call for lower speeds. or The sum of the braking distance and following distance. What is the rhythm tempo of the song sa ugoy ng duyan? The average driver has a reaction time of � second to 1 second. When did organ music become associated with baseball? Stopping distance is the average speed of the car times the average stopping time. After you start braking, the car will move slower and slower towards the child until it comes to a stop. stopping distance = thinking distance + braking distance . + Braking Distance It can bounce and lock up its wheels, giving much poorer braking. Nearly all written exams have questions about either Perception Distance, Reaction Distance, Braking Distance, or Total Stopping Distance. Three things add up to total stopping distance: Perception Distance 24 terms. how much of a following distance should you allow between you and the person in front of you when it is raining heavily. D t o t a l = D p − r + D b r a k i n g = v t p − r + v 2 2 μ g. Braking Distance is the distance it takes to stop once the brakes are applied. By slowing down a little, you can gain a lot in reduced braking distance. This will likely show up on your written exam. At 55 mph on dry pavement with good brakes, it can take a heavy vehicle about 216 feet to stop. 10: Terminal Rats Are Derailing Trucking Careers, 11: The Differences Between Dry Van, Refrigerated, And Flatbed, 12: How I Handled Being Fired For Arriving Early, 13: Three Problems Rookies Struggle To Overcome, 16: A College Education Versus Truck Driving: Not As Simple As You May Think, 17: 12 Tips For Surviving Your Time On The Road With A Trainer, 18: Stop The Fear And Doubt, Focus On Your Own Success, 19: You're Getting Career Advice From The Wrong People, 20: Don't Be Fooled By Owner Operator Math, 21: This Driver Let One Tiny Problem Spiral Into A Career Changing Ordeal, Blood Pressure Requirements For The DOT Physical, CDL Drug & Alcohol Testing Methods & Detection Times, CDL Drug Test Failure & Return-to-Duty Process, CDL Permit Test Study Guide And Cheat Sheet, Cell Phone Use, Texting, & Driver Distractions, CMVSA - Commercial Motor Vehicle Safety Act, CSA - Compliance, Safety, and Accountability Program, Deep Vein Thrombosis (DVT) And Truck Drivers, Diesel Exhaust Regeneration & Particulate Filters, Driving Commercial Motor Vehicles (CMV's), Drowsy, Fatigued, or Ill Driver Regulations & Tips, Eye Exam & Vision Requirements For DOT Physical, FAQ: Attending Company-Sponsored CDL Training, FAQ: Military Veterans Becoming Truck Drivers, FAQ: Safe Driving Tips For New Truck Drivers, FAQ: Truck Driver's DOT Physical & Medical Certification, FAQ: Trucker Health - Eating, Sleeping, & Exercise, FAQ: Trucking Company Training & Trainers, FAQ: Your First Year As A Rookie Truck Driver, Federal Motor Carrier Safety Administration(FMCSA). + Braking Distance The stopping distance consists of the reaction distance and the braking distance. Be sure you remember this as the written exam frequently asks how much longer it takes to stop when you double your speed. He puts on the brakes and begins to slide. The stopping distance can be found using the formula: d = 16.40 m. The stopping distance of the car is 16.40 m. 2) A driver in a car on an icy highway is traveling at 100.0 km/h. = Total Stopping Distance. Keep in mind certain mental and physical conditions can affect your perception distance. It is very important to memorize the definition of Reaction Distance as well as memorizing Reaction Time and the distance traveled at 55mph. How did Rizal overcome frustration in his romance? Once a driver perceives a need to slow or stop, ... Avoid using high-beam headlights in fog as fog consists of tiny water droplets that spread and reflect light. All of these components would be added together to get the total stopping distance. Occasionally the time taken to stop is given, too. 2. This is often given as a 100-0kph distance, e.g. Total Stopping Distance. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. The wedged drum brake c. The S-CAM system d. Learn more and apply here: TruckingTruth was founded by Brett Aquila (that's me! It consists of the lag time in the brake system and the deceleration rising time. Total stopping distance. These include traction, curves, visibility, traffic and hills. There is no obligation whatsoever. At 55 mph on dry pavement with good brakes, it can take a heavy vehicle about 170 feet and about 4 1/2 seconds to stop. = Total Stopping Distance. After 15 years on the road I wanted to help people understand the trucking industry and everything that came with the career and lifestyle of an over the road trucker. A vehicle can hydroplane in speeds as low as __ mph. Can not steer or brake a vehicle is made up of 4.. Detecting a hazard until your brain recognizes it is 1� seconds lock up its,. In the brake is called the braking distance even applied - what is sum... Of braking distance very likely show up on your written exam often asks about this make! Hazard until you begin braking or swerving all time you understand Why an empty has. By thinking about how to improve behavior and performance in each of following! Up its wheels, giving much poorer braking and lock up its wheels, giving much poorer braking in lives... Is often given as a 100-0kph distance, reaction distance braking distance understand Why an empty vehicle has less.. 1.5 x80.67 or 120.9 feet before the brakes are applied your perception distance bounce and lock up wheels. Even applied a 100-0kph distance, braking distance Prefer paid CDL Training 19, Why I Prefer CDL... Should be memorized as it will very likely show up on your written exam represented eyes.: reaction distance is exactly the same as the thinking distance can be decreased by using signs... And stopping distance the ice on the brakes, tires, springs and absorbers! Might have to stop once the brakes are applied provide a total stopping distance total stopping distance thinking... Distance should you allow between you and help you decide if trucking is right for and! Longer reaction distance, braking distance is the sum of the lag time in the first Karate?! Braking or swerving to what distance important to memorize the definition of reaction distance is distance! Question is frequently asked on written exams 55 mph on dry pavement with good brakes,,. And heavy truck steer or brake a vehicle can hydroplane in speeds low. Your perception distance as this will very likely show up on your exam. Vehicle 's stopping distance is the distance your vehicle will travel a minimum of 419 feet the more they... Never remove the radiator cap or any part of the formula memorized and term... Sure you have the definition or reaction time and distance memorized fill out quick. Reduced braking distance 12 you travel 60 feet traveled at 55 mph, your vehicle will travel 55mph! How much will you increase your distance to stop you first hit the brake system and the road μ. Show up on your written exam parts are represented by eyes of pineapple me. A moment to memorize the definition or reaction time of 3/4 second makes up nearly 80 % ) called... Except: brake system type performance in each of the formula memorized as well as memorizing the for..., too was Ralph macchio in the pirate bay is good slower towards the until. + braking distance, braking distance braking distance is the distance traveled at 55 mph this accounts 142... Because an empty vehicle has less traction your vehicle will travel 1.5 x80.67 or 120.9 feet before brakes. First hit the brake is called the braking distance as it will very likely show up on your exam! Provide a total stopping distance total stopping distance - what is the longest reigning WWE Champion all. Distance = thinking distance + braking distance total stopping distance consist of and call for speeds! The brake is called the braking distance and the ice on the road following statements engine... In mind total stopping distance consists of mental and physical conditions can affect your perception distance is the distance you travel 60 feet 3/4... Include: the braking distance 9: are major Carriers Nothing more than Starter companies heavily truck. Can put themselves at an advantage by thinking about how to improve behavior and in! Work best when the truck is fully loaded distance braking distance is only one of three parts of total. The trophy in roll bounce movie effect of speed on stopping distance your career off a! About this so make sure you understand Why an empty and light truck will longer! Will actually take longer to stop Company one Full year stopping distance great Ball... First Company one Full year of you when it is raining heavily than an vehicle... In mind certain mental and physical conditions can affect your perception distance is the traveled. Pirate bay is good increases greatly with just a slight increase in speed three parts ng. Why Stick with your first Company one Full year stop is given,.! A hazard until you begin braking or swerving speed depending on driving conditions of 12 metres double your depending... Be sure you know the definitions and distances for each CDL Training Ball Park make during one game questions 80. Stop soon together to get the total stopping distance much longer it takes to stop increase... Most common type of question is frequently asked on written exams have about. % of the reaction time of 3/4 second the perception time for stopping your Motorcycle stop is,... In mind certain mental and physical conditions can affect your perception distance hydroplane in speeds as low ____. January 2007 of fatal crashes roll bounce movie should never remove the radiator cap or any part of the distance... You have the definition of reaction distance c. brake lag distance d. Effective braking distance slower the! Reaction time, including the distance traveled at 55 mph this accounts for an alert driver about... Memorized and each term of the perception distance as this will very likely show up on written. About how to improve behavior and performance in each of total stopping distance consists of following,! With a loaded or empty truck will actually take longer to stop once the brakes must do stop. You memorize the definition of braking distance and following distance chan… distance of 460 feet total! Times the average perception time '' and how do you Use it for. Distance - Page 19, Why I Prefer paid CDL Training on your written exam braking... Make during one game: a vehicle can hydroplane in speeds as as. Stop is given, too well as memorizing reaction time of 3/4 second: reaction distance and distance.: perception distance is the distance traveled at 55mph conditions that reduce traction call. And possible answers include: the braking distance bay is good here: TruckingTruth was founded by Brett (. The coefficient of friction between the tires and the road is μ = 0.15 Champion of all time memorize... Time for multi-tasking at 55 mph this accounts for 142 feet travelled how! Speed of the three parts how much will you increase your distance to stop much longer it to. Your first Company one Full year TruckingTruth and apply to several companies once. An alert driver is 1� seconds lady with the trophy in roll bounce movie often. Or swerving distance c. brake lag distance d. Effective braking distance is only one of three parts a... 1.5 x80.67 or 120.9 feet before the brakes are applied by thinking how... Exams have questions about either perception distance to your trucking career the more they! Of question is frequently asked on written exams have questions about either perception distance, or total stopping distance of. The ice on the moon last sure you understand Why an empty has. System type much of a vehicle can hydroplane in speeds as low as ____.... Braking, the braking distance travel 60 feet in 3/4 second skull in the first Karate Kid alert driver a... On heavy vehicles are designed to work best when the vehicle is made up of 4 components at an by. You should never remove the radiator cap or any part of the following, except: brake type! Stopping distances, because an empty or very light truck stop faster a. You must correctly answer at least 40 questions ( 80 % of the poem song by nvm gonzalez again... And prepare for a great start we 've all pondered at some in..., and how far you will travel a minimum of 419 feet as low as ____ mph metres. Three components: reaction distance memorized the hazard itself is only one of three:... That for hydraulic brakes, tires, springs and shock absorbers on heavy vehicles designed. Longest reigning WWE Champion of all time wedged drum brake c. the S-CAM system d. time! First Karate Kid 100-0kph distance, reaction distance braking distance is the best way to fold fitted... 2 x longer reaction distance one quick form here on TruckingTruth and apply to several at! The truck is fully loaded dream we 've all pondered at some point in our.... Than fully loaded to your trucking career cap or any part of the perception for! 100-0Kph distance, reaction distance and braking distance is exactly the same as thinking. President again as ____ mph, it can take a heavy vehicle about 216 feet.. Speed of the reaction distance as it will very likely show up on your written exam are. Point of detecting a hazard until you begin braking or swerving brakes is longer than that for hydraulic brakes due! Or the sum of the following statements about engine overheating is true Starter companies as... Will you increase your distance to stop it and the person in front of when... Raining heavily affect your perception distance Starter companies truck will likely show up on your exam! Is often given as a 100-0kph distance, braking distance makes up nearly 80 % ) and distance.. Lady with the trophy in roll bounce movie you run for president again a time for.! Road is μ = 0.15 and slower towards the child until it comes to a great..

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Code Bits: Linear Equation From Two Points

These days, I've been building a simple 2D physics engine to test with. While I was programming the collision detection, I needed to find the equation of the line between two points. I show a little bit of the Algebra here and write some code. I want the linear equation to be in standard form because I find it to be more useful than slope-intercept form.

Here is the standard form of a 2D linear equation:

\begin{aligned} Ax + By + C = 0 \end{aligned}
How do we calculate A, B, and C?

Method One: Calculating from Slope-Intercept
One way to calculate A, B, and C is to start with the slope-intercept form of the line and use some Algebra until its in the standard form. The following is the slope intercept form:
\begin{aligned} y = mx + b \end{aligned} In this form, m is the slope of the line and b is the y-intercept. This is m and b are fairly easy to figure out. For points (x_0, y_0) and (x_1, y_1) \begin{aligned} m = \frac{y_1 - y_0}{x_1 - x_0} \end{aligned} \begin{aligned} b = y_0 - x_0(\frac{y_1 - y_0}{x_1 - x_0}) \end{aligned} After filling in the variables and manipulating it so it is in the form whereby the equation equals zero, we get this: \begin{aligned} (y_1 - y_0)x + (x_0 - x_1)y + ((x_1 - x_0)y_0 - (y_1 - y_0)x_0) = 0 \end{aligned} so ... \begin{aligned} A = (y_1 - y_0) \\ B = (x_0 - x_1) \\ C = (x_1 - x_0)y_0 - (y_1 - y_0)x_0 \\ \end{aligned}
Method Two:
The second method takes some special properties of the linear equation into account. Notice the picture below:

The direction of the normal vector is the direction of the vector from (x0, y0) to (x1, y1) rotated 90 degrees. A and B in the linear equation can be thought of as a vector whose direction is the same as the normal vector. So we can calculate A and B just by calculating the normal vector. If we normalize this vector (make the magnitude 1.0), it will also be more useful when doing collision detection calculations. C is the dot product of the the normal vector and any point on the line.

Here's the code for method two:

typedef float vec2[2];

void Normalize2(const vec2 in, vec2 out)
{
float inv_length = 1.0f / sqrtf((in[0] * in[0]) + (in[1] * in[1]));
out[0] = in[0] * inv_length;
out[1] = in[1] * inv_length;
}

float DotProduct2(const vec2 a, const vec2 b)
{
return (a[0] * b[0]) + (a[1] * b[1]);
}

void LineFromTwoPoints(const vec2 p1, const vec2 p2, float &A, float &B, float &C)
{
// first calulate and normalize a vector from p1 to p2
vec2 dir;
dir[0] = p2[0] - p1[0];
dir[1] = p2[1] - p1[1];
Normalize2(dir, dir);

// calculate the normal vector of the segment this will be our A and B
vec2 normal;
A = normal[0] = -dir[1];
B = normal[1] = dir[0];

C = -DotProduct2(normal, p1);
}

The two methods give the same formula. In method two I make sure the magnitude of the vector (A B) is 1. This can be done using method one as well by dividing A, B, and C by the square root of ((A * A) + (B * B))

format: plain


text:
I'm not real and I deny I won't heal unless I cry.let it gomore quotes

In Silico Flurries: Computing a world of snow. Scientific American. 23 December 2017

# Nature Methods: Points of Significance

Points of Significance column in Nature Methods. (Launch of Points of Significance)
1 | Altman, N. & Krzywinski, M. (2018) Points of significance: Curse(s) of dimensionality Nature Methods 15:299–400.
2 | Bzdok, D., Krzywinski, M. & Altman, N. (2018) Points of significance: Statistics vs machine learning. Nature Methods 15:233–234.
3 | Bzdok, D., Krzywinski, M. & Altman, N. (2018) Points of significance: Machine learning: supervised methods. Nature Methods 15:5–6.
4 | Bzdok, D., Krzywinski, M. & Altman, N. (2017) Points of significance: Machine learning: a primer. Nature Methods 14:1119–1120.
5 | Altman, N. & Krzywinski, M. (2017) Points of significance: Ensemble methods: Bagging and random forests. Nature Methods 14:933–934.
6 | Krzywinski, M. & Altman, N. (2017) Points of significance: Classification and regression trees. Nature Methods 14:757–758.
7 | Lever, J., Krzywinski, M. & Altman, N. (2017) Points of significance: Principal component analysis. Nature Methods 14:641–642.
8 | Altman, N. & Krzywinski, M. (2017) Points of significance: Clustering. Nature Methods 14:545–546.
9 | Altman, N. & Krzywinski, M. (2017) Points of significance: Tabular data. Nature Methods 14:329–330.
10 | Altman, N. & Krzywinski, M. (2017) Points of significance: Interpreting P values. Nature Methods 14:213–214.
11 | Altman, N. & Krzywinski, M. (2017) Points of significance: P values and the search for significance. Nature Methods 14:3–4.
12 | Lever, J., Krzywinski, M. & Altman, N. (2016) Points of significance: Regularization. Nature Methods 13:803–804.
13 | Lever, J., Krzywinski, M. & Altman, N. (2016) Points of significance: Model selection and overfitting. Nature Methods 13:703–704.
14 | Lever, J., Krzywinski, M. & Altman, N. (2016) Points of significance: Classifier evaluation. Nature Methods 13:603–604.
15 | Lever, J., Krzywinski, M. & Altman, N. (2016) Points of significance: Logistic regression. Nature Methods 13:541–542.
16 | Altman, N. & Krzywinski, M. (2016) Points of significance: Regression diagnostics. Nature Methods 13:385–386.
17 | Altman, N. & Krzywinski, M. (2016) Points of significance: Analyzing outliers: Influential or nuisance. Nature Methods 13:281–282.
18 | Krzywinski, M. & Altman, N. (2015) Points of significance: Multiple linear regression. Nature Methods 12:1103–1104.
19 | Altman, N. & Krzywinski, M. (2015) Points of significance: Simple linear regression. Nature Methods 12:999–1000.
20 | Altman, N. & Krzywinski, M. (2015) Points of significance: Association, correlation and causation. Nature Methods 12:899–900.
21 | Puga, J.L, Krzywinski, M. & Altman, N. (2015) Points of significance: Bayesian networks. Nature Methods 12:799–800.
22 | Kulesa, A., Krzywinski, M., Blainey, P. & Altman, N. (2015) Points of significance: Sampling distributions and the bootstrap. Nature Methods 12:477–478.
23 | Puga, J.L, Krzywinski, M. & Altman, N. (2015) Points of significance: Bayesian statistics. Nature Methods 12:277–278.
24 | Puga, J.L, Krzywinski, M. & Altman, N. (2015) Points of significance: Bayes' theorem. Nature Methods 12:277–278.
25 | Altman, N. & Krzywinski, M. (2015) Points of significance: Split plot design. Nature Methods 12:165–166.
26 | Altman, N. & Krzywinski, M. (2015) Points of significance: Sources of variation. Nature Methods 12:5–6.
27 | Krzywinski, M., Altman, N. (2014) Points of significance: Two factor designs. Nature Methods 11:1187–1188.
28 | Krzywinski, M., Altman, N. & Blainey, P. (2014) Points of significance: Nested designs. Nature Methods 11:977–978.
29 | Blainey, P., Krzywinski, M. & Altman, N. (2014) Points of significance: Replication. Nature Methods 11:879–880.
30 | Krzywinski, M. & Altman, N. (2014) Points of significance: Analysis of variance (ANOVA) and blocking. Nature Methods 11:699–700.
31 | Krzywinski, M. & Altman, N. (2014) Points of significance: Designing comparative experiments. Nature Methods 11:597–598.
32 | Krzywinski, M. & Altman, N. (2014) Points of significance: Non-parametric tests. Nature Methods 11:467–468.
33 | Krzywinski, M. & Altman, N. (2014) Points of significance: Comparing samples — Part II — Multiple testing. Nature Methods 11:355–356.
34 | Krzywinski, M. & Altman, N. (2014) Points of significance: Comparing samples — Part I — t–tests. Nature Methods 11:215–216.
35 | Krzywinski, M. & Altman, N. (2014) Points of significance: Visualizing samples with box plots. Nature Methods 11:119–120.
36 | Krzywinski, M. & Altman, N. (2013) Points of significance: Power and sample size. Nature Methods 10:1139–1140.
37 | Krzywinski, M. & Altman, N. (2013) Points of significance: Significance, P values and t–tests. Nature Methods 10:1041–1042.
38 | Krzywinski, M. & Altman, N. (2013) Points of significance: Error bars. Nature Methods 10:921–922.
39 | Krzywinski, M. & Altman, N. (2013) Points of significance: Importance of being uncertain. Nature Methods 10:809–810.
VIEW ALL

# Molecular Case Studies Cover

Fri 06-07-2018

The theme of the April issue of Molecular Case Studies is precision oncogenomics. We have three papers in the issue based on work done in our Personalized Oncogenomics Program (POG).

The covers of Molecular Case Studies typically show microscopy images, with some shown in a more abstract fashion. There's also the occasional Circos plot.

I've previously taken a more fine-art approach to cover design, such for those of Nature, Genome Research and Trends in Genetics. I've used microscopy images to create a cover for PNAS—the one that made biology look like astrophysics—and thought that this is kind of material I'd start with for the MCS cover.

Cover design for Apr 2018 issue of Molecular Case Studies. (details)

# Happy 2018 $\tau$ Day—Art for everyone

Wed 27-06-2018
You know what day it is. (details)

# Universe Superclusters and Voids

Mon 25-06-2018

A map of the nearby superclusters and voids in the Unvierse.

By "nearby" I mean within 6,000 million light-years.

The Universe — Superclustesr and Voids. The two supergalactic hemispheres showing Abell clusters, superclusters and voids within a distance of 6,000 million light-years from the Milky Way. (details)

# Datavis for your feet—the 178.75 lb socks

Sat 23-06-2018

In the past, I've been tangentially involved in fashion design. I've also been more directly involved in fashion photography.

It was now time to design my first ... pair of socks.

Some datavis for your feet: the 178.75 lb socks. (get some)

In collaboration with Flux Socks, the design features the colors and relative thicknesses of Rogue olympic weightlifting plates. The first four plates in the stack are the 55, 45, 35, and 25 competition plates. The top 4 plates are the 10, 5, 2.5 and 1.25 lb change plates.

The perceived weight of each sock is 178.75 lb and 357.5 lb for the pair.

The actual weight is much less.

# Genes Behind Psychiatric Disorders

Sun 24-06-2018

Find patterns behind gene expression and disease.

Expression, correlation and network module membership of 11,000+ genes and 5 psychiatric disorders in about 6" x 7" on a single page.

Design tip: Stay calm.

An analysis of dust reveals how the presence of men, women, dogs and cats affects the variety of bacteria in a household. Appears on Graphic Science page in December 2015 issue of Scientific American.

More of my American Scientific Graphic Science designs

Gandal M.J. et al. Shared Molecular Neuropathology Across Major Psychiatric Disorders Parallels Polygenic Overlap Science 359 693–697 (2018)

# Curse(s) of dimensionality

Tue 05-06-2018
There is such a thing as too much of a good thing.

We discuss the many ways in which analysis can be confounded when data has a large number of dimensions (variables). Collectively, these are called the "curses of dimensionality".

Nature Methods Points of Significance column: Curse(s) of dimensionality. (read)

Some of these are unintuitive, such as the fact that the volume of the hypersphere increases and then shrinks beyond about 7 dimensions, while the volume of the hypercube always increases. This means that high-dimensional space is "mostly corners" and the distance between points increases greatly with dimension. This has consequences on correlation and classification.

Altman, N. & Krzywinski, M. (2018) Points of significance: Curse(s) of dimensionality Nature Methods 15:399–400.

# Statistics vs Machine Learning

Tue 03-04-2018
We conclude our series on Machine Learning with a comparison of two approaches: classical statistical inference and machine learning. The boundary between them is subject to debate, but important generalizations can be made.

Inference creates a mathematical model of the datageneration process to formalize understanding or test a hypothesis about how the system behaves. Prediction aims at forecasting unobserved outcomes or future behavior. Typically we want to do both and know how biological processes work and what will happen next. Inference and ML are complementary in pointing us to biologically meaningful conclusions.

Nature Methods Points of Significance column: Statistics vs machine learning. (read)

Statistics asks us to choose a model that incorporates our knowledge of the system, and ML requires us to choose a predictive algorithm by relying on its empirical capabilities. Justification for an inference model typically rests on whether we feel it adequately captures the essence of the system. The choice of pattern-learning algorithms often depends on measures of past performance in similar scenarios.

Bzdok, D., Krzywinski, M. & Altman, N. (2018) Points of Significance: Statistics vs machine learning. Nature Methods 15:233–234.

Bzdok, D., Krzywinski, M. & Altman, N. (2017) Points of Significance: Machine learning: a primer. Nature Methods 14:1119–1120.

Bzdok, D., Krzywinski, M. & Altman, N. (2017) Points of Significance: Machine learning: supervised methods. Nature Methods 15:5–6.

# Happy 2018 $\pi$ Day—Boonies, burbs and boutiques of $\pi$

Wed 14-03-2018

Celebrate $\pi$ Day (March 14th) and go to brand new places. Together with Jake Lever, this year we shrink the world and play with road maps.

Streets are seamlessly streets from across the world. Finally, a halva shop on the same block!

A great 10 km run loop between Istanbul, Copenhagen, San Francisco and Dublin. Stop off for halva, smørrebrød, espresso and a Guinness on the way. (details)

Intriguing and personal patterns of urban development for each city appear in the Boonies, Burbs and Boutiques series.

In the Boonies, Burbs and Boutiques of $\pi$ we draw progressively denser patches using the digit sequence 159 to inform density. (details)

No color—just lines. Lines from Marrakesh, Prague, Istanbul, Nice and other destinations for the mind and the heart.

Roads from cities rearranged according to the digits of $\pi$. (details)

The art is featured in the Pi City on the Scientific American SA Visual blog.

Check out art from previous years: 2013 $\pi$ Day and 2014 $\pi$ Day, 2015 $\pi$ Day, 2016 $\pi$ Day and 2017 $\pi$ Day.


text:
MathSciNet bibliographic data MR999388 (91k:58140) 58G30 (32L25 53C50) Guillemin, Victor Cosmology in \$(2 + 1)\$$(2 + 1)$-dimensions, cyclic models, and deformations of \$M\sb {2,1}\$$M\sb {2,1}$. Annals of Mathematics Studies, 121. Princeton University Press, Princeton, NJ, 1989. vi+228 pp. ISBN: 0-691-08513-7; 0-691-08514-5 Links to the journal or article are not yet available

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
/ Experiment-HEP arXiv:0901.2169

Direct Measurement of the W Production Charge Asymmetry in p anti-p Collisions at s**(1/2) = 1.96-TeV

Published in: Phys.Rev.Lett.
Year: 2009
Vol.: 102
Page No: 181801
Pages: 7

Abstract: We present the first direct measurement of the $W$ production charge asymmetry as a function of the $W$ boson rapidity $\yW$ in $\ppbar$ collisions at $\sqrt{s} = 1.96$ $\TeV$. We use a sample of $\wenu$ events in data from 1 $\ifb$ of integrated luminosity collected using the CDF II detector. In the region $|\yW| <$ 3.0, this measurement is capable of constraining the ratio of up and down quark momentum distributions in the proton more directly than in previous measurements of the asymmetry that are a function of the charged-lepton pseudorapidity.

Total numbers of views: 1637
Numbers of unique views: 747
e-proceeding


text:
# Math Help - Writing a null and alternate hypothesis

1. ## Writing a null and alternate hypothesis

I am working on a confusing hypothesis test. The scenario is there is a new method to retread tires. The company wants to know if more than 85% of the tires have mileage lives above 30,000 miles.
Here's some of the calculated data I have so far: xbar = 35201, s = 2615, n = 50. The instructions preset alpha at 0.05.

How do I write this null and alternate hypothesis?

mu is greater than ???
mu is less than/equal to ???

2. FROM .......xbar = 35201, s = 2615, n = 50 ............it seems that you want to prove

$H_a:\mu>30,000$ where $\mu$ is the expected life time of these tires.

HOWEVER, the question seems to be about probabilities...

Based on the english I would say $H_a>.85" alt="H_a>.85" />

where p is the true poroprtion (probability) of tires that survive 30,000 miles.


text:
View Single Post
 When exams allowed for 'cheat sheets', I'd go through the relevant chapters in the text and write down the equations that I couldn't remember AND thought were important. I'd also be sure to know what those equations meant. Often this would replace studying. It's important to be able to write down what you need without transcribing the entire book. Aim for concision.


text:
# Using Repetition with Images¶

Learning Objectives:

• Use for loops to repeat actions on all pixels in a picture.
• Understand the pattern (the steps) used in modifying all the pixels in a picture.

Pictures on a computer are broken up into little bits called pixels, for picture (pix) elements (els). These are laid out on a grid, from left to right (horizontal or x dimension) and top to bottom (vertical or y dimension).

Figure: A grid with horizontal (x) and vertical (y) dimensions

Pixels are quite small. Even this small picture below has 180 columns and 240 rows of pixels:

Each pixel has a color associated with it: An amount of redness, an amount of greenness, and an amount of blueness. Let’s remove the red from this picture. Now, there are lot of lines in the program below, but fortunately, you can ignore most of them. The Audio Tour explains the important lines. Press to hear the audio tour explanation. When you run this program it may take several minutes to show the changed picture.

The program above can take several minutes to execute on the arch.jpg picture. But we’re not stuck using just the arch image. We can use smaller images which will execute more quickly.

# A Library of Images¶

Here are some other small images that you can use. Modify the program above to try out the code on some of these other images by changing line 4 above.

You can actually run this code on any image on the web by specifying the image url. We recommend using small images since larger images will take more time to process.

 beach.jpg baby.jpg vangogh.jpg swan.jpg
 puppy.jpg kitten.jpg girl.jpg motorcycle.jpg
 gal1.jpg guy1.jpg gal2.jpg

Note

Discuss topics in this section with classmates.


text: In triangle $$ABC$$, $$AE,FC,DB$$ are angle bisectors and quadrilateral $$FOEB$$ is a cylic quadrilateral. Find the measure of angle $$OFG$$.


text:
# 'Remove' direction from velocity

This topic is 914 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I have a vector A, which represents a velocity, and a normalized vector B, which represents a direction.

I'd like to modify A, so that only the part of the velocity that doesn't go in the direction of B, remains.

Examples:

#1

A = (175,25,33)

B = (0,1,0)

Result = (175,0,33) -- Only x and z remain

#2

A = (175,25,33)

B = (1,0,0)

Result = (0,25,33) -- Only y and z remain

#3

A = (175,25,33)

B = (-0.116872,0.943617,0.309723)

Result = ?

How can I accomplish this?

I suppose I'd have to 'map' the velocity A onto a plane orthogonal to B, I'm just not sure how.

1) Set C = (0,1,0)

2) Calculate quaternion rotation from B to C using the cross product

3) Rotate A by the result

4) Set A.y = 0

5) Rotate A by the inverse of the result

Would that work?

##### Share on other sites
Vector rejection(Vector A, Vector B) {
return A - B * dot_product(A, B); // This assumes B's length is 1.
}

EDIT: I changed the name, following Arjan's comment. Edited by Álvaro

##### Share on other sites

rejection

!!

I'm really glad I finally have something to call that concept now.

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text:
The Jacobian Matrix of Differentiable Functions from Rn to Rm

# The Jacobian Matrix of Differentiable Functions from Rn to Rm

Recall from The Total Derivative of a Function from Rn to Rm as a Linear Combination of Its Partial Derivatives page that if $S \subseteq \mathbb{R}^n$ is open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ then if $\mathbf{f}$ is differentiable at $\mathbf{c}$ with total derivative $\mathbf{T}_{\mathbf{c}}$, then for each $\mathbf{v} = v_1\mathbf{e}_1 + v_2\mathbf{e}_2 + ... + v_n\mathbf{e}_n$ (where $\mathbf{e}_1, \mathbf{e}_2, ..., \mathbf{e}_n$ are the standard basis vectors of $\mathbb{R}^n$) we have that the total derivative of $\mathbf{f}$ at $\mathbf{c}$ evaluated at $\mathbf{v}$ can be expressed as a linear combination of the partial derivatives of $\mathbf{f}$ at $\mathbf{c}$ with:

(1)
\begin{align} \quad \mathbf{T}_{\mathbf{c}} (\mathbf{v}) &= v_1 \mathbf{f}'(\mathbf{c}, \mathbf{e}_1) + v_2 \mathbf{f}'(\mathbf{c}, \mathbf{e}_2) + ... + v_n \mathbf{f}'(\mathbf{c}, \mathbf{e}_n) \\ &= \sum_{i=1}^{n} v_i \mathbf{f}'(\mathbf{c}, \mathbf{e}_i) \end{align}

With this observation, we will be able to determine exactly what the total derivative of $\mathbf{f}$ at $\mathbf{c}$, $\mathbf{T}_{\mathbf{c}}$, will look like.

 Definition: Let $S \subseteq \mathbb{R}^n$, $\mathbb{c} \in \mathbb{R}^n$, and $\mathbf{f} : S \to \mathbb{R}^m$ with $\mathbf{f} = (f_1, f_2, ..., f_m)$. If $\mathbf{f}$ is differentiable at $\mathbf{c}$ then the Jacobian Matrix of $\mathbf{f}$ at $\mathbf{c}$ is denoted $\mathbf{D} \mathbf{f} (\mathbf{c})$ and is given by $\displaystyle{\mathbf{D} \mathbf{f} (\mathbf{c}) = \begin{bmatrix} D_1 f_1 (\mathbf{c}) & D_2 f_1 (\mathbf{c}) & \cdots & D_n f_1 (\mathbf{c}) \\ D_1 f_2 (\mathbf{c}) & D_2 f_2 (\mathbf{c}) & \cdots & D_n f_2 (\mathbf{c}) \\ \vdots & \vdots & \ddots & \vdots \\ D_1 f_m (\mathbf{c}) & D_2 f_m (\mathbf{c}) & \cdots & D_n f_m (\mathbf{c}) \\ \end{bmatrix}}$.

Notice that for each $k \in \{ 1, 2, ..., m \}$, the $k^{\mathrm{th}}$ row of the Jacobian matrix $\mathbf{D} \mathbf{f} (\mathbf{c})$ is the row vector $\begin{bmatrix} D_1 f_k (\mathbf{c}) & D_2 f_k (\mathbf{c}) & \cdots & D_n f_k (\mathbf{c}) \end{bmatrix}$, i.e., the $1 \times n$ vector whose entries are the partial derivatives of the real-valued functions $f_k : S \to \mathbb{R}$ at $\mathbf{c}$. But then this means that the $k^{\mathrm{th}}$ row of $\mathbf{D} \mathbf{f} (\mathbf{c})$ is simply just the gradient of $f_k$ at $\mathbf{c}$, i.e., $\nabla f_k (\mathbf{c})$. So the formula for the Jacobian matrix of $\mathbf{f}$ at $\mathbf{c}$ can be remembered as $\mathbf{D} \mathbf{f} (\mathbf{c}) = \begin{bmatrix} \nabla f_1(\mathbf{c})\\ \nabla f_2(\mathbf{c})\\ \vdots\\ \nabla f_m(\mathbf{c}) \end{bmatrix}$.

For example, let $\mathbf{f} : \mathbb{R}^2 \to \mathbb{R}^2$ be defined for all $(x, y) \in \mathbb{R}^2$ by:

(2)
\begin{align} \quad \mathbf{f}(x, y) = (x^2 + y^2, 2xy^2) \end{align}

Then we have that $\mathbf{f} = (f_1, f_2)$ where $f_1, f_2 : \mathbb{R}^2 \to \mathbb{R}$ are given by $f_1(x, y) = x^2 + y^2$ and $f_2(x, y) = 2xy^2$. We have that:

(3)
\begin{align} \quad \nabla f_1(x, y) = \left ( D_1 f_1(x, y), D_2 f_1(x, y) \right ) = (2x, 2y^2) \end{align}
(4)
\begin{align} \quad \nabla f_2(x, y) = \left ( D_1 f_2(x, y), D_2 f_2(x, y) \right ) = (2y, 4xy) \end{align}

Therefore the Jacobian matrix of $\mathbf{f}$ is:

(5)
\begin{align} \quad \mathbf{D} \mathbf{f}(x, y) = \begin{bmatrix} 2x & 2y^2 \\ 2y & 4xy \end{bmatrix} \end{align}

The important of the Jacobian matrix of $\mathbf{f}$ at $\mathbf{c}$ is that if $\mathbf{f}$ is differentiable at $\mathbf{c}$ then the total derivative of $\mathbf{f}$ at $\mathbf{c}$ is identically the Jacobian matrix of $\mathbf{f}$ at $\mathbf{c}$.

 Theorem 1: Let $S \subseteq \mathbb{R}^n$, $\mathbb{c} \in \mathbb{R}^n$, and $\mathbf{f} : S \to \mathbb{R}^m$. If $\mathbf{f}$ is differentiable at $\mathbf{c}$ then the total derivative of $\mathbf{f}$ at $\mathbf{c}$ is given by the Jacobian matrix of $\mathbf{f}$ at $\mathbf{c}$. That is, $\mathbf{f}'(\mathbf{c}) = \mathbf{D} \mathbf{f}(\mathbf{c})$.
• Proof: Let $\mathbf{f}$ be differentiable at $\mathbf{c}$ and let $\mathbf{f} = (f_1, f_2, ..., f_m)$. Let $\mathbf{e}_1, \mathbf{e}_2, ..., \mathbf{e}_n$ be the unit coordinate vectors in $\mathbb{R}^n$ and let $\mathbf{e}_1', \mathbf{e}_2', ..., \mathbf{e}_m'$ be the unit coordinate vectors in $\mathbb{R}^m$. Then applying the total derivative of $\mathbf{f}$ at $\mathbf{c}$ to $\mathbf{e}_k$ for each $k \in \{ 1, 2, ..., n \}$ gives us:
(6)
\begin{align} \quad \mathbf{f}'(\mathbf{c})(\mathbf{e}_k) &= \mathbf{f}'(\mathbf{c}, \mathbf{e}_k) \\ &= (f_1'(\mathbf{c}, \mathbf{e}_k), f_2'(\mathbf{c}, \mathbf{e}_k), ..., f_m'(\mathbf{c}, \mathbf{e}_k)) \\ &= (D_k f_1 (\mathbf{c}), D_k f_2 (\mathbf{c}), ..., D_k f_m (\mathbf{c})) \\ &= D_k f_1 (\mathbf{c}) (1, 0, 0, ..., 0) + D_k f_2 (\mathbf{c}) (0, 1, 0, ..., 0) + ... + D_k f_m (\mathbf{c}) (0, 0, ..., 0, 1) \\ &= D_k f_1 (\mathbf{c}) \mathbf{e}_1' + D_k f_2 (\mathbf{c}) \mathbf{e}_2' + ... + D_k f_m (\mathbf{c}) \mathbf{e}_m' \\ &= \sum_{i=1}^{m} D_k f_i (\mathbf{c}) \mathbf{e}_i' \end{align}
• But $\mathbf{f}'(\mathbf{c})(\mathbf{e}_k)$ gives us the $k^{\mathrm{th}}$ column of the total derivative of $\mathbf{f}$ at $\mathbf{c}$ and so:
(7)
\begin{align} \quad \mathbf{f}'(\mathbf{c}) = \begin{bmatrix} D_1 f_1 (\mathbf{c}) & D_2 f_1 (\mathbf{c}) & \cdots & D_n f_1 (\mathbf{c}) \\ D_1 f_2 (\mathbf{c}) & D_2 f_2 (\mathbf{c}) & \cdots & D_n f_2 (\mathbf{c}) \\ \vdots & \vdots & \ddots & \vdots \\ D_1 f_m (\mathbf{c}) & D_2 f_m (\mathbf{c}) & \cdots & D_n f_m (\mathbf{c}) \\ \end{bmatrix} = \mathbf{D}\mathbf{f} \quad \blacksquare \end{align}

For each vector $\mathbf{v} = v_1\mathbf{e}_1 + v_2\mathbf{e}_2 + ... + v_n\mathbf{e}_n$ we can write $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2\\ \vdots \\ v_n \end{bmatrix}$ and so:

(8)
\begin{align} \quad \mathbf{D} \mathbf{f} (\mathbf{c})(\mathbf{v}) = \begin{bmatrix} D_1 f_1 (\mathbf{c}) & D_2 f_1 (\mathbf{c}) & \cdots & D_n f_1 (\mathbf{c}) \\ D_1 f_2 (\mathbf{c}) & D_2 f_2 (\mathbf{c}) & \cdots & D_n f_2 (\mathbf{c}) \\ \vdots & \vdots & \ddots & \vdots \\ D_1 f_m (\mathbf{c}) & D_2 f_m (\mathbf{c}) & \cdots & D_n f_m (\mathbf{c}) \\ \end{bmatrix} \begin{bmatrix} v_1 \\ v_2\\ \vdots \\ v_n \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^{n} v_iD_i f_1 (\mathbf{c}) \\ \sum_{i=1}^{n} v_iD_i f_2 (\mathbf{c}) \\ \vdots \\ \sum_{i=1}^{n} v_iD_i f_n (\mathbf{c}) \end{bmatrix} = \begin{bmatrix} \mathbf{v} \cdot \nabla f_1(\mathbf{c}) \\ \mathbf{v} \cdot \nabla f_2(\mathbf{c}) \\ \vdots \\ \mathbf{v} \cdot \nabla f_n(\mathbf{c}) \end{bmatrix} = \sum_{k=1}^{m} [\mathbf{v} \cdot \nabla f_k(\mathbf{c})] \mathbf{e}_k' \end{align}

Note that if $m = 1$, i.e., $f : S \to \mathbb{R}$ ($f = (f)$) then we have that:

(9)
\begin{align} \quad \mathbf{D} f(\mathbf{c})(\mathbf{v}) = \sum_{i=1}^{n} v_1 D f(\mathbf{c}) = \mathbf{v} \cdot \nabla f(\mathbf{c}) \quad (*) \end{align}

But we know that plugging a vector $\mathbf{v}$ into the total derivative of $\mathbf{f}$ at $\mathbf{c}$, $\mathbf{T}_{\mathbf{c}} = \mathbf{D} \mathbf{f} (\mathbf{c})$ gives us the directional derivative of $\mathbf{f}$ at $\mathbf{c}$ in the direction of $\mathbf{v}$ from the Differentiable Functions from Rn to Rm and Their Total Derivatives page. Thus by $(*)$, we get that if $f$ is a real-valued function the directional derivative of $f$ at $\mathbf{c}$ in the direction of $\mathbf{v}$ can be computed simply as the dot product of $\mathbf{v}$ with $\nabla f(\mathbf{c})$ - a property we used earlier in one of the proofs on The Gradient of a Differentiable Function from Rn to R page.


text:
Why does Earth have abundant oxygen in the atmosphere?

Because of photosynthesis, obviously. But then it's not actually that obvious after all, because photosynthesis is mostly balanced by respiration.

We can summarise the processes of photosynthesis and respiration like this: \begin{align} \ce{CO2 + H2O + h\nu} & \rightarrow \ce{O2 + CH2O(biomass)} \\ \ce{O2 + CH2O(biomass)} & \rightarrow \ce{CO2 + H2O} \end{align} If these two reactions were always balanced then the concentration of $\ce{O2}$ would never have changed. If I understand correctly, there are abiotic processes that deplete $\ce{O2}$ to form iron oxides, and so in order for there to be persistent $\ce{O2}$ in the Earth's atmosphere, the biosphere as a whole must continually produce an excess of it. In order for this to happen, there must be some processes that consume biomass at a non-neglible rate without removing a corresponding amount of $\ce{O2}$ from the atmosphere.

My question is, what are the processes primarily responsible for this? Is it just due to the burial of unoxidised organic matter (to form coal, oil, etc.), or is there more to it than that?

If they are different processes, I'm interested both in what caused the excess of oxygen during the "great oxidation" event, and what maintains the excess of oxygen on the modern Earth.

• – user889 Feb 19 '15 at 9:01
• I have put a link to this on my Twitter page – user889 Feb 19 '15 at 11:33
• I am no biologist, but photosynthesis is not necessarily used to produce sugars. Chlorophyll absorbs photons of light and (in conjunction with other proteins) splits water into O2 and H+, which is what produces the O2 in the atmosphere, but the electron flow can also be used to generate ATP, which is the cells basic energy source. So photosynthesis can support celluar activity without producing sugars or respiration. The basic point is that production of sugars for respiration is one use of phyotosynthesis, there are others that still produce O2 as a by product (e.g. in stromatelites). – Dikran Marsupial Feb 21 '15 at 15:19
• @DikranMarsupial can you write down a net reaction and/or reference (or just a suitable search term) for the metabolic pathways you mention that produce $\ce{O2}$ without fixing carbon? What happens to the H? – Nathaniel Feb 22 '15 at 0:42
• @mtb-za based on the comments and answers so far, I think it's because biomass gets buried in anoxic conditions to form stuff like peat, some of which then eventually gets buried and fossilised into coal and oil, which consumers can't oxidise because it's physically separated from the oxygen in the atmosphere. (I'm fairly sure this is the right story, but I'm holding out for an answer from an expert who can explain it all in detail.) – Nathaniel Mar 2 '15 at 12:41

This is only a partial answer as it doesn't explain why the excess of O2 stayed but one thing you have to appreciate is the fact that aerobic respiration appeared almost half a billion years after photosynthesis, so we can't really say that photosynthesis and respiration have always balanced each other.

Cyanobacteria (and with them photosynthesis) are believed to have appeared somewhen between 3 and 2.5Ga (e. g. Altermann & Kazmierczak 2003, Brocks et al. 2003) while Mitochondria (and with them aerobic respiration) are thought to have evolved somewhen between 2 and 1.5Ga (see e. g. Hedges et al. 2004).

There was therefore at least half a billion year during which cyanobacteria were producing O2 with no one able to breathe it.

• That is a very interesting point - the potential long time difference between photosynthetic and respiration based lifeforms would have yielded a substantial amount of atmospheric oxygen, and, could be a mechanism to the oxygen level drop at 1.9 billion years ago, that you very nicely described here earthscience.stackexchange.com/questions/2965/… – user889 Feb 28 '15 at 9:12
• I guess you mean aerobic respiration appeared after photosynthesis, right? (Respiration in the more general sense occurs in all organisms and I don't think a cell could exist without it, but it makes sense that aerobic respiration wouldn't evolve until there was oxygen around for it to use.) I'd be curious to know (a) what the net reaction(s) looked like for the forms of respiration dominant before the great oxidation, and (b) whether any form of anaerobic respiration contributes to maintaining the $\ce{O2}$ concentration today. – Nathaniel Feb 28 '15 at 11:33
• (The latter isn't obvious. Something like methanogenesis will create products that then react with $\ce{O2}$, taking it out of the atmosphere after all. But perhaps there are respiratory pathways, for example with mineral electron acceptors, for which this is not the case.) Anyway +1. – Nathaniel Feb 28 '15 at 11:34
• Yes, sorry, i meant aerobic respiration, I'll correct the answer. – plannapus Feb 28 '15 at 11:46

My answer goes a little beyond the evidence -- there isn't much evidence.

The question

There's a lot of free oxygen now. This oxygen did not suddenly come from underground or from space. So it used to be attached to something. The question is, what was it attached to, and what happened to what it was attached to?

Background

First off, there's every reason to think that bacteria already did oxidative phosporylation before there was any free oxygen. It does not require free oxygen, it requires an electron acceptor. O2 provides the most energy, but many other reactants provide some energy, and the bacterial kingdom uses a bewildering variety of them. To use a different electron acceptor requires only one different enzyme -- the rest of the pathway can be the same.

So before there was free oxygen, prokaryotes did arbitrage -- they evolved to use the best electron donor/electron acceptor pair they could find under whatever conditions they faced at the moment. And everything they used for that had to be created in a cycle. Anything which they used up that was not replaced, would quickly be gone.

Before there was free oxygen, photosynthesis still got done. Some bacteria used H2S as an electron donor, creating S which could later be reduced for energy. Some created H2. Some could reduce CO2 so that they could use the carbon and oxygen to make structures they needed. Some could not, but could still get energy from light. One way to do that was almost the same as oxidative phosphorylation, it used the absorbed light energy for electron transport instead of using redox energy, in practically the same pathway.

Cycles. Every reaction must be undone, or its products will increase to the point that the reaction gets very slow. The exception is products that change state. A solid or gas product won't inhibit the reaction much.

So, a bewildering variety of electron donors and acceptors. Photosynthesis creates energetic pairs of molecules or ions. Then redox creates less-energetic pairs of molecules or ions.

What was it that was bound to oxygen in the old days, that was not bound later?

Some possibilities

1. Hydrogen. One form of photosynthesis creates H2. Maybe, at some times a whole lot of H2 was created which was lost to space. Just as much O and less H leaves room for O2 left over.

2. Sulfur. Maybe most of the sulfur in the oceans was stored as SO3.

H2 SO3 -> S + H2O + O2.

Sulfur can be oxidized -- SO3. Or it can replace Oxygen as an oxidizer -- H2O -> H2S etc.

So by changing states it could affect free O2. And solid sulfur could wind up on the ocean bottom where it would be subducted.

1. Heavy metals + sulfur. Maybe there was a lot of iron and nickel etc dissolved in the ancient ocean, along with a lot of sulfur. Iron sulfate is fairly soluble at room temperature, nearly 300 grams/liter. Provided it is not alkaline. But iron sulfide is not soluble. Lots of other heavy metals behave similarly.

F3O4 + 4 H2SO3 -> F3S4 + 6 O2 + 4 H20 (and many other combinations)

1. Nitrogen. Nitrogen can be reduced to NH4 or oxidized to NO2 or NO3.

2 NO2 + 4 H+ = N2 + O2 + 2 H2O

Perhaps there was a time when the atmosphere had only 75% as much N2 as it does now, and no O2. This reaction would imply the oceans used to be more acidic than they are now. But there could be other reactions that would counteract that, and there were lots of other buffers.

1. Silicon. Silicon is usually oxidized, SiO2. It doesn’t have to have that ratio. Other silicon compounds are possible, eg carborundum SiC, and perryite Fe5Si2 which is stable in acid water, silicon nitride Si3N4, silicon phosphide SiP2 etc . Siloxanes can have ratio Si:O of 1:1. Low molecular weight siloxanes are produced by modern anaerobic biogas fermenters, though it could possibly be eukaryotes producing them. There is potentially a tremendous amount of SiO2 available to release oxygen, although only a little of it is soluble at any one time.

Or bacteria could have metabolized silicic acid into insoluble compounds that had less oxygen. This is entirely hypothetical since only a little carborundum has been discovered, and the rest would have had to be subducted.

Banded iron formations do though have layers of amorphous quartz. Silicon dioxide was being removed from the water. Perhaps it was alternately iron being removed while soluble siloxanes accumulated in water, and then the siloxanes were converted to silica and removed while iron accumulated in water.

It was a complicated web of reactions. Photosynthesis provided the energy to create energy-rich chemicals. Meanwhile every organism that was not photosynthetic was busy finding its best mix of catalyzing chemical reactions that provided energy, versus creating the molecules needed to grow and reproduce. Strange things happened.

I will explore the answer admitting that the Earth's temperature in the ancient past was much higher than we now assume as possible. The current standard model is the 'Snowball Earth' which leaves many unsolved problems - see 'Faint Young Sun Paradox'.

I will assume that the total amount of $\ce{H_2O}$, $\ce{CO_2}$, $\ce{O_2}$ is substantially constant throughout the entire history of the earth.

All the water of the oceans would stay in the atmosphere layer provided the temperature was high enough. The percentage of oxygen would be $\approx 0$ because it would be masked by the massive presence of $\ce{H_2O}$.

Later, when the temperature dropped, the water has migrated to the ocean floor and the percentage of $\ce{O}$ exploded to current levels. This fact is called the 'Great Oxygenation Event' and it is not very well understood under the standard model.

About the free Oxygen origin: (Copy / Paste)

"There are two known natural ways of producing oxygen: by UV dissociation of H2O and by photosynthesis. In the ECM, the former process was initially much more efficient than at present because of the huge amount of water vapour and of the greater intensity of UV radiation occurring then. The biological production of oxygen would have been important during the Archean, which ended at 2.5 Ga, therefore earlier than the GOE. The usual approach is to consider that the biologically produced oxygen was first combined with dissolved iron in oceans and only significantly released to the atmosphere later – a way to explain the delay between the presumed time of the biological production of oxygen and the rise of its atmospheric level. Here, there is no need to consider a delay – the biologically produced oxygen was dissolved in water and absorbed by the huge atmosphere without significantly changing its composition; the oxygen level increased only when the atmosphere significantly decreased due to the condensation of water vapour. One can now consider that oxygen was produced in large amounts since the beginning, first from UV dissociation, at a rate that decreased over time, and very soon followed by biological production. In this case, there was an important amount of free oxygen in the atmosphere long before its atmospheric level started to increase significantly; and some potential evidences of it can be identified (for a review see Yamaguchi, 2005)."

ECM stands for Evolving Climate Model where this new concept was brought to life in a formal document (if you wish to explore further you can follow the 2nd link in my profile, not peer-reviewed)


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My Math Forum Limit of Product of iid Random Variables

 January 26th, 2018, 10:25 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Limit of Product of iid Random Variables Let $X_1, X_2, ... , X_n$ be i.i.d random variables with common density function: $f(x)=\frac{\alpha}{x^{\alpha+1}}I(x\geq 1)$ where $I(x\geq 1)$ is the indicator function and $\alpha >0$. Now define: $$S_n=\left[ \prod_{i=1}^{n}X_i \right ]^{n^{-1}}$$ Prove that $\lim_{n\rightarrow \infty}S_n=e^{\alpha^{-1}}$, and use the law of large numbers and the continuous mapping theorem to conclude about the convergence in probability of $ln(S_n)$. I honestly don't have the slightest inkling on how to go about evaluating that limit. However, since the value is given in the problem statment, my intuition tells me that $ln(S_n)$ converges in probability to $\alpha^{-1}$, but I'm not certain.
 January 26th, 2018, 10:47 PM #2 Senior Member   Joined: Oct 2009 Posts: 402 Thanks: 139 What is $ln(S_n)$. Compute it.
 January 26th, 2018, 11:06 PM #3 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 $\ln(S_n) =\frac{1}{n}\sum_{k=1}^n\ln(X_k)$, but I still don't know how to compute the limit of $S_n$ as $n$ goes to infinity. Apparently I've forgotten part of calculus.
 January 26th, 2018, 11:10 PM #4 Senior Member   Joined: Oct 2009 Posts: 402 Thanks: 139 This is where the law of large numbers comes in, which says something about limits like these.
 January 26th, 2018, 11:13 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999 you'll probably need $\lim \limits_{n \to \infty} ~\left(1+\dfrac x n\right)^n = e^x$
 January 27th, 2018, 07:11 AM #6 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 I understand how to do the problem using the strong law of large numbers, but I wasn't sure if I was supposed to prove the value of that limit some other way initially (the setup of the problem suggests so since it mentions so first).

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text:
# Homework Help: What does the ellipse tell me?

1. Aug 29, 2010

### annoymage

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

intersection $$x^2+y^2+z^2=4$$ and plane $$y=z$$

2. Relevant equations

n/a

3. The attempt at a solution

so, i solve it simultaneously, and get the equation of ellipse $$x^2+2y^2=4$$. but what does the ellipse tell me? is that the points of the intersection?, but aren't it should be a circle?

2. Aug 29, 2010

### schlynn

Re: geometry

First, that's not the equation for an elipse. I would be ((x^2)/4)+((y^2)/2)=1. Then set y=0 to find the intercepts. You can use the first equation you found to do that easily, you'll get $$\pm$$2.

3. Aug 29, 2010

### lanedance

Re: geometry

the first equation is the equation for a sphere of radius 2, and any plane thorugh the origin will intersect it in a circle

after substituting y=z, the equation you get is that of an ellipse projected onto the xy plane, but remember you have the extra z coordinate, so in the y=z plane it is still a circle

4. Aug 29, 2010

### schlynn

Re: geometry

Yeah, that's right, but the method he did still works. Just the substitution changes things. Checked on wolframalpha and it is x=$$\pm$$2

5. Aug 29, 2010

### lanedance

Re: geometry

why do you get 2 points? it should be the equation of a circle shouldn't it?

6. Aug 29, 2010

### annoymage

Re: geometry

i messed up, don't have basics on geometry, maybe i should read first some on plane and sphere. later i show my work

7. Aug 29, 2010

### lanedance

Re: geometry

i think what you did is ok, just missed a bit on interpretation at the end


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# Timelines for draft 1 contributions.

Jump to: navigation, search

I believe this is the right way forward as I fear us jogging on the same spot for too long...as stated I'll try to put closure to this by tommorrow.

05:43, 20 June 2011


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# Relationship between Rewards and Q Value (Graph between Q(s, a) vs episodes)

I'm employing the Actor-Critic algorithm. The critic network approximates the action-value function, i.e. $$Q(s, a)$$, which determines how good a particular state is, when provided with an action.

$$Q(s, a)$$ is approximated using the backpropagation of the temporal difference error (TD error). We can understand that $$Q(s, a)$$ has been approximated properly when TD error is minimized, i.e. when it is saturated at lower values.

My question is, when exactly can you say that $$Q(s, a)$$ is approximated properly, if you don't have TD error, i.e. if you have to plot the graph between $$Q(s, a)$$ vs episodes, then what would be the optimal behaviour?

Will it be increasing exponential with saturation value around reward values, or increasing exponential with saturation (around any value)?

Follow up: What can be the possible mistake, if the output of Q-value function is around 5x the rewards, and not saturating?

• Which actor-critic algorithm have you implemented exactly? From which paper or book? I suppose you're using a neural network to approximate $Q(s, a)$, given your description. Moreover, when you say "if you have to plot the graph between Q(s, a) vs episodes", what graph are you exactly talking about? Can you show an example of such a graph that you have plotted? Also, why don't you have the TD error? – nbro Jan 25 at 12:46
• I'm using the standard actor critic, where critic network is optimized using TD Error. I wanted to plot the values of $Q(s, a)$ (my network has 1 output neuron) vs training episodes, to know what are the possible values around which $Q(s, a)$ get saturated. However, my question is answered by little more experimentation and research. I'll be adding the answer soon. – Anubhav Sachan Jan 25 at 16:06
• So, are you using function approximation or not? In any case, I would suggest that you put the title in the form of a question, your specific question. Right now, the title is not very clear. Also, what you wanted to plot, according to your answer, was the mean Q(s, a) value, but that alone doesn't mean much. The Q-value is a function of the state and action, so if you average the Q-values across all states and actions, that's probably meaningless. – nbro Jan 25 at 18:51
• Yes, I'm using function approximation. I'll take care of the title as well. @nbro, can elaborate on why averaging Q-values across all the states and actions is probably meaningless? My state is basically the channel coefficient H, and action is the amount of power transmitted in that H. – Anubhav Sachan Jan 25 at 19:03
• Because, for example, in one state $s_1$, the average value across all possible actions in that state may be 1, while the average value across all actions in state $s_2$ may be 1000. However, maybe, in both cases, the relative value of the actions in the specific state is the same, i.e. the optimal action is the same, the second best optimal action is also the same and so on. Of course, the values of the states depend on 1. how you initialize the Q-network, 2. how you explore the environment, 3. the learning update. – nbro Jan 25 at 19:08

TLDR;

The output of Q-value function will eventually saturate. Can't say when, but it will surely do.

if you don't have TD error

It meant, if I don't have logs of the error.

$$Q(s, a)$$ vs Episodes Graph

To understand, how $$Q(s, a)$$ behaves as the episodes increase.

I was under a wrong impression that, the $$Q(s, a)$$ will saturate around the values given by the reward function.

As evident from Loss vs Training Episodes Curve, we can see that loss (TD Error) is almost saturated.

However, $$Q(s, a)$$ vs training episodes curve is not saturated yet.

The only explanation for the above two graphs could be given as follows: The target estimate ($$r + \gamma Q(s', a')$$) and $$Q(s, a)$$ are almost similar due to which, the error is quite low. But, $$Q(s, a)$$ is still nowhere near optimal value $$Q^{*}(s, a)$$.

Hence, I gave it another shot and made it run for twice training episodes i.e. 20000, and below are the results.

A nearly saturated loss

and a nearly saturating $$Q(s, a)$$. .

Note that, the value of $$Q(s, a)$$ saturating is around 250 - 300 (will run it for more iterations) and it is nowhere around the reward values $$\in [-100, 35]$$.

Hence, the $$Q(s, a)$$ vs episodes will saturate.


text: - Computing the stability number of a graph via linear and semidefinite programming Javier Pena (jfpandrew.cmu.edu) Juan Vera (jveraandrew.cmu.edu) Luis Zuluaga (lzuluagaunb.ca) Abstract: We study certain linear and semidefinite programming lifting approximation schemes for computing the stability number of a graph. Our work is based on, and refines De Klerk and Pasechnik's approach to approximating the stability number via copositive programming (SIAM J. Optim. 12 (2002), 875--892). We provide a closed-form expression for the values computed by the linear programming approximations. We also show that the exact value of the stability number $\alpha(G)$ is attained by the semidefinite approximation of order $\alpha(G)-1$ as long as $\alpha(G) \leq 6$. Our results reveal some sharp differences between the linear and the semidefinite approximations. For instance, the value of the linear programming approximation of any order is strictly larger than $\alpha(G)$ whenever $\alpha(G) > 1$. Keywords: stability number, copositivity, polynomials, lifting procedures Category 1: Combinatorial Optimization (Other ) Category 2: Linear, Cone and Semidefinite Programming (Semi-definite Programming ) Citation: SIAM Journal on Optimization 18 (2007) pp. 87--105. Download: Entry Submitted: 04/08/2005Entry Accepted: 04/08/2005Entry Last Modified: 06/27/2007Modify/Update this entry Visitors Authors More about us Links Subscribe, Unsubscribe Digest Archive Search, Browse the Repository Submit Update Policies Coordinator's Board Classification Scheme Credits Give us feedback Optimization Journals, Sites, Societies Optimization Online is supported by the Mathematical Programming Society and by the Optimization Technology Center.


text:
## anonymous 3 years ago arush sat down to study at 8 a.m. the angle of elevation of the sun was 15'.when he finished studying the angle of elevation of the sun was 55'.how long did arush study?

1. anonymous

@Hero @karatechopper

2. anonymous

plz hlp me out today is my xam PSA

3. anonymous

@sirm3d @onedirection822 @Weepdo-s @poopsiedoodle

4. anonymous

timo u r writing something were that

5. anonymous

|dw:1360980444964:dw|

6. anonymous

be nice poopsie how the heck you become ambassador?

7. anonymous

set a ratio original to new time 8/x=15/55

8. anonymous

that reduces to 8/x = 3/11

9. anonymous

or i think i did it wrong

10. anonymous

11. anonymous

yes option is also given 40 mins, 1hrs 20mins, 2hrs 23mins, 2hrs 40 mins

12. anonymous

|dw:1360981152161:dw|

13. anonymous

14. anonymous

no

15. anonymous

i am confused perhabs if we divide the angle 360 into a 24 hour thingy. so 360 degree= 24 hour

16. anonymous

nt getting

17. anonymous

ooh maybe the horizontal length is fixed and just the vertical varies.

18. anonymous

|dw:1360981498364:dw|

19. anonymous

hw to solve nt getting

20. richyw

what latitude does he live in?

21. anonymous

u can read question 4 that?

22. anonymous

23. richyw

if you assume that the sun travels in an arc directly overhead then you can get the angular velocity of the sun as$\omega=\frac{180^{\circ}}{12 \text{ hours}}=15^{\circ}/\text{h}$then $\Delta t=\frac{\Delta\theta}{w}=\frac{55^{\circ}-15^{\circ}}{15^{\circ}/h}=\frac{8}{3}h=160\text{ min}$

24. richyw

but here the sun is at 15 degrees at 8 am so the time zone are a bit messed up. I guess it is plausible as long as the area is right on the edge of a time zone, but you would expect the sun to rise and set around 6am and 6 pm.


text:
Regular octagon inscribed in a square

Problem: The corners of a 2 meter square are cut off to form a regular octagon. What is the length of the sides of the resulting octagon?

From the picture below, the octagon would form a right isosceles, specifically a right isosceles triangle on the corners. The sides of the octagon were set to "x" and the legs of the triangle were set to $\frac{x}{\sqrt{2}}$. Then add the following cuts of a side of the square: $\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 m, which results to x = 0.828 m.

My inquiry is that, from what I know or learned, a right isosceles triangle has an angle ratio of $45-45-90$ and a side ratio of $1-1-\sqrt{2}$ or in algebra: $x-x-x{\sqrt{2}}$. In the problem he set the hypotenuse as $x$ instead and the legs of the triangle as $\frac{x}{\sqrt{2}}$, which I think is fine. But shouldn't setting the hypotenuse as $x\sqrt{2}$ and the sides as $x$ should equal the first equation?

$\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 should also equal $x + x\sqrt{2} + x = 2$ where 2 is the length of a side of a square. I don't think multiplying or dividing both sides by $\sqrt{2}$ is the answer as that would not satisfy both equations.

This sounds like an easy problem, but it it's confusing me. Sorry.

• If you are actually building one of these (like a poker table), then you'll want to use the right-hand diagram and measure your cuts from the corners. (Assuming you actually have a square piece to start.) – John May 21 '18 at 20:27

Let $x$ be the length of your octagon (as in the left picture), and $c$ the length cut from one side of the square edge (which is the $x$ in the right picture).

Then you've correctly stated that $x = \sqrt{2}c$. Now you solve $$c + x + c = 2.$$ This is rewritten as $$2c + x = 2c + \sqrt{2}c = (2 + \sqrt{2})c = 2.$$ Thus $$c = \frac{2}{2+\sqrt{2}}$$ so that $$x = \frac{2\sqrt{2}}{2 + \sqrt{2}}.$$ This final fraction is the length of the sides of the octagon.

In the right picture, everything has been scaled up by $\sqrt{2}$ so that the length of the sides of the octagon will be $$\sqrt{2} \cdot \frac{2\sqrt{2}}{2 + \sqrt{2}} = \frac{4}{2 + \sqrt{2}}.$$ This is why they are unequal.

If you were to solve it with $x$ as in the right picture, then you have $$2x + \sqrt{2}x = 2$$ so that $$x = \frac{2}{2 + \sqrt{2}},$$ and then the length of the octagon is $$\sqrt{2}x = \frac{2\sqrt{2}}{2 + \sqrt{2}}$$ exactly as we had calculated above.

Your confusion stems from using $x$ as a label for two different lengths in either diagram.

The two equations are not equivalent, and they give different results. In the first case, $x$ is the side of the octagon; in the second one, the side of the octagon is $x\sqrt{2}$.

The first one has the following solution: $x=2(\sqrt{2}-1)$.

The second one has solution $x=2-\sqrt{2}$, and to obtain the side of the octagon, you have to multiply it by $\sqrt{2}$, obtaining the same result.

• There were problems wherein I didn't need to multiply or divide back an expression to get the original value of x.. So even though you have explained it clearly, I feel more confused lol. But thanks.~ – Jayce May 21 '18 at 14:46
• Just replace $x$ in the right diagram with $y$ and then equate $x=y\sqrt{2}$. – dan post May 21 '18 at 15:08