Mining Equipment Reliability Testing PDF

Reliability of Mining Equipment ENGR 4366 - Fall 2024 Lec10 - Mining Equipment Testing Dr. Ahlam Maremi Bharti School of Engineering Laurentian Univ...

Reliability of Mining Equipment ENGR 4366 - Fall 2024 Lec10 - Mining Equipment Testing Dr. Ahlam Maremi Bharti School of Engineering Laurentian University Office: F215B Email: [email protected] 2 Announcement Representatives from KGHM will be visiting LU The event is on Wed Oct 30 @ 1:00 in F-228 Pizza will be served. Ahlam Maremi  3 Introduction Testing is an important element of any engineering product development program. It is to highlight product’s weaknesses, behavior characteristics and modes of failure. Reliability testing is an important element of testing and is concerned with obtaining information regarding failures and failure consequences A good reliability test program: – Requires a minimal amount of testing, and – Provides the maximum amount of information on failures. 4 Reliability Test Classifications There are three categories of reliability tests based on the objectives: – Reliability development and demonstration testing to: Identify necessary changes in design. Determine the need to improve design. Verify improvements in design reliability. – Qualification and acceptance testing to: Determine if the item should be accepted/rejected individually or on a lot basis. Determine if the design qualifies for its specified objective. – Operational testing to: Verify the results of reliability analyses performed during the equipment design and development phase. Provide data for use in subsequent activities. Provide data indicating necessary changes to operational policies and procedures with respect to reliability and maintainability. Ahlam Maremi  5 Mining Equipment Testing Success testing Accelerated life testing – Perform sudden-death testing. – Increase the sample size. – Increase the test severity. 6 Success Testing The main goal of success testing is to ensure that a specified reliability level is achieved at a given confidence level. – It is used where a no-failure test is specified. In this case for zero failures: RL =  1/ m where RL the lower (the minimum) reliability level m is the total number of items/units placed on test,  is the consumer’s risk or the level of significance (the acceptable error risk). The desired level of confidence, C , is : C = 1− →  = 1− C ln (1 − C ) ln ( ) m = or ln ( Rt ) ln ( Rt ) Rt is the true or actual reliability Ahlam Maremi  7 Success Testing - Example Assume that a manufacturer of a certain part used in a piece of mining equipment is required to demonstrate 95% reliability of that part at a 90% confidence level. Determine the total number of parts to be tested when only zero failures are allowed. What is the minimum reliability level if the number of samples is 15 at a 90% confidence level? 8 Success Testing - Example C = 90% = 0.9 True Reliability ( Rt ) = 95% = 0.95 ln (1 − C ) ln (1 − 0.90 ) m = = 45 parts ln ( Rt ) ln ( 0.95 ) This means a total of 45 parts must be tested to demonstrate 95% reliability at a 90% confidence level. ================================== The minimum reliability level using 15 successful samples: the acceptable error risk ( ) = 1 − C  = 1 − 0.90 = 0.1 RL =  1/ m → 0.1( 1/15 ) = 0.8577 = 85.8% The lower confidence limit on the desired reliability level is 85.8% Ahlam Maremi  9 Accelerated Life Testing (ALT) Conventional testing can take very long time making it impractical. Accelerated testing is concerned with reducing the duration of the test time by changing parameters; There are a number of ways to perform accelerated life testing including: – Perform sudden-death testing. – Increase the sample size. – Increase the test severity. 10 Accelerated Life Testing (ALT) Sudden-death testing: – The test sample is divided into various groups containing an equal number of items or units, – All items/units in each group start their operation simultaneously. – Whenever anyone of the group units fails, all units are considered failed: The testing of non-failed units is terminated immediately. – In the event of failure of the first unit in the last and final group, the entire testing process is terminated. Ahlam Maremi  11 Accelerated Life Testing Increasing the sample size testing: – The size of the sample is increased when the item life distribution does not exhibit a wear-out characteristic during its predicted lifespan. – Note that the increase in the sample size decreases the test time as the test time is inversely proportional to the sample size. 12 Accelerated Life Testing Increasing the test severity testing. It involves: – Time compression (testing is done more intensively than the actual usage). Commercial blender, used 20 times a day, 365 days a year. For its 5-year design life, the switch would be operated 36,500 times. If one on-off application is performed in 10 seconds. Then, 36,500 applications will be completed in 365,000 second or ~101 hours Ahlam Maremi  13 Accelerated Life Testing Increasing the test severity testing. It involves: – Increasing the stress acting on the test unit/item. The stress may be classified under two different areas: application and operational: – The application area includes items such as voltage, self-generated heat and current; – The operational area includes humidity and temperature. – For ex., engine oil cooler (pressure and temperature cycles during engine operation). – To increase the severity and test faster, pressure pulses and heated oil are used. 14 Accelerated Life Testing The following equation can be used to calculate the acceleration factor for this type of accelerated testing:   E  1 1  Fa = exp  −    −    k   Tac Tus  where Fa is the acceleration factor, Tac is the acceleration temperature expressed in degrees Kelvin, Tus is the operating temperature expressed in degrees Kelvin, E is the activation energy, and its value is taken as 0.5 eV, k is the Boltzmann constant, and its value is taken as 0.00008623 eV / degrees Kelvin. Ahlam Maremi  15 Accelerated Life Testing - Example A sample of 20 identical engineering items used in mines were tested to failure at a temperature of 130°C, and their mean time to fail was 2,500 hr. Calculate the MTTF of these items at the normal operating temperature of 85 ◦C. degrees Kelvin (o K) = (o C) + 273 Tac = 403 o K & Tus = 358 o K E = 0.5 eV (constant) k = 0.00008623 eV / degree (constant) MTTF = 2,500 hr (at 130o C or 403 o K)   E  1 1    0.5  1 1  Acceleration factor( Fa ) = exp  −    −  = exp  −   −   = 6.10   k   Tac Tus     0.00008623   403 358  The MTTF if the item is used in mines (normal operating temperature) : MTTF(normal temperature) = MTTF(test temperature) ×acceleration factor MTTF(normal temperature) = ( 2,500 )  ( 6.10 ) = 15, 250 h 16 Confidence Interval Estimates Confidence Interval Estimates for Mining Equipment MTTF: – In reliability studies, times to item failure are assumed to be exponentially distributed. The item failure rate becomes constant. The MTTF is the invers of the item failure rate. The MTTF a point estimate. It provides incomplete picture. – It would be more meaningful if we say: after testing a sample of identical items for (T) hours, (n) number of failures occurred; Failures will not occur at the same time. – Then the actual MTTF lies between upper and lower limits with a certain degree of confidence. Ahlam Maremi  17 Confidence Interval Estimates The confidence interval on MTTF can be calculated using ꭕ2 (chi square) distribution. Confidence intervals are estimated under the following two conditions: 1. Testing is terminated at a pre-assigned number of failures, K∗    2Y 2Y   2Y 2Y   2 , 2 = ,    ( p, df )  ( p, df )    2   , 2 K *   2 1 −  , 2 K *     2   2   2. Testing is terminated at a pre-assigned time, t∗    2Y 2Y   2Y 2Y   2 , 2 = ,    ( p, df )  ( p, df )    2   , 2 K + 2   2 1 −  , 2 K     2   2   18 Confidence Interval Estimates  2 ( chi-square ) distribution it is function of ( p, df ) df is the degrees of freedom p is significance level of the critical value (probability) Values in the attached table  is the acceptable error risk K is the number of failures accumulated by time t  K * is the number of pre-assigned failures t  is the life test termination time C = 1 −  is the confidence level Y is the total success time Ahlam Maremi  19 Confidence Interval Estimates 20 Confidence Interval Estimates Y is the total success time. The value of Y is determined by the type of test, there are four cases: – Replacement of failed units/items test Y = nt  – Non-replacement of failed units/items test K Y = (n − K ) t +  t j j =1  t is the life test termination time n is the number of items that were placed on test at time t = 0 t j is the failure time, 1,2,3,..., k Ahlam Maremi  21 Confidence Interval Estimates – Replacement of failed units/items but non-replacement of withdrawal (loss) items/units test: m Y = ( n − m ) t  +  ti i =1 ti is the ith censorship time, m is the number of censored items/units. – Non-replacement of failed units/items and non-replacement of withdrawal (loss) items/units test: m K Y = ( n − m − K ) t  +  ti +  t j i =1 j =1 22 Confidence Interval Estimates - Examples Assume that a total of 20 identical items used in a mine were placed on test at time t = 0 and none of the failed items were replaced. There is no loss/withdrawal items. The test was terminated after 200 hrs. Five items failed after 50, 70, 120, 130 and 150 hrs of operation. Estimate the items’ MTTF and its upper and lower limits with 90% confidence level. Ahlam Maremi  23 Confidence Interval Estimates - Examples Y is the total success time (all items); Y for non-replacement test: K Y = ( n − K ) t +  t j j =1 n = 20 (the number of items that were placed on test at time t = 0) K = 5 (number of failures accumulated by time) t  = 200 hr ( the life test termination time) t j is the failure time for the five items Confidence level = 90%  = 10% (the acceptable error risk = 100% - 90% = 10%) Y = ( 20 − 5 ) 200 + ( 50 + 70 + 120 + 130 + 150 ) = 3,520 hr Y 3,520 MTTF = = = 704 hr K 5 But, we are looking for the upper and lower limits for the MTBF!! 24 Confidence Interval Estimates - Examples Because the testing is terminated at a pre-assigned time, t * = 200hr The upper and lower limits for the items’ MTTF are: df is the degrees of freedom p is significance level of the critical value (probability)    2Y 2Y   2Y 2Y   2 , 2 = ,    ( p, df )  ( p, df )    2   , 2 K + 2   2 1 −  , 2 K     2   2   2Y 2  3,520 7, 040 7, 040 Lower limit = = = = = 334.9hr 2   2  10%   2 ( 0.05, 12 ) 21.02   , 2K + 2    , 25 + 2 2   2  2Y 2  3,520 7, 040 7, 040 Upper limit = = = 2 = = 1, 786.8hr     1 − , 2 K   1 − 2 2  10% , 25   ( 0.95,10 ) 3.94  2   2  Ahlam Maremi  25 Confidence Interval Estimates 26 Confidence Interval Estimates - Examples The items’ MTTF is 704 hr (point estimate). The upper and lower limits of MTTF with a 90% confidence level are 1,786.8 hrs and 334.9 hrs, respectively. In other words, we can state that the item is expected to last for 1,786.8 hours with a 95% confidence level, acknowledging a 5% risk that it may only last for 334.9 hours. Ahlam Maremi  27 Don’t Forget! Check announcements on D2L regularly Review additional resources available on D2L Design Project due Nov 3rd Last day of classes: Wed Dec 4th Final Exam: Thu Dec 6th (Lec10 – Lec19) Ahlam Maremi  Reliability of Mining Equipment ENGR 4366 - Fall 2024 Lec11 - Mining Equipment Costing - I Dr. Ahlam Maremi Bharti School of Engineering Laurentian University Office: F215B Email: [email protected] 2 Representatives from KGHM will be visiting LU The event is on Wed Oct 30 @ 1:00 in F-228 Pizza will be served. Ahlam Maremi  3 Introduction Over the years cost has been increasing at a significant rate. Like any other engineering equipment, cost estimation in mining equipment is extremely important for making effective decisions. The main elements of the total cost of producing a typical piece of mining equipment are: – Design and development cost, – Materials cost, – Manufacturing cost, – Testing cost, – Operation cost, and – Maintenance cost. 4 Why Mining Equipment Costing? Some of the important reasons for mining equipment costing: – To determine the selling price. – To determine the profitable approach and materials during equipment manufacture. – To control the cost of equipment manufacture. – To determine if it is more economical to produce the parts in-house or to procure them from vendors. – To determine the amount of money required for facilities to manufacture the equipment under consideration. – To determine the efficiency of the equipment manufacturing process. – To perform new equipment feasibility analyses. – To provide inputs to the long-term financial goals of the organization. Ahlam Maremi  5 Mining Equipment Investment Decisions Useful methods used to make mining equipment investment decisions: – Benefit/Cost analysis method, – Return on Investment (ROI) method, – Payback period method. 6 Benefit-Cost Analysis Method This method is used to determine if the benefits from the mining equipment outweigh its cost. – The equipment is considered for development only if its benefits outweigh the investment cost. The benefit/cost ratio is expressed by: UB Rbc = IC where Rbc is the benefit-cost ratio, IC is the investment cost including the operating and maintenance cost, UB is the user benefits. Ahlam Maremi  7 Benefit / Cost Analysis Method - Example A piece of mining equipment is being considered for purchase. Its estimated capital and operating cost $500,000 and expected useful life is 15 years. The total benefits of the equipment over its lifespan are estimated to be around $900,000. Calculate the benefit-cost ratio if the annual maintenance cost of the equipment is $2,000. IC is the investment cost including the operating & maintenance cost IC = Estimated cost + Maintenance cost Maintenance cost = Annual maintenance cost × Expected useful life = 2, 000 15 = $30, 000 UB $900, 000 $900, 000 Rbc = = = =1.67 IC $500, 000 + 30, 000 $530, 000 The value of the benefit-cost ratio > 1 8 Return on Investment (ROI) Method ROI is used to make mining equipment investment decisions. ROI evaluates the efficiency or profitability of an investment or compare the efficiency of a number of different investments. ROI tries to directly measure the amount of return on a particular investment, relative to the investment’s cost. There are two different approaches to calculate the ROI. Ahlam Maremi  9 Return on Investment Method - Approach I & II The return on average The return on original investment: investment: ANP RI a = 100 ANP  a I  RI o = 100  + C w ( I a + Cw ) 2  RI a is the return on the average investment. RI o is the return on the original investment ANP is the annual average net profit, I a is the total amount invested, Cw is the working capital Working capital is the money available to meet your current, short-term obligations. – The difference between current assets and current liabilities. 10 Return on Investment Method - Example A coal mining organization is considering purchasing a piece of equipment to be used in mines that requires $500,000 in investment and $25,000 in working capital. It is estimated that the use of the equipment will generate average net profits of $30,000 per year. Calculate the return on the original and average investment. ANP 30, 000 RI o =  100 =  100 = 5.71% ( I a + Cw ) ( 500, 000 + 25, 000 ) ANP 30, 000 RI a = 100= 100 = 10.91%  Ia   500, 000   + Cw   + 25, 000  2   2  The return on the original investment is 5.71%. The return on the average investment is 10.91% Ahlam Maremi  11 Return on Investment Method Return on original investment: – It calculates ROI based solely on the initial amount invested. – It is simple to calculate – It clearly shows total return relative to initial cost – It doesn't account for additional investments or withdrawals over time – May overstate returns for long-term investments Return on average investment – It uses the average investment amount over the holding period. – It is more accurate for investments with fluctuating amounts – It accounts for additional contributions or withdrawals – More complex to calculate – May understate returns compared to original investment method 12 Return on Investment Method Choosing the right method depends on: – The nature of the investment (static vs. dynamic) – The investment timeframe – The level of detail required for analysis For quick assessments of simple investments, return on original investment may suffice. For more complex or long-term investments, return on average investment often provides a more accurate understanding of performance. Ahlam Maremi  13 Payback Period Method This method is concerned with determining the payback period over which the project capital expenditures can be recovered. The major drawbacks of this method are: – It ignores the time value of money, and – It ignores cash flows beyond the payback period. The method makes use of the following four approaches to compute the payback period. 14 Payback Period Method - Approach I In this case, the payback period is expressed by: TCE PP1 = ACF PP1 is the payback period for approach I, TCE is the total capital expenditure, ACF is the average annual cash flow. The average annual cash flow is expressed by: ACF = ANP + (TCE )( CDR ) ANP is the average annual net profit, CDR is the invested capital ( fixed ) depreciation rate. TCE PP1 = ANP + (TCE  CDR ) Ahlam Maremi  15 Payback Period Method - Approach II In this case, the payback period is expressed by: TCE PP2 = AGP where PP2 is the payback period for approach II, AGP is the average annual gross profit TCE is the total capital expenditure, 16 Payback Period Method - Approach III In this case, the payback period is expressed by: 1 PP3 =    CDR +  (1 − ITR )  where PP3 is the payback period for approach III, ITR is the income tax rate,  is the minimum acceptable rate of return on investment CDR is the invested capital ( fixed ) depreciation rate Ahlam Maremi  17 Payback Period Method - Approach IV In this case, the payback period is expressed by: TCE PP4 = ANP PP4 is the payback period for approach IV. TCE is the total capital expenditure ANP is the average annual net profit; ANP = AGP −  ITR  YY + i (TCE + Cw )  YY = AGP − ( CDR )(TCE ) where i is the interest rate on the borrowed money, Cw is the working capital ITR is the income tax rate CDR is the invested capital ( fixed ) depreciation rate 18 Payback Period Method - Example A mining organization is considering replacing a piece of equipment used in underground operations, at a cost of $400,000. It is estimated that the use of the equipment will generate an average saving of $30,000 per year. Calculate the capital expenditure recovery period. TCE PP2 = AGP PP2 is the payback period for approach II, AGP is the average annual gross profit TCE is the total capital expenditure, 400, 000 PP2 = = 13.3 years 30, 000 The capital expenditure recovery period is 13.3 years. Ahlam Maremi  19 Cost Estimation Models for Mining Equipment Mathematical models used to estimate costs concerning engineering equipment: – Cost-capacity model, – Corrective maintenance labor cost estimation model, – Total maintenance labor cost estimation model, – Production facility downtime cost estimation model, – Motor operation cost estimation model, – Failure mode and effect analysis cost estimation model, – Reliability testing cost estimation model. 20 Cost Estimation Models Cost-capacity Model: – It is quite useful for obtaining quick cost estimates for similar new projects, facilities or equipment of different capacities. – The cost of the new item is expressed as follows:   CP  Cn = Cso  n   CPso  Cn is the cost of the new project, facility, or equipment; Cso is the cost of the old but similar project, facility, or piece of equipment; CPn is the capacity of the new project, facility, or piece of equipment; CPso is the capacity of the old but similar project, facility, or piece of equipment;  is the cost-capacity factor whose value varies for different items or projects. the proposed values of this factor for pumps = 0.6, tasks = 0.7, heaters = 0.8 If no available data for the factor, it is considered to be 0.6 Ahlam Maremi  21 Cost-capacity Model - Example Assume that a mining organization spent a total of $500M to construct a mining site whose annual output is 20,000 tonnes. The organization is considering constructing another similar site whose annual output will be 30,000 tonnes. Estimate the cost of the new site if the value of the cost- capacity factor is 0.6. Cso = $500 million  = 0.6 CPn = 30, 000 tonnes CPso = 20, 000 tonnes   CP  0.6  30, 000  Cn = Cso  n  =500   =$637.71M  so  CP  20, 000  The cost estimate for the new mining site is $637.71 million. 22 Cost Estimation Models Corrective maintenance labor cost estimation model: – This model is concerned with estimating the annual labor cost of corrective maintenance associated with equipment with known mean time to fail and mean time to repair. – The equipment annual corrective maintenance labor cost is defined by:  MTTR  CMLC = ( ASOH )( R )   MTBF  where CMLC is the equipment annual corrective maintenance labor cost, ASOH is the equipment annual scheduled operating hours, R is the hourly corrective maintenance labor cost rate, MTBF is the equipment mean time between failures, MTTR is the equipment mean time to repair. Ahlam Maremi  23 Labor Cost Estimation Model - Example Assume that a piece of mining equipment is scheduled for 6,000 hr of operation in one year and its estimated mean time between fails is 2,000 hr and mean time to repair is 10 hr. Calculate the equipment annual corrective maintenance labor cost if the hourly maintenance labor cost rate is $40. ASOH = 6, 000 hr R = $40 / hr MTTR = 10 hr MTBF = 2, 000 hr  MTTR   10  CMLC = ( ASOH )( R )   = ( 6, 000 )( 40 )   = $1, 200  MTBF   2, 000  The mining equipment annual corrective maintenance labor cost is $1,200. 24 Cost Estimation Models Total maintenance labor cost estimation model: – The total maintenance labor cost is expressed by: MLCt = m ( AH t )( Rl )(1 +  ) where MLCt is the total maintenance labor cost, m is the total number of employees, AH t is the total number of annual hours associated with the maintenance activity, Rl is the labor rate per hour,  is the benefit ratio. Ahlam Maremi  25 Total Maintenance Labor Cost - Example Assume that the following data values are known for the maintenance department of a mining company. Calculate the company’s total maintenance labor cost. The total number of employees, m = 15 The total number of annual hours associated with the maintenance activity, AH t = 2, 000 hr The labor rate per hour, Rl = $40 The benefit ratio,  = 0.4 MLCt = m ( AH t )( Rl )(1 +  ) MLCt = 15 ( 2, 000 )( 40 )(1 + 0.4 ) = $1, 680, 000 The mining company’s total maintenance labor cost is $1,680M 26 Types of Maintenance Maintenance: the act of keeping equipment in good condition by making repairs, correcting problems, etc. Preventive maintenance: includes regular and periodic schedule. Corrective maintenance: occurs when an issue is noticed. Predetermined maintenance: follows factory schedule. Condition-based maintenance: occurs when a situation or condition indicates maintenance is needed. Predictive maintenance: is data-driven and impacted by pre- set parameters. Reactive maintenance: occurs when a total breakdown or failure appears. Ahlam Maremi  27 Don’t Forget! Check announcements on D2L regularly Review additional resources available on D2L Design project due Nov 3rd Presentations start on Nov 21st Last day of classes: Wed Dec 4th Final Exam: Dec (Lec 10 – Lec 19) Ahlam Maremi Reliability of Mining Equipment ENGR 4366 - Fall 2024 Lec13 - Mining Equipment Costing - II Dr. Ahlam Maremi Bharti School of Engineering Laurentian University Office: F215B Email: [email protected]  2 Cost Estimation Models for Mining Equipment Mathematical models can be used to estimate various types of costs concerning engineering equipment: – Cost-capacity model – Corrective maintenance labor cost estimation model – Total maintenance labor cost estimation model – Production facility downtime cost estimation model – Motor operation cost estimation model – Failure mode and effect analysis cost estimation model – Reliability testing cost estimation model  3 Downtime Cost Estimation Model Production facility downtime cost estimation model: – The overall cost is compounded by the inability to produce. This model is concerned with estimating the production facility downtime cost. PFDC = Cr + Crl + Co + C p + Cru + Cti where PFDC is the production facility downtime cost, Cr is the ruined product replacement cost, Crl is the revenue-loss cost, Co is the idle operator salary, C p is the late-penalty cost, Cru is the replacement unit rental cost ( if any ) , Cti is the tangible and intangible costs associated with factors such as loss of good operator and customer dissatisfaction.  4 Downtime Cost Estimation Model Remember that the model can be adjusted based on the specific details of any given manufacturing facility. Consider a front end loader (FEL) manufacturing plant that operates 24 hours a day, 7 days a week. Determine the cost per hour of downtime knowing that the plant produces 60 FELs per hour; the average profit per loader is $5,000; number of workers on shift is 500; average hourly wage including benefits is $50; fixed energy costs per hour is $10,000; lost sales and market share (opportunity cost) is $1,515,000 per hour. Calculate total downtime cost for a 4- hour downtime incident.  5 Downtime Cost Estimation Model Lost production value per hour = 60  $5, 000 = $300, 000 For 4 hours = $300, 000  4 = $1, 200, 000 Labor cost per hour of downtime = 500  $50 = $25, 000 For 4 hours = $25, 000  4 = $100, 000 Fixed energy costs per hour = $10, 000 For 4 hours = $10, 000  4 = $40, 000 Opportunity Cost = $1,515, 000  4 = $6, 060, 000 Total Downtime Cost for 4 hours = $7,400,000 – The example shows how quickly downtime costs can accumulate in a high-value production environment. The model includes both direct costs and indirect costs (opportunity costs). – The actual cost of downtime will vary depending on the specific circumstances.  6 Motor Operation Cost Estimation Model Motor operation cost estimation model: – Used to estimate the cost of operating an alternating-current (AC) motor: MAOC = ( 0.746 )( MAOH )( EC )( MS )  where MAOC is the motor annual operating cost, MAOH is the motor annual operating hours EC is the electricity cost expressed in dollars per kilowatt hour ( $ / KWH ) , MS is the motor size expressed in horsepower,  is the motor efficiency %. – 1 horsepower-hour = 0.7456998714 kilowatt-hour (kWh)  7 Motor Operation Cost Estimation Model - Example A 60-horsepower AC motor is operated for 3,000 hr/year and the cost of electricity is $0.15/KWH. The motor efficiency is 95%. Calculate the motor annual operating cost. MAOH = 3, 000 hr EC = $0.15 / KWH MS = 60 horsepower  = 95% = 0.95 MAOC = ( 0.746 )( MAOH )( EC )( MS )  MAOC = ( 0.746 )( 3, 000 )( 0.15 )( 60 ) = $21, 202.10 0.95 The motor annual operating cost is $21,202.10.  8 Failure Mode and Effect Analysis Cost Failure mode and effect analysis cost estimation model: – The model is concerned with estimating the cost of performing FMEA. – It measures risk of failure in terms of cost by considering the probability of failure and the associated costs. – Expected Failure Cost = Probability of Failure × Cost of Failure – Cost of failure includes: Labor cost Material cost Opportunity cost (due to downtime) Total expected failure cost = labor cost + material cost + opportunity cost  9 Failure Mode and Effect Analysis Cost - Example For a hydraulic excavator, the failure mode is hydraulic hose rupture and the effect is the excavator loses hydraulic pressure and stops operating. Calculate the total expected failure cost knowing that: – Failure rate = 0.5 failures/year – Detection time = 1 hour – Fixing time = 8 hours (including delay time) – Delay time = 0 hours – Recovery time = 9 hours (detection + fixing + delay) – Operational life of the excavator) = 20 years – Labor rate = $100/hour – Number of operators = 2 – Spare parts cost = $1,250 (for a new hydraulic hose) – Opportunity cost = $25,000/hour (due to lost production)  10 Failure Mode and Effect Analysis Cost - Example Labor cost = failure rate×lifespan×labor rate×# of operators ( detection time + fixing time ) Labor cost = 0.5  20  $100  2  (1 + 9 ) = $18, 000 Material cost = failure rate×lifespan× quantity × cost of part Material cost = 0.5  20 1 $1, 250 = $12,500 Opportunity cost = failure rate×lifespan  recovery time × opportunity cost per hour Opportunity cost = 0.5×20  9  $25, 000 = $2, 250, 000 Total expected failure cost = labor cost + material cost + opportunity cost Total expected failure cost = $18, 000 + $12,500 + $2, 250, 000 = $2,280,500 – The high cost is driven by the opportunity cost due to downtime, highlighting the critical nature of hydraulic system failures. – Over a 20-year, the expected cost of this failure mode is substantial. – To minimize the frequency of failures and the associated costs, preventive maintenance measures must be considered (i.e., regular hose inspections and scheduled replacements.  11 Reliability Testing Cost Estimation Model Reliability testing cost estimation model: – It is concerned with estimating the cost of reliability testing: RTC = ( RTCh )(T ) RTC is the reliability testing cost, T is the number of hours needed to perform reliability testing, RTCh is the hourly cost of performing reliability testing. The number of hours needed to perform reliability testing is expressed by: T = (182.07 )   is the factor whose value depends on the complexity of hardware; for more than 25, 000 parts ( = 3) , for between 15, 000 and 25, 000 parts ( = 2 ) for less than 15, 000 parts ( = 1).  12 Life Cycle Costing Concept The life cycle cost of an item is the sum of all costs (purchase and ownership) over its entire life span. This is a widely used concept in the industrial sector because: – Competition, – Rising maintenance costs, – Budget limitations, – Costly products (military systems, aircraft), – Greater ownership costs in comparison to purchase costs, – Increasing cost-effectiveness awareness among product users.  13 Life Cycle Costing Concept This concept can be used for various purposes: – Selecting among options, – Forecasting future budget needs, – Deciding the replacement of aging equipment, – Choosing the most beneficial procurement policy, – Making strategic decisions and design tradeoffs, – Determining cost drivers, – Improving comprehension of basic design-associated parameters in product design and development.  14 Life Cycle Costing Concept Various types of data are required to perform life cycle costing studies of equipment: – Tax, – Discount and escalation rates, – Useful life, – Salvage value/cost, – Periodic operating cost, – Purchase cost, – Periodic maintenance cost, – Transportation and installation cost.  15 Life Cycle Cost Estimation Model In this model, the item/equipment life cycle cost is divided into four areas: research and development cost, production and construction cost, operation and support cost and disposal cost. Mathematically the item/equipment life cycle cost is expressed by: LCC = Crd + C pc + Cos + Cd where Crd is the item/equipment research and development cost, C pc is the item/equipment production and construction cost, Cos is the item/equipment operation and support cost, Cd is the item/equipment disposal cost.  16 Life Cycle Cost Estimation Models - Example A mining company is considering replacing a loader with a better one. Three manufacturers are bidding to sell the equipment. The data given in the table below are available on these three manufacturers’ equipment. Using life cycle costs, determine which of the three pieces of equipment the mining organization would purchase.  17 Life Cycle Cost Estimation Models – Example Discounted value The present value of any item/project over its useful life: 1 − (1 + i )− n  Present Value = Future Value  ,  i  i = Annual interest rate, n = Useful operating life The future value of a car in 10 years $200, i=10% 1 − (1 + 0.1)−10  The present value = 200    = $1,229  0.1   18 Life Cycle Cost Estimation Models - Example Estimate of all involved costs (purchase costs, maintenance costs, failure costs, operating costs …) – All future costs should be discounted to the present. LCC = Selling price + PV of equipment failure costs + PV of equipment operating costs Using the data from the table, we get the following annual expected failure costs of equipment A, B, and C, respectively: The annual expected failure cost (EFC) = Annual failure rate × Cost of a failure EFC A = ( 0.05 )( $8, 000 ) = $400 EFCB = ( 0.06 )( $9, 000 ) = $540 EFCC = ( 0.04 )( $7, 000 ) = $280  19 Life Cycle Cost Estimation Models - Example The present value of the expected failure cost: 1 − (1 + 0.07 )−10  PVEFC A = $400   = $2,809.4  0.07  1 − (1 + 0.07 )−10  PVEFCB = $540   = $3, 792.7  0.07  1 − (1 + 0.07 )−10  PVEFCC = $280   = $1,966.6  0.07  1 − (1 + i )− n  PVEOC = Annual operating cost  ,  i  PVEOC is the present value of any equipment operating costs over its useful life i = Annual interest rate, n = Useful operating life Annual operating c ost is given in the table.  20 Life Cycle Cost Estimation Models - Example The present value of the expected operationg cost: 1 − (1 + 0.07 )−10  PVEOC A = $6, 000   = $42,141.5  0.07  1 − (1 + 0.07 )−10  PVEOCB = $8, 000   = $56,188.7  0.07  1 − (1 + 0.07 )−10  PVEOCC = $4, 000   = $28, 094.3  0.07  Using the given and calculated values, the life cycle costs of equipment A, B, and C are: LCC = Selling price + PV of equipment failure costs + PV of equipment operating costs LCC A = 400, 000 + 2,809.4 + 42,141.5 = $444,950.9 LCCB = 350, 000 + 3, 792.7 + 56,188.7 = $409, 981.4 LCCC = 425, 000 + 1,966.6 + 28, 094.3 = $455, 060.9 As the life cycle cost of equipment B is the lowest, it should be purchased by the mining organization.  21 Life Cycle Costing Steps This study can be performed by following seven steps: 1. Determine useful life of the item/equipment under consideration. 2. Estimates of all involved costs (purchase and maintenance costs, …). 3. Determine item/equipment terminal cost. Loading and unloading at the origin and destination (unavoidable) and intermediate (transshipment) costs (can be avoided). 4. Subtract the item/equipment terminal cost from its ownership cost. 5. Discount the result of the preceding step to the present value. 6. Calculate the item/equipment life cycle cost by adding the result of the previous step to the item/equipment procurement cost. The expenses associated with purchasing supplies and materials to run your item/equipment 7. Repeat the above steps for all items/equipment under consideration for purchase.  22 Life Cycle Cost Estimation Models The item/equipment, research and development cost is made up of seven major components: 1. Research cost, 2. Product planning cost, 3. Engineering design cost, 4. Software cost, 5. Test and evaluation cost, 6. Design documentation cost 7. Life cycle management cost. Three major elements of operation and support cost are: 1. Product distribution cost, 2. Sustaining logistic support cost 3. Product/equipment operation cost.  23 Life Cycle Cost Estimation Models The main components of production and construction cost: 1. Construction cost, 2. Quality control cost, 3. Industrial engineering and operations analysis cost, 4. Manufacturing cost 5. Initial logistics support cost. Three major elements of disposal cost: 1. Ultimate retirement cost 2. Reclamation value 3. Overall equipment disposal cost  24 Don’t Forget! Check announcements on D2L regularly Review additional resources available on D2L Design project due Nov 3rd Presentations start on Nov 21st Last day of classes: Wed Dec 4th Final Exam: Dec 12th @ 14:00 - RC-200 (Lec 10 – Lec 19) Mining Optimization Laboratory Reliability of Mining Equipment ENGR 4366 - Fall 2024 Lec13 – Guest Lecture Natalie Hawkins Dr. Ahlam Maremi Bharti School of Engineering Laurentian University Office: F215B Email: [email protected] Ahlam Maremi Mining Optimization Laboratory 2 Introduction Ahlam Maremi Mining Optimization Laboratory 3 Don’t Forget! Check announcements on D2L regularly Review additional resources available on D2L Presentations start on Nov 21st Last day of classes: Wed Dec 4th Final Exam: Dec 12th @ 14:00 - RC-200 (Lec 10 – Lec 19) Ahlam Maremi Mining Optimization Laboratory Reliability of Mining Equipment ENGR 4366 - Fall 2024 Lec14 - Introduction to Robotics Dr. Ahlam Maremi Bharti School of Engineering Laurentian University Office: F215B Email: [email protected] 2 Introduction The evolution of human technology: – Tool use, – Machine adoption for daily tasks, – Rapid technological advancement, – Computer automation for information handling, – Widespread use of unmanned systems (e.g., ticket consoles), – Integration of robots in various environments. Ahlam Maremi Mining Optimization Laboratory 3 Programmable Electronics (PE) Technology PE technology includes microprocessors, embedded controllers, programmable logic controllers (PLCs), and associated software. It's used to improve safety, increase productivity, and enhance mining's competitive position. 4 Programmable Electronics (PE) Technology Applications in mining: – Automated drilling: PLC-based systems enable tele-monitoring and control of drill equipment. Allows precise programming of drilling instructions (angle, location, depth) from remote locations. Feedback from sensors enables real-time monitoring and adjustments. – Transportation systems: PLCs manage conveyor belts, buckets, and hoists for ore and personnel transport. Enables electronic switching, reducing the use of fault-prone components like large electrical switches and relays. Optimizes transportation with on-time deliveries and reduced downtime. Ahlam Maremi Mining Optimization Laboratory 5 Programmable Electronics (PE) Technology – Equipment positioning and navigation: PLC equipment fitted with laser beams, gyro systems, and accelerometers aids in positioning and navigation. – Energy efficiency monitoring: Digital twin technology used to model rigs and test potential equipment upgrades. Helps identify energy-saving technologies and analyze decarbonization initiatives. – Blasting operations: PLC-based electronic detonators enable precise programming of each blast. Allows for controlled blasts, improving safety and efficiency. 6 Programmable Electronics (PE) Technology Benefits of PE in mining: – Improved safety by removing workers from hazardous areas. – Enhanced efficiency and productivity. – Real-time process information for informed decision-making. – Remote monitoring and control capabilities. – Reduced environmental impact through optimized operations. Ahlam Maremi Mining Optimization Laboratory 7 Programmable Electronics (PE) Technology Challenges and considerations: – PE technology adds complexity that may affect worker safety if not properly implemented. – Requires a systematic approach to safety throughout the system's lifecycle. – Necessitates new skills and training for mining personnel. Future trends: – Increasing automation of drilling processes. – Development of more sophisticated digital twin technologies. – Integration of AI and machine learning for predictive maintenance and optimization. 8 Automation Automation: – Allows humans to enhance the capabilities of their tools and machines. Automaton (a self-operating machine) – It is designed to follow a sequence of operations or respond to predetermined instructions. – Often used to give the illusion of operating under their own power, like mechanical robots, and for entertainment or impressing people with automated puppets resembling humans or animals. – Examples: Bellstrikers in mechanical clocks are automata designed to simulate self-operation. Ahlam Maremi Mining Optimization Laboratory 9 Automation Key functions required for self-operation of tools and machines: – Performance detection: Continuous monitoring of output and process parameters Comparison of actual performance against expected results Use of sensors, cameras, and data analysis to identify deviations – Process correction: Automated adjustment of process variables Feedback loops to maintain desired output Real-time modifications to machine settings or operations – Adjustments due to disturbances: Adaptive control systems to handle unexpected changes Machine learning algorithms to optimize for: – Increased volume – Improved dimensional accuracy – Enhanced overall product quality – Ability to respond to new, unknown disturbances 10 Automation – Enabling autonomous operation: Integration of artificial intelligence and advanced control systems Development of robust decision-making algorithms Implementation of fail-safe mechanisms and error handling Development of these functions evolved in history. – Automation has substituted computer-based control systems for most control systems previously based on human-aided mechanical or pneumatic systems; – Automation has replaced human effort, eliminated significant labor costs and prevented accidents and injuries that might occur. Ahlam Maremi Mining Optimization Laboratory 11 Automation Automation simplified: – It means self-operating systems without human intervention – Origin: Greek "automates" meaning "acting by itself" – Machines, tools, and systems performing tasks independently – It contains its own power source. – Autonomous decision-making based on programs and external inputs – Robots as typical automatons 12 Robots Robots: – Are machines that automate movements and information processing. – Physical machines programmed to carry out a series of actions autonomously or semi-autonomously. Robots are: – Programmable mechanical devices – Perform varied tasks automatically – Capable of manipulation and movement – Designed for flexible, purposeful actions – Specialized for specific domains Examples: – Surgical robots, service robots, welding robots, toy robots, … Ahlam Maremi Mining Optimization Laboratory 13 Robots Advantages of Hiring a robot over hiring a live human: – Never late, – Do not need to take breaks, – Never complain, – Never steal, – Do not call in sick, – Do not serf the web, – Do not make excuses, – Do not come to work drunk or high, – Do the work fast and consistent, – Do not sue the employer. In the near future, robots will control the world. So, learn robotics!! 14 Robots General Motors implemented the first industrial robot, called UNIMATE, in 1961 for die-casting at an automobile factory in New Jersey. By now, millions of robots are routinely employed and integrated throughout the world. Ahlam Maremi Mining Optimization Laboratory 15 Robotics and Automation Automation includes more than just robotics: – Robotics: An important subset of automation – Infrastructure: water supply, power supply networks, – Non-robot devices: timers, locks, valves, and sensors – Automatic and automated machines: flour mills, looms, drills, presses, vehicles and printers, inspection machines, measurement workstations, and testers – Installations: elevators and conveyors, railways, satellites and space stations – Systems: computers and office automation, internet and cellular phones, software packages. 16 Robotics Robotics: – The interdisciplinary field of science and engineering focused on the design, construction, operation, and use of robots. Robotics typically relies on five major components: – Mechanical structure (platform/body) – Actuators (for movement and manipulation) – Sensors (for perception and data collection) – Control system (for autonomy and decision-making) – Power source (for energy supply) Ahlam Maremi Mining Optimization Laboratory 17 Mining Technological Advancement Why mining needs technological advancement: – Competitive pressure: Competition from companies with lower labor costs and better ore deposits – Deeper mining challenges: Increased health and safety risks underground Higher rock stress and hotter environments Rising ventilation costs – Productivity improvement: Need for better efficiency to offset increasing costs – Technological opportunities: Potential to apply advancements from other industries (e.g., automotive) – Automation potential: Leveraging new technologies to address challenges and improve operations 18 Mining Technological Advancement Features to enhance mining technological advancement: – In 1970s: focus on physical capabilities (e.g., 150 lbs weight limit for machine operators) – Now, emphasis on education (college degree) and experience – Key skills for modern mining professionals: Environmental awareness Teamwork abilities Communication skills and technical proficiency – Education and training: Close collaboration with colleges and universities Ongoing professional development through seminars and workshops – Economic factors: Incentives to drive innovation and adoption of new technologies – Interdisciplinary approach: Combining engineering, environmental science, and economics Ahlam Maremi Mining Optimization Laboratory 19 Mining Technological Advancement Jackleg hammers: – The old version: manual operation, high vibration, limited dust control – Now, improved ergonomics, vibration dampening, integrated water injection for dust suppression – https://www.youtube.com/watch?v=U_0bzSV-wDI 20 Mining Technological Advancement Slushers: – The old slusher controlled manually with limited material handling capacity. – Now, automated control systems, remote operation, improved scraper designs. – https://www.bing.com/videos/search?q=slusher+mining&view=detail&mid=4B8D0A5D80F25DE862 7D4B8D0A5D80F25DE8627D&FORM=VIRE Ahlam Maremi Mining Optimization Laboratory 21 Mining Technological Advancement Loaders: – Old loaders operated manually and diesel-powered – Now, autonomous or semi-autonomous operation, electric and battery-powered models, telematics for real-time monitoring. Drilling rigs: – In the past, manual rod handling, limited data collection. – Now, automated rod handling systems, real-time data collection and analysis, remote operation capabilities. 22 Mining Technological Advancement Surveying and mapping: – Used to be manual surveying techniques, 2D maps – Now, drones for aerial surveying, 3D mapping and modeling, GPS technologies – https://www.youtube.com/watch?v=vuh9OX2E6ek Ahlam Maremi Mining Optimization Laboratory 23 Mining Technological Advancement Data collection and analysis: – Data recorded manually, delayed analysis. – Now, real-time data acquisition, artificial intelligence for predictive analytics. Equipment monitoring: – Regular manual inspections (old). – Now, IoT sensors for real-time condition monitoring, predictive maintenance. Safety and training: – In the past, on-site training, limited safety monitoring – Now, virtual and augmented reality for training, real-time personnel tracking 24 Mining Technological Advancement Environmental monitoring: – In the past, periodic manual sampling – Now, there is continuous monitoring with wireless sensors, real-time environmental data analysis. Ore extraction and processing: – It was less precise extraction methods, manual sorting – Now, precision drilling guided by 3D models, automated sorting and processing The integration of digital technologies, automation, and electrification is transforming traditional mining equipment and practices, making operations more productive and environmentally friendly. Ahlam Maremi Mining Optimization Laboratory 25 Automated - Cannot Be Automated!! What can be automated? what cannot be automated? – Anything that can be reduced to an algorithm or computational process can be automated, – Most human thought and most functions of complex adaptive systems, are not reducible to a logical algorithm or a computational process, therefore cannot be automated. Automation is the future of mining: – https://www.youtube.com/watch?v=POqw0rIJe78 to 3:20 26 Programmable Electronics vs. Automation Programmable electronics: – PE Technology is more specific, focusing on programmable electronic devices like microprocessors, embedded controllers, PLCs, and associated software. – PE enable more precise, efficient, and safer operations across various aspects of mining. PE Technology is used to implement precise control and monitoring in automated systems. Automation: – Broader concept involving the use of various technologies to perform tasks with minimal human intervention. – Automation can include PE technology, but it can also include other non-electronic systems. Ahlam Maremi Mining Optimization Laboratory 27 Examples Mechanized pneumatic drilling – https://www.youtube.com/watch?v=EeinapEyEsE&t=110s Development drill rig – https://www.bing.com/videos/riverview/relatedvideo?&q=pneumatic+drill+rigs+mining& &mid=616C55AE728F09C82C09616C55AE728F09C82C09&mmscn=mtsc&aps=12& FORM=VRDGAR LHD COMAND underground – https://www.youtube.com/watch?v=DPZCkqyyoQw XE battery electric LHD – https://www.youtube.com/watch?v=9oFaAjSgQZQ 28 Don’t Forget! Check announcements on D2L regularly Review additional resources available on D2L Presentations start on Nov 21st Last day of classes: Wed Dec 4th Final Exam: Dec 12th @ 14:00 - RC-200 (Lec 10 – Lec 19) Ahlam Maremi  Reliability of Mining Equipment ENGR 4366 - Fall 2024 Lec15 – Classification of Robots Dr. Ahlam Maremi Bharti School of Engineering Laurentian University Office: F215B Email: [email protected] 2 Robots Level of Autonomy A robot is a programmable machine that can complete a task or series of tasks. – The term robotics describes the field of study focused on developing robots and automation. Each robot has a different level of autonomy: – Semi-autonomous robots that carry out tasks (teleoperation) They give the individual the sense of being on location in a remote, dangerous or alien environment, and enable them to interact with it since it continuously provides sensory feedback. – Fully-autonomous robots that perform tasks without any external influences: It perceives its environment, makes decisions based on what it perceives. Ahlam Maremi  3 Classification of Robots Based on the type of trajectory control: – Point-to-point systems. – Continuous systems. Based on the type and number of axes of the manipulator: – Cartesian – Cylindrical – Spherical – Articulated Based on the type of control systems: – Open-loop control systems. – Closed-loop control systems. 4 Point-to-point (PTP-System) Arms of the robot move to a defined number of locations then the motion stops. The robot stays stationary while the end-effector performs the required task. – The end effector is the device at the end of a robotic arm, designed to interact with the environment. – Grippers, magnetic grippers, vacuum cups, cutting tools, tool changers, sensors, … Ahlam Maremi  5 Point-to-point (PTP-System) The robot moves to the location or point upon completion of the current task and the cycle repeats. The end-points are the points which the robot follows during movements to execute a task from the starting position to the finishing point. Applications: – Pick-and-place (to move objects from one place to another). – Assembly lines(to move components from station to station in a fixed sequence). – Material handling (to transport materials in a fixed environment, like warehouses or manufacturing plants). – Medical robotics (very precise - to move between predefined points, like in some forms of surgery or diagnostics). – Underground jumbo drills 6 Jumbo Mining Drill Ahlam Maremi  7 Advantages and Disadvantages Advantages of P2P systems: – It is simple to use and control because it only requires basic motion between set points. – It moves quickly between points (no need for complex path planning). – Less expensive to design, especially for simple tasks. Disadvantages of P2P systems: – The robot can't adapt to changes in the environment or task during motion; it can only move between the predefined points. – If the task requires interaction with an environment or complex maneuvering, P2P robots might not be effective. 8 Continuous-Path Systems (CP-Systems) CP robotic systems move smoothly along a continuous trajectory, adjusting their path in real-time based on feedback from the environment or the task. It can stop at any specified point along the controlled path. All the points along the path must be stored explicitly in the robot's control memory. Ahlam Maremi  9 Continuous-Path Systems (CP-Systems) Characteristics of continuous robot systems: – They maintain motion between points. It does not stop at intermediate positions. – They can adjust their path in response to changes in the environment or task (more flexible and capable of handling more complex tasks). – They require sophisticated path planning algorithms considers the start and end points and the smoothness and efficiency of the path). – They often rely on feedback from various sensors (like cameras, LIDAR, or tactile sensors) to adapt their motion in real time. To avoid collisions, optimize their path, or fine-tune their precision. – They are used in applications that require higher precision and smooth motion between points (autonomous vehicles). 10 Continuous-Path Systems – Applications Applications – Welding, painting, cutting, or assembly, where smooth and continuous motion is required to ensure high precision. – Autonomous vehicles (to navigate along continuous paths while avoiding obstacles, adjusting to road conditions, and making real-time decisions. – Robotic surgery (the robot needs to move smoothly and precisely to avoid damaging delicate tissues or organs). – 3D printers (the robot follows continuous paths to build layers of material in a precise, controlled manner). – To apply paint or coatings to surfaces with consistent and even coverage. – In the mining industry (underground mine – shotcreting). Ahlam Maremi  11 Underground Shotcreting 12 Underground Shotcreting Ahlam Maremi  13 Advantages of CP Systems Advantages of CP Systems: – They can execute smooth, fine-tuned movements, making them ideal for high-precision tasks. – They can dynamically adjust their motion in real-time based on sensory feedback, allowing them to respond to unexpected changes in the environment or task conditions. – They can reduce cycle times in some tasks because it doesn't stop and start at every point. – They can follow complex curves or trajectories. 14 Disadvantages of CP Systems Disadvantages of CP systems: – They are more complex to design and control, requiring advanced algorithms for motion planning, real-time feedback, and path optimization. – More expensive in terms of both hardware and software. – The real-time adjustments and feedback loops required to maintain a continuous path can increase the computational load on the system, demanding more powerful processors and more sophisticated control systems. – Continuous motion requires more energy depending on the complexity of the path and the length of the operation. – Real-time adjustments introduce the potential for errors in decision- making (dynamic environments or where sensor data is unreliable). Ahlam Maremi  15 Number of Axes - The Robotic Manipulator In robotics, a manipulator is a device used to manipulate materials without direct physical contact by the operator. – Manipulators were originally for dealing with radioactive or biohazardous materials or in inaccessible places. The manipulator composed of the body, arm and wrist. The arm and body control overall movement, while the wrist maneuvers the end-effector. 16 Number of Axes - The Robotic Manipulator A Degree of Freedom (DOF): the number of independent movements the manipulator can make, either in a rotational or linear sense. Applications require the manipulator with 4 to 6 degrees of freedom (axes of movement). Example: – Power shovel excavators. Every geometric axis that a joint can rotate around or extend along is counted as a Single Degree of Freedom. 3 degrees of freedom Ahlam Maremi  17 The Robotic Manipulator - Direct and Indirect Drives The joints of the manipulator can be driven directly or indirectly: – Direct drives: The motor is directly connected to the joint without any intermediate gearing or transmission mechanisms. Advantages: – Higher precision and accuracy. – Faster response times and better dynamic performance. – Simple design with fewer components. – Lower maintenance requirements. Disadvantages: – Require larger motors for high-load applications. – Higher cost for high-performance motors. – Potentially higher energy consumption in some applications 18 The Robotic Manipulator - Direct and Indirect Drives – Indirect drives: They use gearing, belts, or other transmission mechanisms between the motor and the joint. Advantages: – Higher torque output for a given motor size due to gear reduction – Ability to use smaller, less expensive motors – Potential for better energy efficiency – Can be used for heavy-duty applications Disadvantages: – Potential gear-related errors – More complex design with additional components – Higher maintenance requirements ?