CHAPTER XVII. POWER CALCULATIONS.Power Units. Power is the rate of doing work. If a force of 10 pounds is applied to a body moving at the rate of 300 feet per minute, the power will be expressed by 10 x 300 = 3000 foot-pounds per minute. As the figures obtained by the foot and pound units are usually inconveniently large, the "Horsepower" unit has been adopted. A horsepower is a unit that represents work done at the rate of 33,000 foot-pounds per minute, or 550 foot-pounds per second. Thus if a certain aeroplane offers a resistance of 200 pounds, and flies at the rate of 6000 feet per minute, then the work done per minute will be equal to 200 x 6000 = 1,200,000 foot-pounds. Since there are 33,000 foot-pounds of work per minute for each horsepower, the horsepower will be: 1200000/33000 = 36.3.As aeroplane speeds are usually given in terms of miles per hour, it will be convenient to convert the foot-minute unit into the mile per hour unit. If H = horsepower, R = resistance of aeroplane, and V = miles per hour, then H = RV/375, the theoretical horsepower, without loss. If an aeroplane flies at 100 miles per hour and requires a propeller thrust of 300 pounds, then the horsepower becomes:H = RV/375 = 300 x 100/375 = 80 horsepower. This is the actual power required to drive the machine, but is not the engine power, as the engine must also supply the losses due to the propeller. The propeller losses are generally expressed as a percentage of the total power supplied. The percentage of useful power is known as the "Efficiency."The efficiency of the average aeroplane propeller will vary from 0.70 to 0.80. If e = propeller efficiency expressed as a decimal, the motor horsepower becomes: H = RV/375e. To obtain the motor horsepower, divide the theoretical horsepower by the efficiency. Using the complete formula for the solution of an example in which the flight speed is 100 M. P. H., the resistance 225 pounds and the efficiency 0.75, we have:H = RV/375e = 225 x 100/375 x 0.75 = 80 horsepower.Power Distribution. Since power depends upon the total resistance to be overcome, part of the power will be used for driving the lifting surfaces and a part for overcoming the parasitic resistance. The power required for driving the wings depends upon the angle of incidence, since the drag varies with every angle. The wing power varies with every flight speed, owing to the changes in angle made necessary to support the constant load. The power for the wings will be least at the speed and angle that corresponds to the greatest lift-drag ratio. Owing to the low value of the L/D at very small and very large angles, the power requirements will be excessive at extremely low and high speeds.As the parasitic resistance increases as the square of the speed, the power for overcoming this resistance will vary as the cube of the speed. It is the parasitic resistance that really limits the higher speeds of the aeroplane, since it increases very rapidly at velocities of over 60 miles per hour.The total power at any speed is the sum of the wing power and power required for the parasitic resistance. Owing to variations in the wing drag and resistance at every point within the flight range, it is exceedingly difficult to directly calculate the total power at any particular speed. The wing drag and the resistance should be calculated for every speed, and then laid out by a graph or curve. The minimum propeller thrust, or the minimum total resistance, occurs approximately at the speed where the body resistance and wing drag are equal. The minimum horsepower occurs at a low speed, but not the lowest speed, and this will differ with every machine.Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section.Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section. The Results Were Taken From Full Size Tests Made by the English Government.Fig. 1 is a set of performance curves drawn from the results of tests on a full size Bleriot monoplane. At the bottom the horizontal row of figures gives the horizontal speed in feet per second. The first column to the left is the horsepower, and the second column is the resistance or drag in pounds. The four curves represent respectively the body resistance, wing or "plane" resistance, horsepower, and total resistance. The horizontal line "AV" shows the available horsepower. It will be noted that the body resistance increases steadily from 9 pounds at 50 feet per second, to 180 pounds at 100 feet per second. The wing resistance, on the other hand, decreases from 350 pounds at 56 feet per second to a minimum of 130 pounds at 83 feet per second. It will be noted that the angles of incidence are marked along the wing-drag curve by small circles. The incidence is 6° at 75 feet per second, and 4° at a little less than 85 feet per second.The available horsepower "AV" is 42. This is shown as a straight line, although in the majority of cases it is slightly curved owing to variations in power at the higher speeds, and to variations in the propeller efficiency. At 90 feet per second the actual horsepower curve crosses the line of available horsepower "AV." Beyond this point horizontal flight is no longer possible, as the power requirements would exceed the available horsepower. It will be noted that the lowest total resistance occurs near the point where the body and wing resistance curves intersect, or in other words, where the body and wing resistance are equal. The minimum horsepower takes place at 63 feet per second, or at a point nearly 1/3 between the lowest flight speed and the highest speed attained by the available horsepower in horizontal flight (90 ft/sec).The actual range of flight speeds is limited to points between the intersection of the "Horsepower required" curve, and the "Available horsepower" curve. By increasing the propeller efficiency, or by increasing the power of the motor, the available horsepower line is raised and the flight range increased.Horsepower For Climbing. Up to the present we have only considered horizontal flight. The power available for climbing is the difference between the power required to maintain horizontal flight at any speed, and the actual horsepower that can be delivered by the propeller. Thus, if the actual power delivered by a motor through the propeller is 85 horsepower, and the power required for horizontal flight at that speed is 45, then we have: 85–45= 40 horsepower available for climbing. Since the difference between the driving power and the power required for horizontal flight is less at extremely low and high speeds, it is evident that we will have a minimum climbing reserve at the high and low speeds. Consulting the power curve for the Bleriot monoplane, we see that the power required at 56 feet per second is 40 horsepower, and at 85 feet per second it is 38 horsepower. At the low speed we have a climbing reserve of 44–40 = 4 H. P., and at the higher speed 44–38 = 6 H. P. The maximum available horsepower "AV" is 44 horsepower. The minimum horizontal power required is found at 63 feet per second, the climbing reserve at this point being 44–28 = 16 horsepower. At 55 feet per second, and at 90, we would not be able to climb, as we would only have sufficient power to maintain horizontal flight.If W = total weight of aeroplane, c = climbing speed, and H = horsepower reserve for climbing, then the climbing speed with a constant air density will be expressed by: c = 33000H/W. Assuming that the weight of the Bleriot monoplane is 800 pounds, and that we are to climb at the speed of the greatest power reserve (16 horsepower), our rate of climb is:c = 33000H/w = 33000 x 16/800 = 660 feet per minute. It should be understood that this is the velocity at the beginning of the climb. After prolonged climbing the rate falls off because of diminishing power and increasing speed. Much depends upon the engine performance at the higher altitudes, so that the reserve power for climb usually diminishes as the machine rises, and hence the rate of climb diminishes in proportion.The following table taken from actual flying tests will show how the rate of climb decreases with the altitude. These machines were equipped with 150 H. P. Hispano-Suisa motors. It will be noted that the S. P. A. D. and the Bleriot hold their rate of climb constant up to 7800 feet altitude, which is a feat that is undoubtedly performed by varying the compression of the engine.Besides increasing the power, the rate of climb can also be increased by decreasing the weight of the aeroplane.Rate of Climb TableMaximum Altitude. The maximum altitude to which a machine can ascend is known as its "Ceiling." This again depends on both the aeroplane and the motor, but principally on the latter. It has been noted that machines having the greatest rate of climb also have the greatest ceiling. Thus the ceiling of a fast climbing scout is higher than that of a larger and slower machine. Based on this principle, a writer in "Flight" has developed the following equation for ceiling, which, of course, assumes a uniform decrease in density. Let H = maximum altitude, h = the altitude at any time t after the start of the climb, and a = the altitude after a time equal to twice the time t, then:H = h / (2 - a/h)Approximate values of h and a may be hadfrom the following table, which are the results of a test on a certain aeroplane:Time(Minutes)0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5Altitude(Feet)0.03300 6150 8730 10760 12610 14190 15530 16650 17600If we assume that the height is 10760 feet after the first 10 minutes, and that the altitude after twice this time (20 minutes) is 16650 feet, then the maximum ceiling attained will be:H = h / (2 - a/h) = 10760 / 2 - 16650/10760 = 23,770 feet. The use of this formula requires that the climb be known for certain time intervals before the ceiling.Gliding Angle. The gliding angle of the wings alone is equal to the lift-drag ratio at the given angle. The best or "Flattest gliding angle" is, of course, the best lift-drag ratio of the wing—say on the average about 1 in 16. The gliding angle of the complete machine is considerably less than this, owing to the resistance of the body and structural parts. This generally reduces the actual angle to less than 12, and in most cases between 6 and 8. Expressed in terms of degrees, tan ø = R/W where R = head resistance and W = weight in pounds.Fig. 2 is a diagram giving the gliding force diagram. The plane descends along the gliding path AC, making the angle of incidence (ø). When in horizontal flight, the lift is along OL and the weight is OW. When descending on the gliding path the lift maintains the same relation with the wing, but the relative angle of the weight is altered. The weight now acts along OG. The drag is represented by OD, with the propeller thrust OP equal and opposite to it. With the weight constant, the lift OL is decreased by the angle so that the total life = L = W cos ø. The action of the weight W produces the propelling component OP that gives forward velocity. The line AB is the horizontal ground line. If the total lift-drag ratio is 8, then the gliding angle will be 1 in 8, or measured in degrees, tan ø = R/W = 1/8 = 0.125. From a trigonometric table it will be found that this tangent corresponds to an angle of 7° – 10'. It should be noted that R is the total resistance and not the wing-drag.Fig. 2. Gliding Angle DiagramFig. 2. Gliding Angle Diagram Showing Component of Gravity That Causes Forward Motion. The Gliding Angle Depends Upon the Ratio of the Resistance to the Weight.Complete Power Calculations. Knowing the total weight and the desired speed, we must determine the wing section and area before we start on the actual power calculations. This can either be determined by empirical rules in the case of a preliminary investigation, or by actual calculation by means of the lift coefficients after the approximate values are known. Sustaining a given weight, we can vary the angle, area, wing section, or the speed, the choice of these items being regulated principally by the power. Given a small area and a great angle of incidence, we can support the load, but the power consumption will be excessive because of the low value of the L/D ratio at high angles. If small area is desired, a large value of Ky due to a high lift-wing section is preferable to a low lift wing at high angles. In general, the area should be so arranged that the wing is at the angle of the maximum lift-drag ratio at the rated speed. A low angle means a smaller motor, less fuel, and hence a lighter machine. This selection involves considerable difficulty, and a number of wing sections and areas must be tried by the trial and error method until the most economical combination is discovered.Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.The first consideration being the total weight, we must first estimate this from the required live load. This can be estimated from previous examples of nearly the same type. Say that our required live load is 660 pounds, and that a live load factor of 0.30 is used. The total weight now becomes 660/0.30=2200 pounds. To make a preliminary estimate of the area we must find the load per square foot. An empirical formula for biplane loading reads: w = 0.065V - 0.25 where V = maximum speed in M. P. H., and w = load per square foot. If we assume a maximum speed of 90 M. P. H. for our machine, the unit loading is w = (0.065 x 90) - 0.25 = 5.6 pounds per square foot. The approximate area can now be found from 2200/5.6 = 393 square feet. (Call 390.) The minimum speed is about 48 per cent of the maximum, or 43 M. P. H. We can now choose one or more wing sections that will come approximately to our requirements by the use of the basic formula, Ky = w/V².At high speed, Ky = 5.6/(90 x 90) = 0.000691. At low speed, Ky = 5.6/(40 x 40) = 0.003030. We must choose the most economical wing between these limits of lift, and on reference to our wing section tables we find:Wing Section TableIt would seem from the above that the chosen area is a little too large, as the majority of the L/D ratios at high speed are poor, the best being 11.00 of the U.S.A.-1. The angles are small, being negative in most cases at high speed. While the lift-drag of the R.A.F-3 is very good at low speed, it is very poor at high, hence the area for this section should be reduced to increase the loading. The R.A.F.-6 and the U.S.A.-1 show up the best, for they are both near the maximum lift at low speed and have fair L/D ratios at high speed. It will be seen that for the best results there should be a series of power curves drawn for the various wings and areas. This method is too complicated and tedious to take up here, and so we will use U.S.A.-1, which does not really show up so bad at this stage. Both the R.A.F.-6 and the U.S.A.-1 have been used extensively on machines of the size and type under consideration. While we require Ky = 0.003030, and U.S.A-1 gives 0.003165, we will not attempt to utilize this excess, as it will be remembered that we should not assume the maximum lift for reasons of stability.The wing-drag at high speed will be 2200/11.0 = 200 pounds, and at low speed it will be: 2200/10.4 = 211 pounds. Since the maximum L/D is 17.8 at 3°, where Ky is 0.00133, the least drag will be: 2200/17.8 = 124 pounds. This least drag will occur at V = V5.6/000133 = 65 M. P. H.The wing drag for each speed must now be divided by the correction factor 0.85, which converts the monoplane values of drag into biplane values. Since this is practically constant it does not affect the relative values of Kx in comparing wings, but it should be used in final results. For this type of machine we will take the total parasitic resistance as r = 0.036V². At 90 M.P.H., r = 0.036 x 90 x 90 = 291.6 pounds. At 65 M. P. H., the resistance is: 0.036 x 65 x 65 = 152.1. At the extreme low speed of 43 M. P.H. we have r - 0.036x43 x 43 = 66.56 pounds. The total resistance (R) is equal to the sum of the wing-drag and the parasitic resistance. At 90 M. P. H. the total resistance becomes 200 + 291.6 = 491.6 pounds. At 65 M.P.H. the total is 124 + 152.1 = 176.1, and at 43 M.P.H. it is 211 + 66.56 = 277.56 pounds. The horsepower is computed from H = RV/375e, and at 90 M. P. H. this is : H = 491.6 x 90/375 x 0.80 = 147.5 H. P. where 0.80 is the assumed propeller efficiency. At 65 M. P. H. the horsepower drops to H = 176.1 x 65/375 x 0.8 = 38.1 H. P., assuming the same efficiency. In the same way the H. P. at 43 M. P. H. r is 39.8.A table and power chart should be worked out for a number of sections and areas according to the following table. The calculations should be computed at intervals of 5 M. P. H., at least the lower speeds. Wing drag is not corrected for biplane interference:Power Chart TableWeight and Power. The weight lifted per horsepower varies in the different types of aeroplanes, this difference lying principally in the reserve allowed for climbing and horizontal speed. A speed scout may carry as little as 8 pounds per horsepower, while a slow two-seater may exceed 20 pounds per horsepower. A rough estimate of the horsepower required may be had by dividing the total weight by the weight per horsepower ratio for that particular type. Thus if the unit H. P. loading is 16 pounds and the total weight is 3200, then the horsepower will equal 3200/16 = 200 horsepower. Assuming that the live load w' is 0.32 of the total weight W, then W = w'/0.32. If m = lbs. per H.P., then H = W/m or H = w'/0.32m. Taking the case of a training machine where m = 20, and the live load is 640 pounds, the approximate horsepower will be: H = w'/0.32m = 640/0.32 x 20 = 100 horsepower. A speed scout carrying 320 pounds useful load, with m = 10, will require H = 320/0.32 x 10 = 100 horsepower.
CHAPTER XVII. POWER CALCULATIONS.Power Units. Power is the rate of doing work. If a force of 10 pounds is applied to a body moving at the rate of 300 feet per minute, the power will be expressed by 10 x 300 = 3000 foot-pounds per minute. As the figures obtained by the foot and pound units are usually inconveniently large, the "Horsepower" unit has been adopted. A horsepower is a unit that represents work done at the rate of 33,000 foot-pounds per minute, or 550 foot-pounds per second. Thus if a certain aeroplane offers a resistance of 200 pounds, and flies at the rate of 6000 feet per minute, then the work done per minute will be equal to 200 x 6000 = 1,200,000 foot-pounds. Since there are 33,000 foot-pounds of work per minute for each horsepower, the horsepower will be: 1200000/33000 = 36.3.As aeroplane speeds are usually given in terms of miles per hour, it will be convenient to convert the foot-minute unit into the mile per hour unit. If H = horsepower, R = resistance of aeroplane, and V = miles per hour, then H = RV/375, the theoretical horsepower, without loss. If an aeroplane flies at 100 miles per hour and requires a propeller thrust of 300 pounds, then the horsepower becomes:H = RV/375 = 300 x 100/375 = 80 horsepower. This is the actual power required to drive the machine, but is not the engine power, as the engine must also supply the losses due to the propeller. The propeller losses are generally expressed as a percentage of the total power supplied. The percentage of useful power is known as the "Efficiency."The efficiency of the average aeroplane propeller will vary from 0.70 to 0.80. If e = propeller efficiency expressed as a decimal, the motor horsepower becomes: H = RV/375e. To obtain the motor horsepower, divide the theoretical horsepower by the efficiency. Using the complete formula for the solution of an example in which the flight speed is 100 M. P. H., the resistance 225 pounds and the efficiency 0.75, we have:H = RV/375e = 225 x 100/375 x 0.75 = 80 horsepower.Power Distribution. Since power depends upon the total resistance to be overcome, part of the power will be used for driving the lifting surfaces and a part for overcoming the parasitic resistance. The power required for driving the wings depends upon the angle of incidence, since the drag varies with every angle. The wing power varies with every flight speed, owing to the changes in angle made necessary to support the constant load. The power for the wings will be least at the speed and angle that corresponds to the greatest lift-drag ratio. Owing to the low value of the L/D at very small and very large angles, the power requirements will be excessive at extremely low and high speeds.As the parasitic resistance increases as the square of the speed, the power for overcoming this resistance will vary as the cube of the speed. It is the parasitic resistance that really limits the higher speeds of the aeroplane, since it increases very rapidly at velocities of over 60 miles per hour.The total power at any speed is the sum of the wing power and power required for the parasitic resistance. Owing to variations in the wing drag and resistance at every point within the flight range, it is exceedingly difficult to directly calculate the total power at any particular speed. The wing drag and the resistance should be calculated for every speed, and then laid out by a graph or curve. The minimum propeller thrust, or the minimum total resistance, occurs approximately at the speed where the body resistance and wing drag are equal. The minimum horsepower occurs at a low speed, but not the lowest speed, and this will differ with every machine.Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section.Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section. The Results Were Taken From Full Size Tests Made by the English Government.Fig. 1 is a set of performance curves drawn from the results of tests on a full size Bleriot monoplane. At the bottom the horizontal row of figures gives the horizontal speed in feet per second. The first column to the left is the horsepower, and the second column is the resistance or drag in pounds. The four curves represent respectively the body resistance, wing or "plane" resistance, horsepower, and total resistance. The horizontal line "AV" shows the available horsepower. It will be noted that the body resistance increases steadily from 9 pounds at 50 feet per second, to 180 pounds at 100 feet per second. The wing resistance, on the other hand, decreases from 350 pounds at 56 feet per second to a minimum of 130 pounds at 83 feet per second. It will be noted that the angles of incidence are marked along the wing-drag curve by small circles. The incidence is 6° at 75 feet per second, and 4° at a little less than 85 feet per second.The available horsepower "AV" is 42. This is shown as a straight line, although in the majority of cases it is slightly curved owing to variations in power at the higher speeds, and to variations in the propeller efficiency. At 90 feet per second the actual horsepower curve crosses the line of available horsepower "AV." Beyond this point horizontal flight is no longer possible, as the power requirements would exceed the available horsepower. It will be noted that the lowest total resistance occurs near the point where the body and wing resistance curves intersect, or in other words, where the body and wing resistance are equal. The minimum horsepower takes place at 63 feet per second, or at a point nearly 1/3 between the lowest flight speed and the highest speed attained by the available horsepower in horizontal flight (90 ft/sec).The actual range of flight speeds is limited to points between the intersection of the "Horsepower required" curve, and the "Available horsepower" curve. By increasing the propeller efficiency, or by increasing the power of the motor, the available horsepower line is raised and the flight range increased.Horsepower For Climbing. Up to the present we have only considered horizontal flight. The power available for climbing is the difference between the power required to maintain horizontal flight at any speed, and the actual horsepower that can be delivered by the propeller. Thus, if the actual power delivered by a motor through the propeller is 85 horsepower, and the power required for horizontal flight at that speed is 45, then we have: 85–45= 40 horsepower available for climbing. Since the difference between the driving power and the power required for horizontal flight is less at extremely low and high speeds, it is evident that we will have a minimum climbing reserve at the high and low speeds. Consulting the power curve for the Bleriot monoplane, we see that the power required at 56 feet per second is 40 horsepower, and at 85 feet per second it is 38 horsepower. At the low speed we have a climbing reserve of 44–40 = 4 H. P., and at the higher speed 44–38 = 6 H. P. The maximum available horsepower "AV" is 44 horsepower. The minimum horizontal power required is found at 63 feet per second, the climbing reserve at this point being 44–28 = 16 horsepower. At 55 feet per second, and at 90, we would not be able to climb, as we would only have sufficient power to maintain horizontal flight.If W = total weight of aeroplane, c = climbing speed, and H = horsepower reserve for climbing, then the climbing speed with a constant air density will be expressed by: c = 33000H/W. Assuming that the weight of the Bleriot monoplane is 800 pounds, and that we are to climb at the speed of the greatest power reserve (16 horsepower), our rate of climb is:c = 33000H/w = 33000 x 16/800 = 660 feet per minute. It should be understood that this is the velocity at the beginning of the climb. After prolonged climbing the rate falls off because of diminishing power and increasing speed. Much depends upon the engine performance at the higher altitudes, so that the reserve power for climb usually diminishes as the machine rises, and hence the rate of climb diminishes in proportion.The following table taken from actual flying tests will show how the rate of climb decreases with the altitude. These machines were equipped with 150 H. P. Hispano-Suisa motors. It will be noted that the S. P. A. D. and the Bleriot hold their rate of climb constant up to 7800 feet altitude, which is a feat that is undoubtedly performed by varying the compression of the engine.Besides increasing the power, the rate of climb can also be increased by decreasing the weight of the aeroplane.Rate of Climb TableMaximum Altitude. The maximum altitude to which a machine can ascend is known as its "Ceiling." This again depends on both the aeroplane and the motor, but principally on the latter. It has been noted that machines having the greatest rate of climb also have the greatest ceiling. Thus the ceiling of a fast climbing scout is higher than that of a larger and slower machine. Based on this principle, a writer in "Flight" has developed the following equation for ceiling, which, of course, assumes a uniform decrease in density. Let H = maximum altitude, h = the altitude at any time t after the start of the climb, and a = the altitude after a time equal to twice the time t, then:H = h / (2 - a/h)Approximate values of h and a may be hadfrom the following table, which are the results of a test on a certain aeroplane:Time(Minutes)0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5Altitude(Feet)0.03300 6150 8730 10760 12610 14190 15530 16650 17600If we assume that the height is 10760 feet after the first 10 minutes, and that the altitude after twice this time (20 minutes) is 16650 feet, then the maximum ceiling attained will be:H = h / (2 - a/h) = 10760 / 2 - 16650/10760 = 23,770 feet. The use of this formula requires that the climb be known for certain time intervals before the ceiling.Gliding Angle. The gliding angle of the wings alone is equal to the lift-drag ratio at the given angle. The best or "Flattest gliding angle" is, of course, the best lift-drag ratio of the wing—say on the average about 1 in 16. The gliding angle of the complete machine is considerably less than this, owing to the resistance of the body and structural parts. This generally reduces the actual angle to less than 12, and in most cases between 6 and 8. Expressed in terms of degrees, tan ø = R/W where R = head resistance and W = weight in pounds.Fig. 2 is a diagram giving the gliding force diagram. The plane descends along the gliding path AC, making the angle of incidence (ø). When in horizontal flight, the lift is along OL and the weight is OW. When descending on the gliding path the lift maintains the same relation with the wing, but the relative angle of the weight is altered. The weight now acts along OG. The drag is represented by OD, with the propeller thrust OP equal and opposite to it. With the weight constant, the lift OL is decreased by the angle so that the total life = L = W cos ø. The action of the weight W produces the propelling component OP that gives forward velocity. The line AB is the horizontal ground line. If the total lift-drag ratio is 8, then the gliding angle will be 1 in 8, or measured in degrees, tan ø = R/W = 1/8 = 0.125. From a trigonometric table it will be found that this tangent corresponds to an angle of 7° – 10'. It should be noted that R is the total resistance and not the wing-drag.Fig. 2. Gliding Angle DiagramFig. 2. Gliding Angle Diagram Showing Component of Gravity That Causes Forward Motion. The Gliding Angle Depends Upon the Ratio of the Resistance to the Weight.Complete Power Calculations. Knowing the total weight and the desired speed, we must determine the wing section and area before we start on the actual power calculations. This can either be determined by empirical rules in the case of a preliminary investigation, or by actual calculation by means of the lift coefficients after the approximate values are known. Sustaining a given weight, we can vary the angle, area, wing section, or the speed, the choice of these items being regulated principally by the power. Given a small area and a great angle of incidence, we can support the load, but the power consumption will be excessive because of the low value of the L/D ratio at high angles. If small area is desired, a large value of Ky due to a high lift-wing section is preferable to a low lift wing at high angles. In general, the area should be so arranged that the wing is at the angle of the maximum lift-drag ratio at the rated speed. A low angle means a smaller motor, less fuel, and hence a lighter machine. This selection involves considerable difficulty, and a number of wing sections and areas must be tried by the trial and error method until the most economical combination is discovered.Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.The first consideration being the total weight, we must first estimate this from the required live load. This can be estimated from previous examples of nearly the same type. Say that our required live load is 660 pounds, and that a live load factor of 0.30 is used. The total weight now becomes 660/0.30=2200 pounds. To make a preliminary estimate of the area we must find the load per square foot. An empirical formula for biplane loading reads: w = 0.065V - 0.25 where V = maximum speed in M. P. H., and w = load per square foot. If we assume a maximum speed of 90 M. P. H. for our machine, the unit loading is w = (0.065 x 90) - 0.25 = 5.6 pounds per square foot. The approximate area can now be found from 2200/5.6 = 393 square feet. (Call 390.) The minimum speed is about 48 per cent of the maximum, or 43 M. P. H. We can now choose one or more wing sections that will come approximately to our requirements by the use of the basic formula, Ky = w/V².At high speed, Ky = 5.6/(90 x 90) = 0.000691. At low speed, Ky = 5.6/(40 x 40) = 0.003030. We must choose the most economical wing between these limits of lift, and on reference to our wing section tables we find:Wing Section TableIt would seem from the above that the chosen area is a little too large, as the majority of the L/D ratios at high speed are poor, the best being 11.00 of the U.S.A.-1. The angles are small, being negative in most cases at high speed. While the lift-drag of the R.A.F-3 is very good at low speed, it is very poor at high, hence the area for this section should be reduced to increase the loading. The R.A.F.-6 and the U.S.A.-1 show up the best, for they are both near the maximum lift at low speed and have fair L/D ratios at high speed. It will be seen that for the best results there should be a series of power curves drawn for the various wings and areas. This method is too complicated and tedious to take up here, and so we will use U.S.A.-1, which does not really show up so bad at this stage. Both the R.A.F.-6 and the U.S.A.-1 have been used extensively on machines of the size and type under consideration. While we require Ky = 0.003030, and U.S.A-1 gives 0.003165, we will not attempt to utilize this excess, as it will be remembered that we should not assume the maximum lift for reasons of stability.The wing-drag at high speed will be 2200/11.0 = 200 pounds, and at low speed it will be: 2200/10.4 = 211 pounds. Since the maximum L/D is 17.8 at 3°, where Ky is 0.00133, the least drag will be: 2200/17.8 = 124 pounds. This least drag will occur at V = V5.6/000133 = 65 M. P. H.The wing drag for each speed must now be divided by the correction factor 0.85, which converts the monoplane values of drag into biplane values. Since this is practically constant it does not affect the relative values of Kx in comparing wings, but it should be used in final results. For this type of machine we will take the total parasitic resistance as r = 0.036V². At 90 M.P.H., r = 0.036 x 90 x 90 = 291.6 pounds. At 65 M. P. H., the resistance is: 0.036 x 65 x 65 = 152.1. At the extreme low speed of 43 M. P.H. we have r - 0.036x43 x 43 = 66.56 pounds. The total resistance (R) is equal to the sum of the wing-drag and the parasitic resistance. At 90 M. P. H. the total resistance becomes 200 + 291.6 = 491.6 pounds. At 65 M.P.H. the total is 124 + 152.1 = 176.1, and at 43 M.P.H. it is 211 + 66.56 = 277.56 pounds. The horsepower is computed from H = RV/375e, and at 90 M. P. H. this is : H = 491.6 x 90/375 x 0.80 = 147.5 H. P. where 0.80 is the assumed propeller efficiency. At 65 M. P. H. the horsepower drops to H = 176.1 x 65/375 x 0.8 = 38.1 H. P., assuming the same efficiency. In the same way the H. P. at 43 M. P. H. r is 39.8.A table and power chart should be worked out for a number of sections and areas according to the following table. The calculations should be computed at intervals of 5 M. P. H., at least the lower speeds. Wing drag is not corrected for biplane interference:Power Chart TableWeight and Power. The weight lifted per horsepower varies in the different types of aeroplanes, this difference lying principally in the reserve allowed for climbing and horizontal speed. A speed scout may carry as little as 8 pounds per horsepower, while a slow two-seater may exceed 20 pounds per horsepower. A rough estimate of the horsepower required may be had by dividing the total weight by the weight per horsepower ratio for that particular type. Thus if the unit H. P. loading is 16 pounds and the total weight is 3200, then the horsepower will equal 3200/16 = 200 horsepower. Assuming that the live load w' is 0.32 of the total weight W, then W = w'/0.32. If m = lbs. per H.P., then H = W/m or H = w'/0.32m. Taking the case of a training machine where m = 20, and the live load is 640 pounds, the approximate horsepower will be: H = w'/0.32m = 640/0.32 x 20 = 100 horsepower. A speed scout carrying 320 pounds useful load, with m = 10, will require H = 320/0.32 x 10 = 100 horsepower.
CHAPTER XVII. POWER CALCULATIONS.Power Units. Power is the rate of doing work. If a force of 10 pounds is applied to a body moving at the rate of 300 feet per minute, the power will be expressed by 10 x 300 = 3000 foot-pounds per minute. As the figures obtained by the foot and pound units are usually inconveniently large, the "Horsepower" unit has been adopted. A horsepower is a unit that represents work done at the rate of 33,000 foot-pounds per minute, or 550 foot-pounds per second. Thus if a certain aeroplane offers a resistance of 200 pounds, and flies at the rate of 6000 feet per minute, then the work done per minute will be equal to 200 x 6000 = 1,200,000 foot-pounds. Since there are 33,000 foot-pounds of work per minute for each horsepower, the horsepower will be: 1200000/33000 = 36.3.As aeroplane speeds are usually given in terms of miles per hour, it will be convenient to convert the foot-minute unit into the mile per hour unit. If H = horsepower, R = resistance of aeroplane, and V = miles per hour, then H = RV/375, the theoretical horsepower, without loss. If an aeroplane flies at 100 miles per hour and requires a propeller thrust of 300 pounds, then the horsepower becomes:H = RV/375 = 300 x 100/375 = 80 horsepower. This is the actual power required to drive the machine, but is not the engine power, as the engine must also supply the losses due to the propeller. The propeller losses are generally expressed as a percentage of the total power supplied. The percentage of useful power is known as the "Efficiency."The efficiency of the average aeroplane propeller will vary from 0.70 to 0.80. If e = propeller efficiency expressed as a decimal, the motor horsepower becomes: H = RV/375e. To obtain the motor horsepower, divide the theoretical horsepower by the efficiency. Using the complete formula for the solution of an example in which the flight speed is 100 M. P. H., the resistance 225 pounds and the efficiency 0.75, we have:H = RV/375e = 225 x 100/375 x 0.75 = 80 horsepower.Power Distribution. Since power depends upon the total resistance to be overcome, part of the power will be used for driving the lifting surfaces and a part for overcoming the parasitic resistance. The power required for driving the wings depends upon the angle of incidence, since the drag varies with every angle. The wing power varies with every flight speed, owing to the changes in angle made necessary to support the constant load. The power for the wings will be least at the speed and angle that corresponds to the greatest lift-drag ratio. Owing to the low value of the L/D at very small and very large angles, the power requirements will be excessive at extremely low and high speeds.As the parasitic resistance increases as the square of the speed, the power for overcoming this resistance will vary as the cube of the speed. It is the parasitic resistance that really limits the higher speeds of the aeroplane, since it increases very rapidly at velocities of over 60 miles per hour.The total power at any speed is the sum of the wing power and power required for the parasitic resistance. Owing to variations in the wing drag and resistance at every point within the flight range, it is exceedingly difficult to directly calculate the total power at any particular speed. The wing drag and the resistance should be calculated for every speed, and then laid out by a graph or curve. The minimum propeller thrust, or the minimum total resistance, occurs approximately at the speed where the body resistance and wing drag are equal. The minimum horsepower occurs at a low speed, but not the lowest speed, and this will differ with every machine.Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section.Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section. The Results Were Taken From Full Size Tests Made by the English Government.Fig. 1 is a set of performance curves drawn from the results of tests on a full size Bleriot monoplane. At the bottom the horizontal row of figures gives the horizontal speed in feet per second. The first column to the left is the horsepower, and the second column is the resistance or drag in pounds. The four curves represent respectively the body resistance, wing or "plane" resistance, horsepower, and total resistance. The horizontal line "AV" shows the available horsepower. It will be noted that the body resistance increases steadily from 9 pounds at 50 feet per second, to 180 pounds at 100 feet per second. The wing resistance, on the other hand, decreases from 350 pounds at 56 feet per second to a minimum of 130 pounds at 83 feet per second. It will be noted that the angles of incidence are marked along the wing-drag curve by small circles. The incidence is 6° at 75 feet per second, and 4° at a little less than 85 feet per second.The available horsepower "AV" is 42. This is shown as a straight line, although in the majority of cases it is slightly curved owing to variations in power at the higher speeds, and to variations in the propeller efficiency. At 90 feet per second the actual horsepower curve crosses the line of available horsepower "AV." Beyond this point horizontal flight is no longer possible, as the power requirements would exceed the available horsepower. It will be noted that the lowest total resistance occurs near the point where the body and wing resistance curves intersect, or in other words, where the body and wing resistance are equal. The minimum horsepower takes place at 63 feet per second, or at a point nearly 1/3 between the lowest flight speed and the highest speed attained by the available horsepower in horizontal flight (90 ft/sec).The actual range of flight speeds is limited to points between the intersection of the "Horsepower required" curve, and the "Available horsepower" curve. By increasing the propeller efficiency, or by increasing the power of the motor, the available horsepower line is raised and the flight range increased.Horsepower For Climbing. Up to the present we have only considered horizontal flight. The power available for climbing is the difference between the power required to maintain horizontal flight at any speed, and the actual horsepower that can be delivered by the propeller. Thus, if the actual power delivered by a motor through the propeller is 85 horsepower, and the power required for horizontal flight at that speed is 45, then we have: 85–45= 40 horsepower available for climbing. Since the difference between the driving power and the power required for horizontal flight is less at extremely low and high speeds, it is evident that we will have a minimum climbing reserve at the high and low speeds. Consulting the power curve for the Bleriot monoplane, we see that the power required at 56 feet per second is 40 horsepower, and at 85 feet per second it is 38 horsepower. At the low speed we have a climbing reserve of 44–40 = 4 H. P., and at the higher speed 44–38 = 6 H. P. The maximum available horsepower "AV" is 44 horsepower. The minimum horizontal power required is found at 63 feet per second, the climbing reserve at this point being 44–28 = 16 horsepower. At 55 feet per second, and at 90, we would not be able to climb, as we would only have sufficient power to maintain horizontal flight.If W = total weight of aeroplane, c = climbing speed, and H = horsepower reserve for climbing, then the climbing speed with a constant air density will be expressed by: c = 33000H/W. Assuming that the weight of the Bleriot monoplane is 800 pounds, and that we are to climb at the speed of the greatest power reserve (16 horsepower), our rate of climb is:c = 33000H/w = 33000 x 16/800 = 660 feet per minute. It should be understood that this is the velocity at the beginning of the climb. After prolonged climbing the rate falls off because of diminishing power and increasing speed. Much depends upon the engine performance at the higher altitudes, so that the reserve power for climb usually diminishes as the machine rises, and hence the rate of climb diminishes in proportion.The following table taken from actual flying tests will show how the rate of climb decreases with the altitude. These machines were equipped with 150 H. P. Hispano-Suisa motors. It will be noted that the S. P. A. D. and the Bleriot hold their rate of climb constant up to 7800 feet altitude, which is a feat that is undoubtedly performed by varying the compression of the engine.Besides increasing the power, the rate of climb can also be increased by decreasing the weight of the aeroplane.Rate of Climb TableMaximum Altitude. The maximum altitude to which a machine can ascend is known as its "Ceiling." This again depends on both the aeroplane and the motor, but principally on the latter. It has been noted that machines having the greatest rate of climb also have the greatest ceiling. Thus the ceiling of a fast climbing scout is higher than that of a larger and slower machine. Based on this principle, a writer in "Flight" has developed the following equation for ceiling, which, of course, assumes a uniform decrease in density. Let H = maximum altitude, h = the altitude at any time t after the start of the climb, and a = the altitude after a time equal to twice the time t, then:H = h / (2 - a/h)Approximate values of h and a may be hadfrom the following table, which are the results of a test on a certain aeroplane:Time(Minutes)0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5Altitude(Feet)0.03300 6150 8730 10760 12610 14190 15530 16650 17600If we assume that the height is 10760 feet after the first 10 minutes, and that the altitude after twice this time (20 minutes) is 16650 feet, then the maximum ceiling attained will be:H = h / (2 - a/h) = 10760 / 2 - 16650/10760 = 23,770 feet. The use of this formula requires that the climb be known for certain time intervals before the ceiling.Gliding Angle. The gliding angle of the wings alone is equal to the lift-drag ratio at the given angle. The best or "Flattest gliding angle" is, of course, the best lift-drag ratio of the wing—say on the average about 1 in 16. The gliding angle of the complete machine is considerably less than this, owing to the resistance of the body and structural parts. This generally reduces the actual angle to less than 12, and in most cases between 6 and 8. Expressed in terms of degrees, tan ø = R/W where R = head resistance and W = weight in pounds.Fig. 2 is a diagram giving the gliding force diagram. The plane descends along the gliding path AC, making the angle of incidence (ø). When in horizontal flight, the lift is along OL and the weight is OW. When descending on the gliding path the lift maintains the same relation with the wing, but the relative angle of the weight is altered. The weight now acts along OG. The drag is represented by OD, with the propeller thrust OP equal and opposite to it. With the weight constant, the lift OL is decreased by the angle so that the total life = L = W cos ø. The action of the weight W produces the propelling component OP that gives forward velocity. The line AB is the horizontal ground line. If the total lift-drag ratio is 8, then the gliding angle will be 1 in 8, or measured in degrees, tan ø = R/W = 1/8 = 0.125. From a trigonometric table it will be found that this tangent corresponds to an angle of 7° – 10'. It should be noted that R is the total resistance and not the wing-drag.Fig. 2. Gliding Angle DiagramFig. 2. Gliding Angle Diagram Showing Component of Gravity That Causes Forward Motion. The Gliding Angle Depends Upon the Ratio of the Resistance to the Weight.Complete Power Calculations. Knowing the total weight and the desired speed, we must determine the wing section and area before we start on the actual power calculations. This can either be determined by empirical rules in the case of a preliminary investigation, or by actual calculation by means of the lift coefficients after the approximate values are known. Sustaining a given weight, we can vary the angle, area, wing section, or the speed, the choice of these items being regulated principally by the power. Given a small area and a great angle of incidence, we can support the load, but the power consumption will be excessive because of the low value of the L/D ratio at high angles. If small area is desired, a large value of Ky due to a high lift-wing section is preferable to a low lift wing at high angles. In general, the area should be so arranged that the wing is at the angle of the maximum lift-drag ratio at the rated speed. A low angle means a smaller motor, less fuel, and hence a lighter machine. This selection involves considerable difficulty, and a number of wing sections and areas must be tried by the trial and error method until the most economical combination is discovered.Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.The first consideration being the total weight, we must first estimate this from the required live load. This can be estimated from previous examples of nearly the same type. Say that our required live load is 660 pounds, and that a live load factor of 0.30 is used. The total weight now becomes 660/0.30=2200 pounds. To make a preliminary estimate of the area we must find the load per square foot. An empirical formula for biplane loading reads: w = 0.065V - 0.25 where V = maximum speed in M. P. H., and w = load per square foot. If we assume a maximum speed of 90 M. P. H. for our machine, the unit loading is w = (0.065 x 90) - 0.25 = 5.6 pounds per square foot. The approximate area can now be found from 2200/5.6 = 393 square feet. (Call 390.) The minimum speed is about 48 per cent of the maximum, or 43 M. P. H. We can now choose one or more wing sections that will come approximately to our requirements by the use of the basic formula, Ky = w/V².At high speed, Ky = 5.6/(90 x 90) = 0.000691. At low speed, Ky = 5.6/(40 x 40) = 0.003030. We must choose the most economical wing between these limits of lift, and on reference to our wing section tables we find:Wing Section TableIt would seem from the above that the chosen area is a little too large, as the majority of the L/D ratios at high speed are poor, the best being 11.00 of the U.S.A.-1. The angles are small, being negative in most cases at high speed. While the lift-drag of the R.A.F-3 is very good at low speed, it is very poor at high, hence the area for this section should be reduced to increase the loading. The R.A.F.-6 and the U.S.A.-1 show up the best, for they are both near the maximum lift at low speed and have fair L/D ratios at high speed. It will be seen that for the best results there should be a series of power curves drawn for the various wings and areas. This method is too complicated and tedious to take up here, and so we will use U.S.A.-1, which does not really show up so bad at this stage. Both the R.A.F.-6 and the U.S.A.-1 have been used extensively on machines of the size and type under consideration. While we require Ky = 0.003030, and U.S.A-1 gives 0.003165, we will not attempt to utilize this excess, as it will be remembered that we should not assume the maximum lift for reasons of stability.The wing-drag at high speed will be 2200/11.0 = 200 pounds, and at low speed it will be: 2200/10.4 = 211 pounds. Since the maximum L/D is 17.8 at 3°, where Ky is 0.00133, the least drag will be: 2200/17.8 = 124 pounds. This least drag will occur at V = V5.6/000133 = 65 M. P. H.The wing drag for each speed must now be divided by the correction factor 0.85, which converts the monoplane values of drag into biplane values. Since this is practically constant it does not affect the relative values of Kx in comparing wings, but it should be used in final results. For this type of machine we will take the total parasitic resistance as r = 0.036V². At 90 M.P.H., r = 0.036 x 90 x 90 = 291.6 pounds. At 65 M. P. H., the resistance is: 0.036 x 65 x 65 = 152.1. At the extreme low speed of 43 M. P.H. we have r - 0.036x43 x 43 = 66.56 pounds. The total resistance (R) is equal to the sum of the wing-drag and the parasitic resistance. At 90 M. P. H. the total resistance becomes 200 + 291.6 = 491.6 pounds. At 65 M.P.H. the total is 124 + 152.1 = 176.1, and at 43 M.P.H. it is 211 + 66.56 = 277.56 pounds. The horsepower is computed from H = RV/375e, and at 90 M. P. H. this is : H = 491.6 x 90/375 x 0.80 = 147.5 H. P. where 0.80 is the assumed propeller efficiency. At 65 M. P. H. the horsepower drops to H = 176.1 x 65/375 x 0.8 = 38.1 H. P., assuming the same efficiency. In the same way the H. P. at 43 M. P. H. r is 39.8.A table and power chart should be worked out for a number of sections and areas according to the following table. The calculations should be computed at intervals of 5 M. P. H., at least the lower speeds. Wing drag is not corrected for biplane interference:Power Chart TableWeight and Power. The weight lifted per horsepower varies in the different types of aeroplanes, this difference lying principally in the reserve allowed for climbing and horizontal speed. A speed scout may carry as little as 8 pounds per horsepower, while a slow two-seater may exceed 20 pounds per horsepower. A rough estimate of the horsepower required may be had by dividing the total weight by the weight per horsepower ratio for that particular type. Thus if the unit H. P. loading is 16 pounds and the total weight is 3200, then the horsepower will equal 3200/16 = 200 horsepower. Assuming that the live load w' is 0.32 of the total weight W, then W = w'/0.32. If m = lbs. per H.P., then H = W/m or H = w'/0.32m. Taking the case of a training machine where m = 20, and the live load is 640 pounds, the approximate horsepower will be: H = w'/0.32m = 640/0.32 x 20 = 100 horsepower. A speed scout carrying 320 pounds useful load, with m = 10, will require H = 320/0.32 x 10 = 100 horsepower.
Power Units. Power is the rate of doing work. If a force of 10 pounds is applied to a body moving at the rate of 300 feet per minute, the power will be expressed by 10 x 300 = 3000 foot-pounds per minute. As the figures obtained by the foot and pound units are usually inconveniently large, the "Horsepower" unit has been adopted. A horsepower is a unit that represents work done at the rate of 33,000 foot-pounds per minute, or 550 foot-pounds per second. Thus if a certain aeroplane offers a resistance of 200 pounds, and flies at the rate of 6000 feet per minute, then the work done per minute will be equal to 200 x 6000 = 1,200,000 foot-pounds. Since there are 33,000 foot-pounds of work per minute for each horsepower, the horsepower will be: 1200000/33000 = 36.3.
As aeroplane speeds are usually given in terms of miles per hour, it will be convenient to convert the foot-minute unit into the mile per hour unit. If H = horsepower, R = resistance of aeroplane, and V = miles per hour, then H = RV/375, the theoretical horsepower, without loss. If an aeroplane flies at 100 miles per hour and requires a propeller thrust of 300 pounds, then the horsepower becomes:
H = RV/375 = 300 x 100/375 = 80 horsepower. This is the actual power required to drive the machine, but is not the engine power, as the engine must also supply the losses due to the propeller. The propeller losses are generally expressed as a percentage of the total power supplied. The percentage of useful power is known as the "Efficiency."
The efficiency of the average aeroplane propeller will vary from 0.70 to 0.80. If e = propeller efficiency expressed as a decimal, the motor horsepower becomes: H = RV/375e. To obtain the motor horsepower, divide the theoretical horsepower by the efficiency. Using the complete formula for the solution of an example in which the flight speed is 100 M. P. H., the resistance 225 pounds and the efficiency 0.75, we have:
H = RV/375e = 225 x 100/375 x 0.75 = 80 horsepower.
Power Distribution. Since power depends upon the total resistance to be overcome, part of the power will be used for driving the lifting surfaces and a part for overcoming the parasitic resistance. The power required for driving the wings depends upon the angle of incidence, since the drag varies with every angle. The wing power varies with every flight speed, owing to the changes in angle made necessary to support the constant load. The power for the wings will be least at the speed and angle that corresponds to the greatest lift-drag ratio. Owing to the low value of the L/D at very small and very large angles, the power requirements will be excessive at extremely low and high speeds.
As the parasitic resistance increases as the square of the speed, the power for overcoming this resistance will vary as the cube of the speed. It is the parasitic resistance that really limits the higher speeds of the aeroplane, since it increases very rapidly at velocities of over 60 miles per hour.
The total power at any speed is the sum of the wing power and power required for the parasitic resistance. Owing to variations in the wing drag and resistance at every point within the flight range, it is exceedingly difficult to directly calculate the total power at any particular speed. The wing drag and the resistance should be calculated for every speed, and then laid out by a graph or curve. The minimum propeller thrust, or the minimum total resistance, occurs approximately at the speed where the body resistance and wing drag are equal. The minimum horsepower occurs at a low speed, but not the lowest speed, and this will differ with every machine.
Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section.Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section. The Results Were Taken From Full Size Tests Made by the English Government.
Fig. 1. Power Chart of Bleriot Monoplane, With Outline of Wing Section. The Results Were Taken From Full Size Tests Made by the English Government.
Fig. 1 is a set of performance curves drawn from the results of tests on a full size Bleriot monoplane. At the bottom the horizontal row of figures gives the horizontal speed in feet per second. The first column to the left is the horsepower, and the second column is the resistance or drag in pounds. The four curves represent respectively the body resistance, wing or "plane" resistance, horsepower, and total resistance. The horizontal line "AV" shows the available horsepower. It will be noted that the body resistance increases steadily from 9 pounds at 50 feet per second, to 180 pounds at 100 feet per second. The wing resistance, on the other hand, decreases from 350 pounds at 56 feet per second to a minimum of 130 pounds at 83 feet per second. It will be noted that the angles of incidence are marked along the wing-drag curve by small circles. The incidence is 6° at 75 feet per second, and 4° at a little less than 85 feet per second.
The available horsepower "AV" is 42. This is shown as a straight line, although in the majority of cases it is slightly curved owing to variations in power at the higher speeds, and to variations in the propeller efficiency. At 90 feet per second the actual horsepower curve crosses the line of available horsepower "AV." Beyond this point horizontal flight is no longer possible, as the power requirements would exceed the available horsepower. It will be noted that the lowest total resistance occurs near the point where the body and wing resistance curves intersect, or in other words, where the body and wing resistance are equal. The minimum horsepower takes place at 63 feet per second, or at a point nearly 1/3 between the lowest flight speed and the highest speed attained by the available horsepower in horizontal flight (90 ft/sec).
The actual range of flight speeds is limited to points between the intersection of the "Horsepower required" curve, and the "Available horsepower" curve. By increasing the propeller efficiency, or by increasing the power of the motor, the available horsepower line is raised and the flight range increased.
Horsepower For Climbing. Up to the present we have only considered horizontal flight. The power available for climbing is the difference between the power required to maintain horizontal flight at any speed, and the actual horsepower that can be delivered by the propeller. Thus, if the actual power delivered by a motor through the propeller is 85 horsepower, and the power required for horizontal flight at that speed is 45, then we have: 85–45= 40 horsepower available for climbing. Since the difference between the driving power and the power required for horizontal flight is less at extremely low and high speeds, it is evident that we will have a minimum climbing reserve at the high and low speeds. Consulting the power curve for the Bleriot monoplane, we see that the power required at 56 feet per second is 40 horsepower, and at 85 feet per second it is 38 horsepower. At the low speed we have a climbing reserve of 44–40 = 4 H. P., and at the higher speed 44–38 = 6 H. P. The maximum available horsepower "AV" is 44 horsepower. The minimum horizontal power required is found at 63 feet per second, the climbing reserve at this point being 44–28 = 16 horsepower. At 55 feet per second, and at 90, we would not be able to climb, as we would only have sufficient power to maintain horizontal flight.
If W = total weight of aeroplane, c = climbing speed, and H = horsepower reserve for climbing, then the climbing speed with a constant air density will be expressed by: c = 33000H/W. Assuming that the weight of the Bleriot monoplane is 800 pounds, and that we are to climb at the speed of the greatest power reserve (16 horsepower), our rate of climb is:
c = 33000H/w = 33000 x 16/800 = 660 feet per minute. It should be understood that this is the velocity at the beginning of the climb. After prolonged climbing the rate falls off because of diminishing power and increasing speed. Much depends upon the engine performance at the higher altitudes, so that the reserve power for climb usually diminishes as the machine rises, and hence the rate of climb diminishes in proportion.
The following table taken from actual flying tests will show how the rate of climb decreases with the altitude. These machines were equipped with 150 H. P. Hispano-Suisa motors. It will be noted that the S. P. A. D. and the Bleriot hold their rate of climb constant up to 7800 feet altitude, which is a feat that is undoubtedly performed by varying the compression of the engine.
Besides increasing the power, the rate of climb can also be increased by decreasing the weight of the aeroplane.
Rate of Climb Table
Maximum Altitude. The maximum altitude to which a machine can ascend is known as its "Ceiling." This again depends on both the aeroplane and the motor, but principally on the latter. It has been noted that machines having the greatest rate of climb also have the greatest ceiling. Thus the ceiling of a fast climbing scout is higher than that of a larger and slower machine. Based on this principle, a writer in "Flight" has developed the following equation for ceiling, which, of course, assumes a uniform decrease in density. Let H = maximum altitude, h = the altitude at any time t after the start of the climb, and a = the altitude after a time equal to twice the time t, then:
H = h / (2 - a/h)
Approximate values of h and a may be had
from the following table, which are the results of a test on a certain aeroplane:
Time(Minutes)
0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5
Altitude(Feet)
0.03300 6150 8730 10760 12610 14190 15530 16650 17600
If we assume that the height is 10760 feet after the first 10 minutes, and that the altitude after twice this time (20 minutes) is 16650 feet, then the maximum ceiling attained will be:
H = h / (2 - a/h) = 10760 / 2 - 16650/10760 = 23,770 feet. The use of this formula requires that the climb be known for certain time intervals before the ceiling.
Gliding Angle. The gliding angle of the wings alone is equal to the lift-drag ratio at the given angle. The best or "Flattest gliding angle" is, of course, the best lift-drag ratio of the wing—say on the average about 1 in 16. The gliding angle of the complete machine is considerably less than this, owing to the resistance of the body and structural parts. This generally reduces the actual angle to less than 12, and in most cases between 6 and 8. Expressed in terms of degrees, tan ø = R/W where R = head resistance and W = weight in pounds.
Fig. 2 is a diagram giving the gliding force diagram. The plane descends along the gliding path AC, making the angle of incidence (ø). When in horizontal flight, the lift is along OL and the weight is OW. When descending on the gliding path the lift maintains the same relation with the wing, but the relative angle of the weight is altered. The weight now acts along OG. The drag is represented by OD, with the propeller thrust OP equal and opposite to it. With the weight constant, the lift OL is decreased by the angle so that the total life = L = W cos ø. The action of the weight W produces the propelling component OP that gives forward velocity. The line AB is the horizontal ground line. If the total lift-drag ratio is 8, then the gliding angle will be 1 in 8, or measured in degrees, tan ø = R/W = 1/8 = 0.125. From a trigonometric table it will be found that this tangent corresponds to an angle of 7° – 10'. It should be noted that R is the total resistance and not the wing-drag.
Fig. 2. Gliding Angle DiagramFig. 2. Gliding Angle Diagram Showing Component of Gravity That Causes Forward Motion. The Gliding Angle Depends Upon the Ratio of the Resistance to the Weight.
Fig. 2. Gliding Angle Diagram Showing Component of Gravity That Causes Forward Motion. The Gliding Angle Depends Upon the Ratio of the Resistance to the Weight.
Complete Power Calculations. Knowing the total weight and the desired speed, we must determine the wing section and area before we start on the actual power calculations. This can either be determined by empirical rules in the case of a preliminary investigation, or by actual calculation by means of the lift coefficients after the approximate values are known. Sustaining a given weight, we can vary the angle, area, wing section, or the speed, the choice of these items being regulated principally by the power. Given a small area and a great angle of incidence, we can support the load, but the power consumption will be excessive because of the low value of the L/D ratio at high angles. If small area is desired, a large value of Ky due to a high lift-wing section is preferable to a low lift wing at high angles. In general, the area should be so arranged that the wing is at the angle of the maximum lift-drag ratio at the rated speed. A low angle means a smaller motor, less fuel, and hence a lighter machine. This selection involves considerable difficulty, and a number of wing sections and areas must be tried by the trial and error method until the most economical combination is discovered.
Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.
Graphical Gliding Diagrams of Several Aeroplanes Recorded in British Army Contest of 1912.
The first consideration being the total weight, we must first estimate this from the required live load. This can be estimated from previous examples of nearly the same type. Say that our required live load is 660 pounds, and that a live load factor of 0.30 is used. The total weight now becomes 660/0.30=2200 pounds. To make a preliminary estimate of the area we must find the load per square foot. An empirical formula for biplane loading reads: w = 0.065V - 0.25 where V = maximum speed in M. P. H., and w = load per square foot. If we assume a maximum speed of 90 M. P. H. for our machine, the unit loading is w = (0.065 x 90) - 0.25 = 5.6 pounds per square foot. The approximate area can now be found from 2200/5.6 = 393 square feet. (Call 390.) The minimum speed is about 48 per cent of the maximum, or 43 M. P. H. We can now choose one or more wing sections that will come approximately to our requirements by the use of the basic formula, Ky = w/V².
At high speed, Ky = 5.6/(90 x 90) = 0.000691. At low speed, Ky = 5.6/(40 x 40) = 0.003030. We must choose the most economical wing between these limits of lift, and on reference to our wing section tables we find:
Wing Section Table
It would seem from the above that the chosen area is a little too large, as the majority of the L/D ratios at high speed are poor, the best being 11.00 of the U.S.A.-1. The angles are small, being negative in most cases at high speed. While the lift-drag of the R.A.F-3 is very good at low speed, it is very poor at high, hence the area for this section should be reduced to increase the loading. The R.A.F.-6 and the U.S.A.-1 show up the best, for they are both near the maximum lift at low speed and have fair L/D ratios at high speed. It will be seen that for the best results there should be a series of power curves drawn for the various wings and areas. This method is too complicated and tedious to take up here, and so we will use U.S.A.-1, which does not really show up so bad at this stage. Both the R.A.F.-6 and the U.S.A.-1 have been used extensively on machines of the size and type under consideration. While we require Ky = 0.003030, and U.S.A-1 gives 0.003165, we will not attempt to utilize this excess, as it will be remembered that we should not assume the maximum lift for reasons of stability.
The wing-drag at high speed will be 2200/11.0 = 200 pounds, and at low speed it will be: 2200/10.4 = 211 pounds. Since the maximum L/D is 17.8 at 3°, where Ky is 0.00133, the least drag will be: 2200/17.8 = 124 pounds. This least drag will occur at V = V5.6/000133 = 65 M. P. H.
The wing drag for each speed must now be divided by the correction factor 0.85, which converts the monoplane values of drag into biplane values. Since this is practically constant it does not affect the relative values of Kx in comparing wings, but it should be used in final results. For this type of machine we will take the total parasitic resistance as r = 0.036V². At 90 M.P.H., r = 0.036 x 90 x 90 = 291.6 pounds. At 65 M. P. H., the resistance is: 0.036 x 65 x 65 = 152.1. At the extreme low speed of 43 M. P.H. we have r - 0.036x43 x 43 = 66.56 pounds. The total resistance (R) is equal to the sum of the wing-drag and the parasitic resistance. At 90 M. P. H. the total resistance becomes 200 + 291.6 = 491.6 pounds. At 65 M.P.H. the total is 124 + 152.1 = 176.1, and at 43 M.P.H. it is 211 + 66.56 = 277.56 pounds. The horsepower is computed from H = RV/375e, and at 90 M. P. H. this is : H = 491.6 x 90/375 x 0.80 = 147.5 H. P. where 0.80 is the assumed propeller efficiency. At 65 M. P. H. the horsepower drops to H = 176.1 x 65/375 x 0.8 = 38.1 H. P., assuming the same efficiency. In the same way the H. P. at 43 M. P. H. r is 39.8.
A table and power chart should be worked out for a number of sections and areas according to the following table. The calculations should be computed at intervals of 5 M. P. H., at least the lower speeds. Wing drag is not corrected for biplane interference:
Power Chart Table
Weight and Power. The weight lifted per horsepower varies in the different types of aeroplanes, this difference lying principally in the reserve allowed for climbing and horizontal speed. A speed scout may carry as little as 8 pounds per horsepower, while a slow two-seater may exceed 20 pounds per horsepower. A rough estimate of the horsepower required may be had by dividing the total weight by the weight per horsepower ratio for that particular type. Thus if the unit H. P. loading is 16 pounds and the total weight is 3200, then the horsepower will equal 3200/16 = 200 horsepower. Assuming that the live load w' is 0.32 of the total weight W, then W = w'/0.32. If m = lbs. per H.P., then H = W/m or H = w'/0.32m. Taking the case of a training machine where m = 20, and the live load is 640 pounds, the approximate horsepower will be: H = w'/0.32m = 640/0.32 x 20 = 100 horsepower. A speed scout carrying 320 pounds useful load, with m = 10, will require H = 320/0.32 x 10 = 100 horsepower.