Thanks a lot for this! I find myself studying a course on Aircraft Design and these videos did help me a lot in understanding the initial sizing of the airplane.
Dear Dr. Gudmundsson, how comes that the curve for service ceiling in your constraint diagramm changes its shape so strongly after tranforming the chart fom T/W to BHP? It looks like you multiplied it somehow by the factor (-1)? Allthough the transformation to BHP only always comes with a positive factor. I unfortunately can not understand your procedure in this case. But apart from this fact the explainations in your video and book are really helpful.
Hi Gregor. Sorry for the delay in responding. Your question sounds exactly like a question I received on Feb. 28 on my ERAU email. I responded to that question, but forgot about answering the same question here until now. Of course I suspect you may the other person and if true, you've already been responded to. But if not, please let me know and I'll respond asap. Cheers.
Hi Ignacio. Here's the response: I think the effect you are seeing is driven by a change in Vy (which increases with W/S and altitude). You can split the function for T/W for the service ceiling into two components; one is due to the contribution D/W (constant) and the other Vv/Vy (reduces with increase in W/S because Vy increases with W/S). The sum of the two dictates the reduction in T/W with W/S. Take a look at the derivation. However, this does not mean that power will reduce with W/S. It should (and will) increase. To convert T/W into required engine power at altitude (Palt) and then at sea-level, follow the procedure below. (1) Get required thrust at altitude (Talt) for the weight at altitude (alt): Talt = (T/W)∙Walt (2) Convert Talt into required engine power at altitude (Palt): Palt = Talt∙Vy/(eta∙550) (3) Convert Peng into corresponding S-L power (Psl): Psl = Palt/r Remember that you have to use eta during the conversion, because the engine must generate more power than the product T∙V indicates (thrustpower). Other variables used are: Walt = Weight at condition (i.e. altitude) Talt = Thrust required to climb at 100 fpm at altitude Palt = Power required to climb at 100 fpm at altitude Psl = Power the engine must be capable of delivering at S-L to be able to deliver Palt at altitude. r = Gagg and Farrar factor = 1.132∙sigma - 0.132 sigma = Density ratio D = Drag W = weight Vv = vertical speed (100 fpm) Vy = airspeed for best rate of climb (changes with W/S) - For now, you can estimate Vy = 57.1 + 1.768(W/S) eta = propeller efficiency at Vy (expect 0.6 to 0.65 for a fixed pitch prop) You see that in step (2), you have to multiply Talt by Vy to get the power. Since Vy increases with sqr(W/S), so does power required (as one would expect). I hope this sheds some light on what's going on in the formulation.
Thank you Prof. Gudmundsson for these great videos. I am going to start next Fall2016 my senior design class. In addition of buying the book I am going to use in the class, I also bought your GA book which I think is really good. My question is that some of the equations for initial sizing in my class book are a little bit different than your equations in your text book. I came to the conclusion that your equations are for GA Aircrafts while my class book focus more on general aircraft design. I am using Craigner AIAA series. Can you please clarify me this?
Thank you for the kind words, elnica0941 and for buying my book. The equations in my book are for GA=General Aviation aircraft. I am not familiar with a design book by "Craigner" . Are you spelling it correctly? At any rate, my equations are derived in my text, so you can see for yourself. Have fun designing!
Hi! Since this video is related to Aircraft Performance, how do we approximate a ducted fan; a propeller with approximately zero induced velocity or a jet engine?
A ducted fan would be treated in a similar way to the piston propeller (assuming that's how you're driving the propeller). You need to know how to convert the required thrust (which is obtained from the T/W versus W/S part - in other words T = W(T/W) ) into the power the engine must deliver in order to produce that thrust. If normally aspirated, you have to normalize this power to S-L. A jet engine does ordinarily not need normalization to S-L, as long as you have an "engine deck" (which is a program written by the engine manufacturer and provided to the customer). Typical engine decks calculate the thrust of the engine at altitude and Mach number (among many other parameters) specified by the user and, thus, can be compared directly to the value of T = W(T/W). I hope this helps. Best wishes.
@@dr.gudmundssonaircraftdesign Actually I am using a ducted fan driven by electric motor, to be used in radio controlled aircraft. Will there still be a need for conversion of T/W to power?
Technically, you should do that, although you should also evaluate whether the effort is warranted considering this is for an RC (with full respect for RC flying). The thrust of the duct (in part) depends on how much motor power is converted to thrust power (i.e. T∙V). The problem is that there is not much literature out there on small-fan propulsive efficiencies. A typical RC prop will have efficiency below 0.6, sometimes well below that (due to low Reynolds number effects). I am not sure what this is for ducted fans at low Reynolds numbers. If you are doing FPV with an OSD, then you can calculate the motor power by multiplying your Voltage (V) with Current (I) (e.g. if V=11.9 V and I=10.5 amps, then the power is P=VI=125 Watts). Even though this is power delivered by the battery (i.e. before it reaches the motor), typical motor converts over 95% of this into motor power (so actual motor power might be 119 Watts). Therefore, you may have to invert the process and instead ask: If I know the motor power (P in Watts), weight of the aircraft (in N), and airspeed (in m/s), what propulsive efficiency (eta) is required at that flight condition? It is possible to show that eta = (T/W) W∙V/P. So, imagine your airplane weighs 3 kg (=29.4 N), power is 119 Watts, desired airspeed is 15 m/s (= 54 kmh), and the constraint analysis calls for a T/W = 0.2, then the ducted fan must provide propulsive efficiency of eta = (0.2)∙(29.4)∙(15)/(119) = 0.74. In this case, I'd say this required efficiency is too high and thus, I would not expect the fan being capable of delivering enough thrust at that flight condition. However, had the resulting eta been, say 0.45, I would have considered it possible the motor can generate enough thrust. I suspect the max propulsive efficiency of this system is around 0.5 (quite possibly less). Anyway, that's one way to work around the lack of info on this type of propulsive devices. I hope this helps. Have fun designing.
Hello Dr. Gudmundsson, thank you for the videos. I´ve got a question regarding the formulas used, do you recommend using them for the analysis of a RPAS? Or will it be more accurate by first doing the conversion of T/W to P/W because of the use of a propeller? If that´s the case, how can we modify the formulas from the different flight stages?
Hi Arleth. I apologize for the late response. I should check my channel's message more frequently, but projects keep getting in the way. Anyway, let my try and answer you question. Yes, I recommend the formulas for most design project, including Remotely Piloted Aircraft System (RPAS). Converting T/W to P/W is a requirement for propeller aircraft because of the thrust's dependency on power and prop geometry. If you mount two different propellers on the same engine you'll get different thrust at the same engine power (one cannot make such changes to a jet engine). Your question about how to modify formulas for different flight stages is excellent. Unfortunately, the answer it demands is too long to discuss here. All the formulas are derived in my book and when you see the derivations, you'll know how to modify them. The 2nd edition of my book, which is coming out in Nov. 2021, has an enlarged section on constraint analysis with more formulas (e.g. a revised T-O formula and landing constraint). Perhaps some of those are what you're looking for to serve as a seed for modification. Best wishes with your design project.
@SainathReddy-n6x Please rephrase your question. This is because the W/S, per se, does not depend on ROC as much as ROC depends on W/S. You can see this by noting that on a day when a Cessna 172 is light (low W/S) it climbs faster than when it is heavy (high W/S). In contrast, climbing at different rates does not change the W/S (assuming we can neglect weight reduction due to fuel burn). Does this response confuse you? If so, welcome to the club. Your question confuses me. Thus, I invite you to rephrase it to make it more understandable. In return, I promise to try and explain (providing my confusion abates). Best wishes.
@rkxff9593 If I am too fast for you, it tells you that your understanding is not yet at a university level. Please solidify your foundation and, then, please revisit.
Thanks a lot for this! I find myself studying a course on Aircraft Design and these videos did help me a lot in understanding the initial sizing of the airplane.
+jibeneyto You're welcome. Glad to read it helped you. :-)
Great videos!
Thank you Daniel.
Dear Dr. Gudmundsson, how comes that the curve for service ceiling in your constraint diagramm changes its shape so strongly after tranforming the chart fom T/W to BHP? It looks like you multiplied it somehow by the factor (-1)? Allthough the transformation to BHP only always comes with a positive factor. I unfortunately can not understand your procedure in this case. But apart from this fact the explainations in your video and book are really helpful.
Hi Gregor. Sorry for the delay in responding. Your question sounds exactly like a question I received on Feb. 28 on my ERAU email. I responded to that question, but forgot about answering the same question here until now. Of course I suspect you may the other person and if true, you've already been responded to. But if not, please let me know and I'll respond asap. Cheers.
Dr. Gudmundsson, I have the same question! Could you please answer it? Thank you. Greetings from México!
Hi Ignacio. Here's the response:
I think the effect you are seeing is driven by a change in Vy (which increases with W/S and altitude). You can split the function for T/W for the service ceiling into two components; one is due to the contribution D/W (constant) and the other Vv/Vy (reduces with increase in W/S because Vy increases with W/S). The sum of the two dictates the reduction in T/W with W/S. Take a look at the derivation. However, this does not mean that power will reduce with W/S. It should (and will) increase. To convert T/W into required engine power at altitude (Palt) and then at sea-level, follow the procedure below.
(1) Get required thrust at altitude (Talt) for the weight at altitude (alt): Talt = (T/W)∙Walt
(2) Convert Talt into required engine power at altitude (Palt): Palt = Talt∙Vy/(eta∙550)
(3) Convert Peng into corresponding S-L power (Psl): Psl = Palt/r
Remember that you have to use eta during the conversion, because the engine must generate more power than the product T∙V indicates (thrustpower). Other variables used are:
Walt = Weight at condition (i.e. altitude)
Talt = Thrust required to climb at 100 fpm at altitude
Palt = Power required to climb at 100 fpm at altitude
Psl = Power the engine must be capable of delivering at S-L to be able to deliver Palt at altitude.
r = Gagg and Farrar factor = 1.132∙sigma - 0.132
sigma = Density ratio
D = Drag
W = weight
Vv = vertical speed (100 fpm)
Vy = airspeed for best rate of climb (changes with W/S) - For now, you can estimate Vy = 57.1 + 1.768(W/S)
eta = propeller efficiency at Vy (expect 0.6 to 0.65 for a fixed pitch prop)
You see that in step (2), you have to multiply Talt by Vy to get the power. Since Vy increases with sqr(W/S), so does power required (as one would expect).
I hope this sheds some light on what's going on in the formulation.
Thank you Prof. Gudmundsson for these great videos. I am going to start next Fall2016 my senior design class. In addition of buying the book I am going to use in the class, I also bought your GA book which I think is really good. My question is that some of the equations for initial sizing in my class book are a little bit different than your equations in your text book. I came to the conclusion that your equations are for GA Aircrafts while my class book focus more on general aircraft design. I am using Craigner AIAA series. Can you please clarify me this?
Thank you for the kind words, elnica0941 and for buying my book. The equations in my book are for GA=General Aviation aircraft. I am not familiar with a design book by "Craigner" . Are you spelling it correctly? At any rate, my equations are derived in my text, so you can see for yourself. Have fun designing!
It is "Raymer" not "Craimer".
Hi!
Since this video is related to Aircraft Performance, how do we approximate a ducted fan; a propeller with approximately zero induced velocity or a jet engine?
A ducted fan would be treated in a similar way to the piston propeller (assuming that's how you're driving the propeller). You need to know how to convert the required thrust (which is obtained from the T/W versus W/S part - in other words T = W(T/W) ) into the power the engine must deliver in order to produce that thrust. If normally aspirated, you have to normalize this power to S-L. A jet engine does ordinarily not need normalization to S-L, as long as you have an "engine deck" (which is a program written by the engine manufacturer and provided to the customer). Typical engine decks calculate the thrust of the engine at altitude and Mach number (among many other parameters) specified by the user and, thus, can be compared directly to the value of T = W(T/W). I hope this helps. Best wishes.
@@dr.gudmundssonaircraftdesign Actually I am using a ducted fan driven by electric motor, to be used in radio controlled aircraft. Will there still be a need for conversion of T/W to power?
Technically, you should do that, although you should also evaluate whether the effort is warranted considering this is for an RC (with full respect for RC flying). The thrust of the duct (in part) depends on how much motor power is converted to thrust power (i.e. T∙V). The problem is that there is not much literature out there on small-fan propulsive efficiencies. A typical RC prop will have efficiency below 0.6, sometimes well below that (due to low Reynolds number effects). I am not sure what this is for ducted fans at low Reynolds numbers. If you are doing FPV with an OSD, then you can calculate the motor power by multiplying your Voltage (V) with Current (I) (e.g. if V=11.9 V and I=10.5 amps, then the power is P=VI=125 Watts). Even though this is power delivered by the battery (i.e. before it reaches the motor), typical motor converts over 95% of this into motor power (so actual motor power might be 119 Watts). Therefore, you may have to invert the process and instead ask: If I know the motor power (P in Watts), weight of the aircraft (in N), and airspeed (in m/s), what propulsive efficiency (eta) is required at that flight condition? It is possible to show that eta = (T/W) W∙V/P. So, imagine your airplane weighs 3 kg (=29.4 N), power is 119 Watts, desired airspeed is 15 m/s (= 54 kmh), and the constraint analysis calls for a T/W = 0.2, then the ducted fan must provide propulsive efficiency of eta = (0.2)∙(29.4)∙(15)/(119) = 0.74. In this case, I'd say this required efficiency is too high and thus, I would not expect the fan being capable of delivering enough thrust at that flight condition. However, had the resulting eta been, say 0.45, I would have considered it possible the motor can generate enough thrust. I suspect the max propulsive efficiency of this system is around 0.5 (quite possibly less). Anyway, that's one way to work around the lack of info on this type of propulsive devices. I hope this helps. Have fun designing.
Hello Dr. Gudmundsson, thank you for the videos. I´ve got a question regarding the formulas used, do you recommend using them for the analysis of a RPAS? Or will it be more accurate by first doing the conversion of T/W to P/W because of the use of a propeller? If that´s the case, how can we modify the formulas from the different flight stages?
Hi Arleth. I apologize for the late response. I should check my channel's message more frequently, but projects keep getting in the way. Anyway, let my try and answer you question. Yes, I recommend the formulas for most design project, including Remotely Piloted Aircraft System (RPAS). Converting T/W to P/W is a requirement for propeller aircraft because of the thrust's dependency on power and prop geometry. If you mount two different propellers on the same engine you'll get different thrust at the same engine power (one cannot make such changes to a jet engine). Your question about how to modify formulas for different flight stages is excellent. Unfortunately, the answer it demands is too long to discuss here. All the formulas are derived in my book and when you see the derivations, you'll know how to modify them. The 2nd edition of my book, which is coming out in Nov. 2021, has an enlarged section on constraint analysis with more formulas (e.g. a revised T-O formula and landing constraint). Perhaps some of those are what you're looking for to serve as a seed for modification. Best wishes with your design project.
can anyone please tell me what happens to the W/S if the rate of climb increases and the design space plot changes?
@SainathReddy-n6x Please rephrase your question. This is because the W/S, per se, does not depend on ROC as much as ROC depends on W/S. You can see this by noting that on a day when a Cessna 172 is light (low W/S) it climbs faster than when it is heavy (high W/S). In contrast, climbing at different rates does not change the W/S (assuming we can neglect weight reduction due to fuel burn). Does this response confuse you? If so, welcome to the club. Your question confuses me. Thus, I invite you to rephrase it to make it more understandable. In return, I promise to try and explain (providing my confusion abates). Best wishes.
very fast in explaining and need little more examplly way to understand
@rkxff9593 If I am too fast for you, it tells you that your understanding is not yet at a university level. Please solidify your foundation and, then, please revisit.