Timeline for Is the MTOW determined by the wheel maximum supported weight?
Current License: CC BY-SA 3.0
28 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2017 at 17:37 | answer | added | PHChilly | timeline score: 0 | |
| Jun 28, 2017 at 7:43 | vote | accept | CalvT | ||
| Jun 28, 2017 at 5:17 | comment | added | vasin1987 | English for picture description : latam.com/vamos/get-inspired/… | |
| Jun 28, 2017 at 2:10 | answer | added | TomMcW | timeline score: 7 | |
| Jun 28, 2017 at 0:35 | comment | added | Devil07 | @CalvT븃 I noticed that you divided the MTOW by 12 tires, but there are actually 14 tires. Unfortunately, the nose wheel tires appear to be different size, so it wouldn't be accurate to divide by 14, but since some of the weight is on the nose wheel, the total MTOW will not be on the main wheels. | |
| Jun 27, 2017 at 20:33 | answer | added | tj1000 | timeline score: 0 | |
| Jun 27, 2017 at 15:36 | comment | added | Devil07 | @CalvT븃, I suspect you are correct. They jumped the gun (assumed) that is max of each tire, because I find it hard to believe that the tires don't have some redundancy built in to them, so that if one blows while taxing, they all wouldn't then blow out because of the increased load. | |
| Jun 27, 2017 at 14:55 | comment | added | mins | I can't answer, and maybe it's difficult to find the information in the technical specs, but for sure there is a safety margin that may be 100% or more of the static load (the weight on the wheel, which depends on the position of the center of gravity). Effect of wind while on the ground (negative lift) must also be taken into account. | |
| Jun 27, 2017 at 14:52 | comment | added | CalvT | @mins I'm calm don't worry :) - my question in a nutshell for you - "are these figures correct, as they don't make sense to me, and if they aren't, what are the correct figures?" | |
| Jun 27, 2017 at 14:49 | comment | added | mins | I feel your initial question was whether the figures for the wheel were accurate in the brochure or significantly underestimated by the marketing guy doing a simplistic division MTOW/# of wheels (a good question imho), but the title was a bit remote... and now you are overwhelmed with details and almost asked to justify your question :-) Keep cool! | |
| Jun 27, 2017 at 14:14 | comment | added | CalvT | Let me rephrase that: any numbers specific to the 777 which help answer this question? | |
| Jun 27, 2017 at 14:12 | comment | added | Ron Beyer | 14 CFR 25.301: Loads, 14 CFR 25.303: Factor of Safety, 14 CFR 25.473: Landing Load Conditions, 14 CFR 25.479: Level Landing Conditions, Basically anything in 25.471-25.519 | |
| Jun 27, 2017 at 14:07 | comment | added | CalvT | @RonBeyer that's what I thought. Do you have any numbers on that? | |
| Jun 27, 2017 at 14:06 | comment | added | Ron Beyer | @CalvT븃 Not so much "jumped the gun", but trying to impress passengers with numbers that they probably don't fully understand. The gear system itself (tires included) can withstand landing pressures significantly higher than that. | |
| Jun 27, 2017 at 13:45 | history | tweeted | twitter.com/StackAviation/status/879697308480393216 | ||
| Jun 27, 2017 at 13:23 | history | edited | CalvT | CC BY-SA 3.0 |
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| Jun 27, 2017 at 13:05 | history | edited | Federico | CC BY-SA 3.0 |
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| Jun 27, 2017 at 13:04 | comment | added | CalvT | @RonBeyer so, to answer my question, Marketing People jumped the gun? | |
| Jun 27, 2017 at 13:03 | history | edited | CalvT | CC BY-SA 3.0 |
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| Jun 27, 2017 at 13:02 | comment | added | Ron Beyer | Seat back brochures aren't an accurate representation of the engineering behind this. The wheels are not at the maximum loading at MTOW. The weight is more determined by stopping the aircraft prior to V1 during take off given the configuration of the aircraft. It isn't really about how much the tires hold. There is no one "formula" to cover this. | |
| Jun 27, 2017 at 12:57 | comment | added | Federico | ok, point taken :D (but be wary, one thing is the MTOW, another is the maximum landing weight) | |
| Jun 27, 2017 at 12:56 | comment | added | CalvT | @Federico I'm asking you guys for the formula :P | |
| Jun 27, 2017 at 12:56 | comment | added | Federico | "surely"? have you run the formula? or is "surely I am right in thinking this and I can't be wrong?" :P If you run the formula you'll see that one element is the weight, the other the speed of impact. You could land with 1% of the MTOW and still completely destroy your aircraft if the speed is excessive. | |
| Jun 27, 2017 at 12:53 | comment | added | CalvT | @Sanchises right. I get that. But according to this, when the plane is at MTOW and stationary, the wheels are at their max. So surely the impact of landing with 70% of that weight would exceed that max? | |
| Jun 27, 2017 at 12:48 | history | edited | CalvT | CC BY-SA 3.0 |
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| Jun 27, 2017 at 12:46 | comment | added | Ron Beyer | Maximum take off weight is often higher than maximum landing weight specifically for that reason. There are quite a few aircraft that can take off but have to burn fuel before it can land again. | |
| Jun 27, 2017 at 12:36 | history | edited | Federico | CC BY-SA 3.0 |
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| Jun 27, 2017 at 12:28 | history | asked | CalvT | CC BY-SA 3.0 |