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Bug M4, M5 & M6: Target Size


galvedro

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Hi,

 

The briefing for missions 4, 5 & 6 state that the targets are enclosed in a 30 m circle and that the fictitious target base must be set to 17m.

 

This is wrong. The circles are 60m in diameter, and the fictitious target base should be set to 33m instead.

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Hi,

 

The briefing for missions 4, 5 & 6 state that the targets are enclosed in a 30 m circle and that the fictitious target base must be set to 17m.

 

This is wrong. The circles are 60m in diameter, and the fictitious target base should be set to 33m instead.

You're right. This will be fixed. Thank.

 

Unjqhc7.png

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  • 2 weeks later...

Thank you. While you are at it, maybe you can also check M7. There are also a few errors in the briefing:

 

The target building is about 60m wide. The slant range at the bomb release point should be 1711 m. That makes for a fictitious target size of about 28 m (not 15 m as stated in the briefing).

 

Further down, in the common errors: If Hact > Hcalc, or Vact > Vcalc, the bombs will fall long, not short.

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Thank you. While you are at it, maybe you can also check M7. There are also a few errors in the briefing:

 

The target building is about 60m wide. The slant range at the bomb release point should be 1711 m. That makes for a fictitious target size of about 28 m (not 15 m as stated in the briefing).

 

Further down, in the common errors: Vact > Vcalc, the bombs will fall long, not short.

 

Thanks a lot, these errors will be fixed.

 

 

 

Further down, in the common errors: If Hact > Hcalc, the bombs will fall long, not short.

 

This is not clear to me.

 

IflNG6c.png

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Aha, I see. Those plots show a deviation in H, as well as in slant range. Basically, they are releasing with the target centred in the reticle which, for an Hact>Hcalc release means releasing further away from the target. That can indeed result in a short drop, depending on the reticle depression.

 

But that's an F5 in the drawing, isn't it?, and I believe the F5 does not allow you to adjust the reticle for a calculated slant range, nor does the calculation for you either.

 

Anyway, what I was thinking when I wrote my comment was that, if you consider the velocity vector imparted to the bomb at the release point, and consider that all parameters: initial speed, distance to the target and dive angle remain the same, and only consider altitude as a source of error (which is what I thought you where trying to illustrate in the lesson), then if Hact > Hcalc, the bombs will tend to fall long, and this is why: if released from a higher altitude, the bomb will spend more time in the air traveling at the same speed with respect to the ground (also true if you consider horizontal deceleration due to aerodynamic forces). Therefore, it will travel more distance over the ground and fall farther away.

 

In practice, when I did this lesson, I had a tendency to release 100-200 meters lower than I should, and my bombs tended to fall short every time. All other parameters where about right (dive angle, speed).

 

EDIT: Actually, looking at the diagrams you posted, I realise that my numbers are also slightly off. Slant range should be 1435 m, giving a fictitious target size of 33.4. Hopefully these ones are correct.


Edited by galvedro
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If I should trust the manual in that the maximum distance is 800m for the target scale in the reticle, I stand by my calculation of 33.4 m for the target base.

 

I checked the spreadsheet, and I didn't manage to understand how the number is derived in there. This is how I am doing it, is it wrong?

ReticleSetup-M7.thumb.png.e667dbd4f5874d4ea296fcaa1c14cf4d.png

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I believe I understand what's going on.

 

The fictitious target size you are getting with the spreadsheet (28m) is the same I was getting before I realised we have to introduce a correction for sight depression. Since the depression is 10°, it affects the result significantly, so it cannot be ignored if you want to do the calculation properly.

 

You and I are getting the same slant range, after I reviewed my numbers as described above, just because the sight depression happens to be 10°, which makes it so DiveAngle + Depression = 90° - DiveAngle. This is a coincidence.

 

If I am understanding this correctly, you are entering the dive angle 40° in cell B41. This is wrong. The formulation in that row is using the second equation in the diagram, which expects the angle phi = 90° - DiveAngle as input, in other words, the angle with respect to the vertical, not with respect to the horizontal. We get the same result because making phi = 40° results in the same triangle I am resolving when introducing the correction for sight depression :).


Edited by galvedro
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I talked to the pilot. He says your formula for the use of unguided rockets. For bombs, the trajectory of the bomb must be considered. The formula has not yet been found.

 

If I am understanding this correctly, you are entering the dive angle 40° in cell B41. This is wrong.

You're right. I was wrong.


Edited by SL PAK
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The trajectory of the bomb is already accounted for when you are given the release parameters: H=1100m, DiveAngle 40°, speed=500km/h, Depression=10°. These parameters define the release vector for a weapon of certain characteristics.

 

Here, we are only deducting the slant range as a function of those given parameters. We are not calculating the release point itself.

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  • 2 years later...

Maybe just an extra note. In all your calculation above you consider 800m (maximum distance in the gunsight) however in each briefing there is "Enter the minimum distance to the gunsight" which is 100m. What works for 60m diameter and max distance 800m: Target base 33° for 20° dive angle or target base 40° for 30° dive angle. So the formula from the manual works. This is how it looks like:

image.png

My other note is about using GYRO sight for the attack which is also in the briefing. Is it used IRL? If I use it, I'm hunting the target all over the place. 

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